Two very large, nonconducting plastic sheets, each 10.0 cm
thick, carry uniform charge densities σ1,σ2,σ3
and σ4
on their surfaces, as shown in the following figure(Figure 1). These surface charge densities have the values σ1 = -7.30 μC/m2 , σ2=5.00μC/m2, σ3= 1.90 μC/m2 , and σ4=4.00μC/m2. Use Gauss's law to find the magnitude and direction of the electric field at the following points, far from the edges of these sheets.
A:What is the magnitude of the electric field at point A , 5.00 cm
from the left face of the left-hand sheet?(Express your answer with the appropriate units.)
B:What is the direction of the electric field at point A, 5.00 cm
from the left face of the left-hand sheet?(LEFT,RIGHT,UPWARDS,DOWNWARDS)
C:What is the magnitude of the electric field at point B, 1.25 cm
from the inner surface of the right-hand sheet?(Express your answer with the appropriate units.)
D:What is the direction of the electric field atpoint B, 1.25 cm
from the inner surface of the right-hand sheet?(LEFT,RIGHT,UPWARDS,DOWNWARDS)
E:What is the magnitude of the electric field at point C , in the middle of the right-hand sheet?(Express your answer with the appropriate units.)
F:What is the direction of the electric field at point C, in the middle of the right-hand sheet?(LEFT,RIGHT,UPWARDS,DOWNWARDS)
Answers
Answer:
Explanation:
To use Gauss's Law, we need to choose a Gaussian surface that encloses the point of interest and has symmetry such that the electric field is constant over the surface. For all points in this problem, we can choose a cylinder as our Gaussian surface with its axis perpendicular to the sheets.
Let's assume that the cylinders are tall enough such that the electric field at the top and bottom faces of the cylinder is negligible. The electric flux through the curved part of the cylinder is constant and equal to Φ_E = E*A, where A is the surface area of the curved part of the cylinder.
Using Gauss's Law, Φ_E = Q_in / ε0, where Q_in is the net charge enclosed by the Gaussian surface and ε0 is the permittivity of free space.
A: The Gaussian surface is a cylinder with radius r = 5.00 cm and height h = the distance between the sheets (20.0 cm). The net charge enclosed is Q_in = σ1 * A_top + σ2 * A_bottom, where A_top and A_bottom are the areas of the top and bottom faces of the cylinder, respectively. Since the electric field is perpendicular to the faces, the flux through them is zero. So, Q_in = (σ1 - σ2) * A, where A is the surface area of the curved part of the cylinder. Thus,
Φ_E = E * A = Q_in / ε0
E = (σ1 - σ2) / (ε0 * r) = (-7.30 μC/m^2 - 5.00 μC/m^2) / (8.85 x 10^-12 C^2/Nm^2 * 0.0500 m) = -2.31 x 10^5 N/C
The magnitude of the electric field at point A is 2.31 x 10^5 N/C.
B: The electric field points from higher potential to lower potential. Since the left-hand sheet has a negative charge density and the right-hand sheet has a positive charge density, the potential decreases from left to right. Thus, the electric field at point A points from left to right.
The direction of the electric field at point A is RIGHT.
C: The Gaussian surface is a cylinder with radius r = 1.25 cm and height h = the thickness of the right-hand sheet (10.0 cm). The net charge enclosed is Q_in = σ4 * A, where A is the surface area of the curved part of the cylinder. Thus,
Φ_E = E * A = Q_in / ε0
E = σ4 / (ε0 * r) = 4.00 μC/m^2 / (8.85 x 10^-12 C^2/Nm^2 * 0.0125 m) = 3.77 x 10^7 N/C
The magnitude of the electric field at point B is 3.77 x 10^7 N/C.
D: The electric field points from higher potential to lower potential. Since the right-hand sheet has a positive charge density, the potential decreases from the right-hand sheet to the left. Thus, the electric field at point B points from right to left.
The direction of the electric field at point B is LEFT.
E:
Since point C is in the middle of the right-hand sheet, the electric field due to this sheet alone cancels out due to symmetry. Thus, the only electric field present is due to the left-hand sheet. The Gaussian surface is a cylinder with radius r = the radius of the sheet (10.0 cm) and height h = the thickness of the sheet (10.0 cm). The net charge enclosed is Q
The net charge enclosed within this Gaussian surface is:
Q = σ1 × (2πrh)
where h is the thickness of the left-hand sheet, r is the distance from the left-hand sheet to point C, and σ1 is the surface charge density of the left-hand sheet. Plugging in the given values, we get:
Q = (-7.30 × 10^-6 C/m^2) × (2π × 0.1 m × 0.1 m) = -4.60 × 10^-8 C
Using Gauss's law, we can find the electric field at point C:
E × (2πrh) = Q/ε0
where ε0 is the permittivity of free space. Solving for E, we get:
E = Q / (2πε0rh)
Plugging in the values, we get:
E = (-4.60 × 10^-8 C) / (2π × 8.85 × 10^-12 C^2/(N·m^2) × 0.1 m × 0.1 m) = -1.64 × 10^5 N/C
Therefore, the magnitude of the electric field at point C is 1.64 × 10^5 N/C.
To find the electric field at point C, we need to consider both sheets since point C is equidistant from both sheets. Thus, we can use Gauss's law to find the total electric field due to both sheets.
The net charge enclosed by a cylindrical Gaussian surface of radius r = 1.25 cm and height h = 20.0 cm is given by:
qenc = σ2 * (2πrh) + σ4 * (2πrh) = (σ2 + σ4) * (2πrh)
where σ2 is the charge density on the inner surface of the right-hand sheet, σ4 is the charge density on the outer surface of the left-hand sheet, and h is the distance between the two sheets.
Substituting the given values, we get:
qenc = (5.00 μC/m^2 + 4.00 μC/m^2) * (2π * 1.25 cm * 20.0 cm) = 628.32 nC
Using Gauss's law, we have:
E * 2πrh = qenc/ε0
where ε0 is the permittivity of free space.
Solving for E, we get:
E = qenc / (2πrhε0) = 2.22 × 10^4 N/C
Therefore, the magnitude of the electric field at point C is 2.22 × 10^4 N/C.
F:
The direction of the electric field at point C is perpendicular to the surface of the sheet, pointing away from the positive charge density and towards the negative charge density. Since the positive charge density is on the outer surface of the left-hand sheet and the negative charge density is on the inner surface of the right-hand sheet, the direction of the electric field at point C is from left to right. Therefore, the direction of the electric field at point C is RIGHT.
The net flux of an electric field in a closed surface is directly proportionate to the charge contained, according to Gauss' equation.
State Gauss’s lawTo use Gauss's Law, we need to choose a Gaussian surface that encloses the point of interest and has symmetry such that the electric field is constant over the surface. For all points in this problem, we can choose a cylinder as our Gaussian surface with its axis perpendicular to the sheets.
Let's assume that the cylinders are tall enough such that the electric field at the top and bottom faces of the cylinder is negligible. The electric flux through the curved part of the cylinder is constant and equal to Φ_E = E*A, where A is the surface area of the curved part of the cylinder.
Using Gauss's Law, Φ_E = Q_in / ε0, where Q_in is the net charge enclosed by the Gaussian surface and ε0 is the permittivity of free space.
A: The Gaussian surface is a cylinder with radius r = 5.00 cm and height h = the distance between the sheets (20.0 cm). The net charge enclosed is Q_in = σ1 * A_top + σ2 * A_bottom, where A_top and A_bottom are the areas of the top and bottom faces of the cylinder, respectively.
Φ_E = E * A = Q_in / ε0
E = (σ1 - σ2) / (ε0 * r) = (-7.30 μC/m^2 - 5.00 μC/m^2) / (8.85 x 10^-12 C^2/Nm^2 * 0.0500 m) = -2.31 x 10^5 N/C
The magnitude of the electric field at point A is 2.31 x 10^5 N/C.
B: The electric field points from higher potential to lower potential. Since the left-hand sheet has a negative charge density and the right-hand sheet has a positive charge density, the potential decreases from left to right. Thus, the electric field at point A points from left to right.
The direction of the electric field at point A is RIGHT.
C: The Gaussian surface is a cylinder with radius r = 1.25 cm and height h = the thickness of the right-hand sheet (10.0 cm). The net charge enclosed is Q_in = σ4 * A, where A is the surface area of the curved part of the cylinder. Thus,
Φ_E = E * A = Q_in / ε0
E = σ4 / (ε0 * r) = 4.00 μC/m^2 / (8.85 x 10^-12 C^2/Nm^2 * 0.0125 m) = 3.77 x 10^7 N/C
The magnitude of the electric field at point B is 3.77 x 10^7 N/C.
D: The electric field points from higher potential to lower potential. Since the right-hand sheet has a positive charge density, the potential decreases from the right-hand sheet to the left. Thus, the electric field at point B points from right to left.
The direction of the electric field at point B is LEFT.
E:Since point C is in the middle of the right-hand sheet, the electric field due to this sheet alone cancels out due to symmetry. Thus, the only electric field present is due to the left-hand sheet. The Gaussian surface is a cylinder with radius r = the radius of the sheet (10.0 cm) and height h = the thickness of the sheet (10.0 cm). The net charge enclosed is Q
The net charge enclosed within this Gaussian surface is:
Q = σ1 × (2πrh)
where h is the thickness of the left-hand sheet, r is the distance from the left-hand sheet to point C, and σ1 is the surface charge density of the left-hand sheet. Plugging in the given values, we get:
Q = (-7.30 × 10^-6 C/m^2) × (2π × 0.1 m × 0.1 m) = -4.60 × 10^-8 C
Using Gauss's law, we can find the electric field at point C:
E × (2πrh) = Q/ε0
where ε0 is the permittivity of free space. Solving for E, we get:
E = Q / (2πε0rh)
Plugging in the values, we get:
E = (-4.60 × 10^-8 C) / (2π × 8.85 × 10^-12 C^2/(N·m^2) × 0.1 m × 0.1 m) = -1.64 × 10^5 N/C
Therefore, the magnitude of the electric field at point C is 1.64 × 10^5 N/C.
To find the electric field at point C, we need to consider both sheets since point C is equidistant from both sheets. Thus, we can use Gauss's law to find the total electric field due to both sheets.
The net charge enclosed by a cylindrical Gaussian surface of radius r = 1.25 cm and height h = 20.0 cm is given by:
qenc = σ2 * (2πrh) + σ4 * (2πrh) = (σ2 + σ4) * (2πrh)
where σ2 is the charge density on the inner surface of the right-hand sheet, σ4 is the charge density on the outer surface of the left-hand sheet, and h is the distance between the two sheets.
Substituting the given values, we get:
qenc = (5.00 μC/m^2 + 4.00 μC/m^2) * (2π * 1.25 cm * 20.0 cm) = 628.32 nC
Using Gauss's law, we have:
E * 2πrh = qenc/ε0
where ε0 is the permittivity of free space.
Solving for E, we get:
E = qenc / (2πrhε0) = 2.22 × 10^4 N/C
Therefore, the magnitude of the electric field at point C is 2.22 × 10^4 N/C.
F:The direction of the electric field at point C is perpendicular to the surface of the sheet, pointing away from the positive charge density and towards the negative charge density. Since the positive charge density is on the outer surface of the left-hand sheet and the negative charge density is on the inner surface of the right-hand sheet, the direction of the electric field at point C is from left to right. Therefore, the direction of the electric field at point C is RIGHT.
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Related Questions
A girl with a mass of 32 kg is playing on a swing. There are three main forces
acting on her at any time: gravity, force due to centripetal acceleration, and
the tension in the swing's chain (ignore the effects of air resistance). At the
instant shown in the image below, she is at the bottom of the swing and is
traveling at a constant speed of 4 m/s. What is the tension in the swing's
chain at this time? (Recall that g = 9.8 m/s²)
Tension
Weight
A. 333.6 N
OB. 817.8 N
C. 562.8 N
D. 441.6 N
4 m/s
Answers
The tension in the swing's chain at the instant shown in the image is 441.6 N.
option D
What is the tension at bottom swing?At the bottom of the swing, the girl is traveling at a constant speed, so her acceleration is zero. Therefore, the net force acting on her is also zero.
Thus, we have:
0 = T - mg - mv²/r
where;
T is the tension in the swing's chain, m is the girl's mass, g is the acceleration due to gravity, v is her speed, and r is the radius of the swing.At the bottom of the swing, the radius is equal to the length of the chain, so we have:
r = L = 4.0 m
Substituting the values we have:
T = (32 kg)(9.8 m/s²) + (32 kg)(4 m/s)²/4.0 m
Solving for T, we get:
T = (32 kg)(9.8 m/s²) + (32 kg)(4 m/s)²/4.0 m
T = 441.6 N.
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Which of the following does a scientist NOT need to calculate the age of something?
Amount of radioactive isotope within a sample
Half-life of radioactive isotope
Abundance on Earth
100g of the sample
Answers
Answer:
Abundance on Earth is not necessary for calculating the age of something using radioactive dating methods.
In contrast, the amount of radioactive isotope within a sample and its half-life are crucial for determining the age of the sample using radioactive dating methods. The amount of sample, such as 100g, is also important for determining the quantity of radioactive isotopes present, which is used in the calculation of the age of the sample.
Select in the ticker-timer a frequency of 25 Hz or 50 Hz. Determine the period of the ticker-timer.
Answers
Answer:
The period of a ticker-timer is the time interval between two consecutive dots made by the ticker.
If the frequency of the ticker-timer is 25 Hz, then it makes 25 dots in one second. Therefore, the period of the ticker-timer can be calculated as:
Period = 1/frequency = 1/25 Hz = 0.04 seconds
If the frequency of the ticker-timer is 50 Hz, then it makes 50 dots in one second. Therefore, the period of the ticker-timer can be calculated as:
Period = 1/frequency = 1/50 Hz = 0.02 seconds
So, the period of the ticker-timer is 0.04 seconds for a frequency of 25 Hz and 0.02 seconds for a frequency of 50 Hz
Explanation:
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When heal flows between systems Entropy
Answers
Answer: increases
Explanation:
Which type of marco molecules help a cell break down food?
Answers
Answer: Each macromolecule is broken down by a specific enzyme. For instance, carbohydrates are broken down by amylase, sucrase, lactase, or maltase. Proteins are broken down by the enzymes pepsin and peptidase, and by hydrochloric acid. Lipids are broken down by lipases.
Explanation:
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13 Which of these fitness events happened LAST? OA. PE becomes part of American school curriculum. OB. Title IX forbids gender discrimination in sports. O C. The YMCA opens a gym in America. O D. The adjustable plate-loaded barbell is invented.
Answers
The answer is A. PE becomes part of American school curriculum.
Title IX, which forbids gender discrimination in sports, was enacted in 1972.
The YMCA opened its first gym in America in Boston in 1851.
The adjustable plate-loaded barbell was invented in the late 1940s by a man named George Snyder.
Physical Education (PE) has been a part of American school curriculums for well over a century, with the first school gymnasiums being built in the 1800s.
What is an american school curriculum?
The American school curriculum is the set of educational standards and guidelines used by schools in the United States to structure their academic programs. The curriculum typically includes courses in core academic subjects such as English, math, science, and social studies, as well as elective courses in areas such as the arts, foreign languages, and physical education.
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A rocket takes off from a space station, where there is no gravity other than the negligible gravity due to the space station, and reaches a speed of 110 m/s in 10.0 s. If the exhaust speed is 1,600 m/s and the mass of fuel burned is 118 kg, what was the initial mass (in kg) of the rocket?
Answers
The initial mass of the rocket was 106 kg.
What is the initial mass of the rocket?We can use the principle of conservation of momentum to solve this problem.
The momentum of the rocket before takeoff is zero, since it is at rest, and the momentum after takeoff is the product of the mass of the rocket and its velocity.
However, during the takeoff, the rocket ejects a mass of fuel at a certain velocity, which creates a backward force (thrust) that propels the rocket forward.
This thrust can be calculated using the equation:
Thrust = (mass flow rate) x (exhaust velocity)
mass flow rate = (mass of fuel burned) / (burn time)
The mass of the rocket at any given time can be calculated using the equation:
mass = (initial mass) - (mass of fuel burned)
Using these equations, we can solve for the initial mass of the rocket:
Calculate the thrust:
Thrust = (118 kg / 10.0 s) x 1600 m/s = 1,888 N
Calculate the mass of the rocket at the end of the burn:
mass(end) = (initial mass) - (mass of fuel burned) = (initial mass) - 118 kg
Use the principle of conservation of momentum to find the initial mass:
momentum before = momentum after
0 = (mass(end) + 118 kg) x 110 m/s
mass(end) = -118 kg / 110 m/s = -1.07 kg/s
mass(end) = (initial mass) - 118 kg
(initial mass) = mass(end) + 118 kg
(initial mass) = (-1.07 kg/s x 10.0 s) + 118 kg
(initial mass) = 106 kg
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Describe the change to the graph of Y= X +3 when Y=2X -3 is graphed 
Answers
Answer: a stretch of 2
Explanation: because it 2 (x) - 3
Please HELP me answer these three questions!! (show work)
Answers
Answer:
2) 42.6
3) 1500000 m
4) no he will not be late, it will take 3s
Explanation:
#2:
Formula for average velocity:
⇒ [tex]v=\frac{d}{t}[/tex]
Substitute
⇒ [tex]\frac{2560}{60}[/tex]
⇒ [tex]42.6[/tex]
#3:
1st convert 60km to m/s (correct si unit must always come first)
to convert kilometer to meter, multiply by 1000 (prefix 'kilo' = 1000)
60 km x 1000 = 60000 m
use the formula for speed, distance, & time
60000 = d/25
to find displacement just multiply the speed and time together
60000 x 25
1500000m is the displacement
#4:
s = d/t
given are displacement and speed/velocity
d = 25m
v = 8.5 m/s
to find the time just divide displacement by speed
8.5 = 25/t
25/8.5
≈ 3 seconds
so he will not be late for class.
A carpenter tosses a shingle off a 9.4 m high roof, giving it an initial horizontal velocity of 7.2 m/s.] How far does it move horizontally in this time
Answers
Answer:
Explanation:
Assuming negligible air resistance, the horizontal velocity of the shingle will remain constant and the vertical motion will be influenced by gravity.
We can use the kinematic equations of motion to determine the horizontal distance traveled by the shingle. The relevant equation is:
d = v * t
where d is the distance, v is the initial horizontal velocity, and t is the time of flight.
To find the time of flight, we can use the equation for the vertical displacement of an object under constant acceleration:
y = v0t + (1/2)at^2
where y is the vertical displacement, v0 is the initial vertical velocity (which is zero), a is the acceleration due to gravity (-9.8 m/s^2), and t is the time of flight. Solving for t, we get:
t = sqrt((2y)/a)
where sqrt means square root.
Substituting the given values, we have:
y = 9.4 m
a = -9.8 m/s^2
t = sqrt((2*9.4 m) / -9.8 m/s^2) = 1.45 s (using the positive root since time cannot be negative)
Now, we can use the horizontal velocity to find the distance traveled in this time:
d = v * t = 7.2 m/s * 1.45 s = 10.44 m
Therefore, the shingle moves a horizontal distance of 10.44 meters in this time.
(a) Find the frequency ratio between the two frequencies fi =256 Hz and f2 = 320 Hz. (b) Add the interval of a fifth to f2 to obtain fs, and find the frequency ratio fs/fi. (c) Find the frequency of f3.
Answers
(a) The frequency ratio between the two frequencies fi = 256 Hz and f2 = 320 Hz is:
[tex]\frac{f_2}{f_i} = \frac{320}{256} = \frac{5}{4} = 1.25[/tex]
So the frequency ratio is 1.25.
(b) Adding the interval of a fifth to f2 = 320 Hz gives:
fs = f2 * (3/2) = 320 * (3/2) = 480 Hz
The frequency ratio fs/fi is:
[tex]\frac{f_s}{f_i} = \frac{480}{256} = \frac{15}{8} = 1.875[/tex]
So the frequency ratio is 1.875.
(c) To find the frequency of f3, we need to add the interval of a fourth to f2:
f3 = f2 * (4/3) = 320 * (4/3) = 426.67 Hz
Therefore, the frequency of f3 is 426.67 Hz.
how long will be required for a car to go from a speed of 20.0 m/s to a speed of 25.0 m/s if the acceleration is 3.0 m/s2?
Answers
Answer:
Explanation:
We can use the following kinematic equation to solve for the time required:
v_f = v_i + at
where:
v_f = final velocity = 25.0 m/s
v_i = initial velocity = 20.0 m/s
a = acceleration = 3.0 m/s^2
t = time
Rearranging the equation to solve for t, we get:
t = (v_f - v_i) / a
Substituting the given values, we get:
t = (25.0 m/s - 20.0 m/s) / 3.0 m/s^2
t = 1.67 s
Therefore, it will take 1.67 seconds for the car to go from a speed of 20.0 m/s to a speed of 25.0 m/s, assuming a constant acceleration of 3.0 m/s^2.
Is this statement is correct ?? According to Newton's 4rd law, Action and reaction never start from the same point.
help me...
Answers
Explanation:
I'm sorry, but the statement you provided is incorrect. There is no such thing as Newton's 4th law. Newton's laws of motion consist of three laws, which are:
An object at rest tends to stay at rest, and an object in motion tends to stay in motion with the same speed and in the same direction unless acted upon by an unbalanced force.
The acceleration of an object is directly proportional to the force applied to it and inversely proportional to its mass.
For every action, there is an equal and opposite reaction.
None of these laws state that action and reaction never start from the same point. However, it is true that the action and reaction forces act on different objects, not necessarily at the same point. This is because Newton's third law states that every action has an equal and opposite reaction, which means that when one object exerts a force on another object, the second object exerts an equal and opposite force back on the first object.
When a disrupted part of a wetland ecosystem is left alone so that nature can help restore it to what it once was, what are people counting on occurring? explain..
Answers
Answer: When a disrupted part of the ecosystem is left alone so that nature can help restore itself what people are counting on happening is secondary succession
Explanation:
Work Energy Theorem Question: You apply 50 N to a 10 kg object to cause it to move from rest to 2.5 m/s. What distance was the object moved?
Answers
Answer:
0.625 meters
Explanation:
We can use the work-energy that the work done on an object is equal to the change in its kinetic energy:
Work = ΔK = Kf - Ki
Where:
Work is the work done on the object
ΔK is the change in kinetic energy of the object
Kf is the final kinetic energy of the object
Ki is the initial kinetic energy of the object (which is zero since the object is at rest)
The work done on the object is equal to the force applied to the object multiplied by the distance over which the force is applied:
Work = F × d
Where:
F is the force applied to the object (50 N)
d is the distance over which the force is applied (unknown)
So we can write:
F × d = Kf - Ki
Substituting the given values:
50 N × d = 1/2 × 10 kg × (2.5 m/s)^2 - 0
Simplifying:
50 N × d = 31.25 J
Solving for d:
d = 31.25 J / 50 N = 0.625 m
Therefore, the object was moved a distance of 0.625 meters.
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A 0.5kg wooden block is placed on top of a 1.0kgwooden block. The coefficient static friction between the two blocks is 0.35. The coefficient of kinetic friction between the lower block and the level table is 0.20 wht is the maximum horizontal force that can be applied to the lower block
Answers
A block of 0.5 kg is placed on top of another wooden block which weighs 1.0 kg. The coefficient of static friction between the two blocks is 0.35, whereas the coefficient of kinetic friction between the lower block and the level table is 0.20.
To calculate the maximum horizontal force that can be applied to the lower block, we need to determine the limiting frictional force between the two blocks.
Since the upper block is not moving, the force of static friction is acting on it. We can calculate this force as follows:
`F_static = friction coefficient * normal force`
where, normal force = weight of upper block = 0.5 kg * 9.81 m/s^2 = 4.905 N
`F_static = 0.35 * 4.905 = 1.718 N`
Therefore, the static frictional force acting on the upper block is 1.718 N.
Now, we need to find the maximum force that can be applied to the lower block before it starts moving. This force is equal to the force of static friction acting on the lower block.
Since the upper block is not moving, the force of static friction acting on the lower block is equal to the force of static friction acting on the upper block.
`F_static(lower block) = F_static(upper block) = 1.718 N`
This means that the maximum horizontal force that can be applied to the lower block is 1.718 N.
However, if the applied force exceeds this value, the lower block will start moving and the force of kinetic friction will be acting on it, which is equal to:
`F_kinetic = friction coefficient * normal force`
`F_kinetic = 0.20 * 4.905 = 0.981 N`
Hence, if the applied force exceeds 1.718 N, the lower block will start moving and the force of kinetic friction will act on it, which is 0.981 N.
Therefore, the maximum horizontal force that can be applied to the lower block is 1.718 N.
Answer:
Explanation:
To determine the maximum horizontal force that can be applied to the lower block without causing the blocks to move, we need to calculate the maximum static friction force between the two blocks. This force is given by:
F_friction = coefficient of static friction * normal force
where the normal force is the force perpendicular to the surface of contact between the blocks. Since the blocks are resting on a level table, the normal force acting on the lower block is equal to the weight of both blocks, which is:
N = (m1 + m2) * g
where m1 is the mass of the lower block, m2 is the mass of the upper block, and g is the acceleration due to gravity (9.81 m/s^2).
Plugging in the given values, we have:
N = (1.0 kg + 0.5 kg) * 9.81 m/s^2 = 14.715 N
The maximum static friction force is then:
F_friction = 0.35 * 14.715 N = 5.15025 N
Therefore, the maximum horizontal force that can be applied to the lower block without causing the blocks to move is 5.15025 N. If a greater force is applied, the blocks will start to move and the kinetic friction force will take effect, which is given by:
F_kinetic = coefficient of kinetic friction * normal force
where the coefficient of kinetic friction is 0.20 in this case.
460miles per hour with the wind nd 420 per hour gainst the wind
Answers
The speed of the wind is 20 miles per hour.
To solve this problem, we can use the formula:
Speed = Distance/Time
Let's assume that the speed of the wind is x miles per hour.
With the wind, the plane travels at a speed of 460 miles per hour. This means that its speed relative to the ground is the sum of its airspeed and the speed of the wind:
460 = Airspeed + x
Against the wind, the plane travels at a speed of 420 miles per hour. This means that its speed relative to the ground is the difference between its airspeed and the speed of the wind:
420 = Airspeed - x
We can solve this system of equations to find the airspeed of the plane:
460 = Airspeed + x
420 = Airspeed - x
Adding the two equations gives:
880 = 2Airspeed
Dividing both sides by 2 gives:
Airspeed = 440 miles per hour
Now that we know the airspeed of the plane, we can find the speed of the wind by substituting this value into one of the equations we obtained earlier:
460 = Airspeed + x
460 = 440 + x
x = 20
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Liam pulls a box with a horizontal force of 17 N over a distance of 19 m. What is the change in energy of the box after the 19 m?
Answers
So the change in energy of the box after being pulled by Liam over 19 m is 323 J.
What are some illustrations of energy change?Energy is capable of changing its forms. For instance, electrical energy transforms into thermal and light energy when a lightbulb is turned on. A automobile transforms the gasoline's molecular bonds' stored energy into a variety of other forms. Chemical energy is converted to light in the engine through a chemical reaction.
The work done on the box by Liam's force is:
work = force x distance x cos(theta)
where theta is the angle between the force and the direction of motion. Since the force is horizontal and the motion is also horizontal, theta is 0 degrees and cos(theta) is 1. Therefore, the work done on the box is:
work = 17 N x 19 m x 1 = 323 J
The change in energy of the box is equal to the work done on it, according to the work-energy principle. Therefore, the change in energy of the box is:
change in energy = work = 323 J
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Suppose the angles shown in Fig. 5.31 are 52° and 25°. If the left-hand mass is 2.5 kg, what should the right-hand mass be so that it accelerates (a) downslope at 0.64 m/s2 and (b) upslope at 0.76 m/s2?
Answers
Downslope at 0.64 m/s², m = 12.4 kg
Upslope at 0.76 m/s², m = 6.35 kg
Define Mass?In Physics, mass is the most basic property of matter, and it is one of the fundamental quantities. Mass is defined as the amount of matter present in a body. The SI unit of mass is the kilogram (kg). The formula of mass can be written as:
Mass = Density × VolumePart A)
The sum of forces on the left-hand mass
T - mgsin(angle) = ma
T - (2.6) (9.8) (sin 70) = 2.6(.64)
T = 25.6 N
m = left mass.........M = right mass
T - mg×sin70 = ma
Mg×sin16 - T = Ma
Mg×sin16 - mg×sin70 = a×(M+m)
M×g×sin16 - mg×sin70 = Ma + ma
M× (g×sin16 -a) = m× (a + gsin70)
M = m× (a + gsin70) / (g×sin16 -a)
a) a = 0.64
M = 10.98 Kg
b) M = 11.82 kg
For the right-hand mass, the sum of forces...
mgsin(angle) - T = ma
m (9.8) (sin 16) - 25.6 = m (.64)
2.7m - 25.6 = .64m
m = 12.4 kg
Part B)
For the left-hand mass
mgsin(70) - T = ma
(2.6) (9.8) (sin 70) - T = (2.6) (.76)
T = 21.97 N
Then for the right-hand mass
T - mgsin(16) = ma
21.97 - m (9.8) (sin 16) = m (.76)
21.97 - 2.7m = .76m
m = 6.35 kg
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The right-hand mass should be 3.3 kg to accelerate up the slope at 0.76 m/s². To solve this problem, we need to use the principles of Newton's laws of motion and trigonometry.
We know that the force of gravity acting on the mass is equal to its weight, which can be calculated using the formula Fg = mg, where Fg is the force of gravity, m is the mass, and g is the acceleration due to gravity (which is approximately 9.8 m/s²).
To find the force acting down the slope, we need to calculate the component of the weight that acts down the slope, which is given by Fg sin θ, where θ is the angle of the slope. Using the given angle of 52°, we can calculate the force acting down the slope for the left-hand mass as:
Fdown = Fg sin θ
Fdown = (2.5 kg)(9.8 m/s²) sin 52°
Fdown = 18.9 N
To find the required mass for the right-hand mass to accelerate at 0.64 m/s^2 down the slope, we can use Newton's second law of motion, which states that the force acting on an object is equal to its mass times its acceleration (F = ma). Therefore, we can calculate the required force for the right-hand mass as:
F = ma
F = (m)(0.64 m/s²)
Since the force acting down the slope is 18.9 N, we can set these two equations equal to each other and solve for the mass:
F = Fdown
(m)(0.64 m/s²) = 18.9 N
m = 29.5 kg
Therefore, the right-hand mass should be 29.5 kg to accelerate down the slope at 0.64 m/s².
To find the required mass for the right-hand mass to accelerate at 0.76 m/s² up the slope, we can use the same approach, but this time we need to use the component of the weight that acts up the slope, which is given by Fg cos θ, where θ is the angle of the slope. Using the given angle of 25°, we can calculate the force acting up the slope for the right-hand mass as:
Fup = Fg cos θ
Fup = (m)(9.8 m/s²) cos 25°
Setting this equal to the force required for the right-hand mass to accelerate up the slope, we get:
Fup = ma
(m)(0.76 m/s²) = (m)(9.8 m/s²) cos 25°
m = 3.3 kg
Therefore, the right-hand mass should be 3.3 kg to accelerate up the slope at 0.76 m/s².
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A 4.0-kg mass is moving to the right at 3.0 m/s. An 8.0 kg mass is moving to the left at 2.0 m/s. If after collision the two
masses join together, what is their velocity after collision?
O-0.33 m/s
O-0.20 m/s
O +1.4 m/s
O +2.3 m/s
Answers
Answer:
- 0.33 m/s
Explanation:
An illustration is shown above,
In this case, since the two objects move in opposite directions before collision, then move together, the formula to be used is,
m1u1 - m2u2 = (m1 + m2)v
Where,
m1 = mass of the first object
u1 = initial velocity of the first object
v1 = final velocity of the first object
m2 = mass of the second object
u2 = initial velocity of the second object
v2 = final velocity of the second object
Therefore,
(4.0 • 3.0) - (8.0 • 2.0) = (4.0 + 8.0)v
12 - 16 = 12v
-4 = 12v
Divide both sides by 12,
-4 / 12 = 12v / 12
-1 / 3 = v
v = -0.33 m/s
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What might happen if people did not have the rights established in Miranda v. Arizona
Answers
Answer:
If people did not have the rights established in Miranda v. Arizona, they could be subjected to police coercion and forced confessions without a lawyer present. This could lead to wrongful convictions and denial of due process, which are essential components of the American justice system
A block of mass m1=3.0kg rests on a frictionless horizontal surface. A second block of m2=2.0kg hangs from an ideal cord of negligible mass that runs over an ideal pulley and then is connected to the first block . the blocks are released from rest . determine the displacement of the velocityA block of mass m1=3.0kg rests on a frictionless horizontal surface. A second block of m2=2.0kg hangs from an ideal cord of negligible mass that runs over an ideal pulley and then is connected to the first block . the blocks are released from rest . Determine how far has block 1 moved during the 1.2-s interval? A) 13.4 m B) 2.1m C) 28.2m D) 7.6mA block of mass m1=3.0kg rests on a frictionless horizontal surface. A second block of m2=2.0kg hangs from an ideal cord of negligible mass that runs over an ideal pulley and then is connected to the first block . the blocks are released from rest . determine the displacement of the velocityA block of mass m1=3.0kg rests on a frictionless horizontal surface. A second block of m2=2.0kg hangs from an ideal cord of negligible mass that runs over an ideal pulley and then is connected to the first block . the blocks are released from rest . Determine how far has block 1 moved during the 1.2-s interval?
Answers
To solve this problem, we can use the conservation of mechanical energy principle. When the blocks are released from rest, the potential energy of the system is converted to kinetic energy. Since the surface is frictionless, the mechanical energy of the system is conserved.
Using the principle of mechanical energy conservation, we can write:
m1*g*h = (m1+m2)*v^2/2
where m1 is the mass of the first block, m2 is the mass of the second block, g is the acceleration due to gravity, h is the height that the second block falls, and v is the velocity of the system after the blocks have moved a distance x.
The displacement of the first block can be found by using the time it takes the system to reach this velocity. The time t can be found using the formula:
x = (1/2) * a * t^2
where a is the acceleration of the first block.
The acceleration of the first block is equal to the acceleration of the system, which can be found by using the equation:
m1*a = m2*g - m1*g
Substituting the value of a in the previous formula, we get:
x = (1/2) * (m2*g - m1*g) * t^2 / m1
Substituting the values we get:
x = (1/2) * (2.0 kg * 9.81 m/s^2 - 3.0 kg * 9.81 m/s^2) * (1.2 s)^2 / 3.0 kg
x ≈ 7.6 m
Therefore, the correct answer is D) 7.6 m.
A loudspeaker of mass 15.0 kg is suspended a distance of h = 1.00 m below the ceiling by two cables that make equal angles with the ceiling. Each cable has a length of l = 2.70 m .
1. What is the tension T in each of the cables?
Use 9.80 m/s2 for the magnitude of the free-fall acceleration.
Answers
Answer:
To solve the problem, we use the equations of equilibrium to find the tension in each cable holding up the loudspeaker. Since the loudspeaker is in equilibrium, the sum of the forces acting on it is zero. The weight of the loudspeaker is calculated first, and then we use trigonometry to find the horizontal and vertical components of the tension in one of the cables. We then apply the equation of equilibrium in the y direction to find the tension in each cable. The final answer is that the tension in each cable is approximately 81.1 N, which balances the weight of the loudspeaker.
___
Given:
Mass of loudspeaker (m) = 15.0 kg
Distance from ceiling (h) = 1.00 m
Length of cable (l) = 2.70 m
Acceleration due to gravity (g) = 9.80 m/s^2
Weight of loudspeaker:
Fg = mg
Fg = (15.0 kg)(9.80 m/s^2)
Fg = 147 N
Horizontal and vertical components of tension in one cable:
sinθ = h/l
sinθ = 1.00 m / 2.70 m
θ = sin^(-1)(1.00/2.70)
θ ≈ 21.6°
T_x = T sinθ
T_y = T cosθ
where T is the tension in the cable.
Equation of equilibrium in y direction:
ΣF_y = 2T cosθ - Fg = 0
Solving for T:
2T cosθ = Fg
T = Fg / (2 cosθ)
Plugging in the values:
T = (147 N) / (2 cos(21.6°))
T ≈ 81.1 N
A metal filament Lamp rated at 750w, 100v into be connected in series with a capacitor across a 230v, 60hz supply. Calculate the capacitance required
Answers
Answer : 0.00885 farads or 8.85 microfarads
Explanation: To calculate the capacitance required, we can use the following formula:
C = 1 / [2 * pi * f * ((V^2 - Vlamp^2)/P)]
where:
C = capacitance in farads (F)
pi = 3.14159...
f = frequency in hertz (Hz)
V = voltage in volts (V)
P = power in watts (W)
Vlamp = voltage of the lamp in volts (V)
Using the given values, we have:
C = 1 / [2 * pi * 60 * ((230^2 - 100^2)/750)]
C = 1 / [2 * 3.14159 * 60 * ((230^2 - 100^2)/750)]
C = 1 / [113.09724]
C = 0.00885 farads (F)
Therefore, the capacitance required is approximately 0.00885 farads or 8.85 microfarads.
A stuntman of mass 55 kg is to be launched horizontally out of a spring- loaded cannon. The spring that will launch the stuntman has a spring coefficient of 266N / m and is compressed 5 m prior to launching the stuntman. If friction and air resistance can be ignored, what will be the approximate velocity of the stuntman once he has left the cannon?
Answers
The approximate velocity of the stuntman, once he has left the cannon, is 11 m/s.
StepsWe can use the conservation of energy, where the potential energy stored in the compressed spring is converted into the kinetic energy of the stuntman as he is launched out of the cannon.
The potential energy stored in the spring is given by:
PE = (1/2)kx²
where k is the spring constant and x is the distance the spring is compressed.
PE = (1/2)(266 N/m)(5 m)² = 3325 J
This potential energy is then converted into kinetic energy:
KE = (1/2)mv²
where m is the mass of the stuntman and v is his velocity.
3325 J = (1/2)(55 kg)v²
v² = (2*3325 J) / 55 kg
v² = 121 m²/s²
v = √(121 m²/s²) = 11 m/s
Therefore, the approximate velocity of the stuntman, once he has left the cannon, is 11 m/s.
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A 300 g football is kicked with an initial velocity of 140 m/s in a direction that
makes a 30° angle with the horizon. Find the peak height of the football.
Answers
Answer:
Explanation:
Assuming that air resistance is negligible, we can use the following kinematic equations to solve for the peak height:
v_f^2 = v_i^2 + 2ad
where v_f = 0 m/s (at the peak height) and a = -9.8 m/s^2 (acceleration due to gravity)
and
d = v_i t + (1/2)at^2
where d is the displacement or the peak height we want to find, v_i is the initial velocity, t is the time it takes to reach the peak height.
First, we need to resolve the initial velocity into its vertical and horizontal components:
v_i_x = v_i cos(30°) = 121.1 m/s
v_i_y = v_i sin(30°) = 70.0 m/s
Next, we can use the vertical component of the initial velocity to find the time it takes to reach the peak height:
v_f = v_i_y + at
0 m/s = 70.0 m/s + (-9.8 m/s^2)t
t = 7.14 s
Finally, we can use the time we found and the kinematic equation for displacement to find the peak height:
d = v_i_y t + (1/2)at^2
d = (70.0 m/s)(7.14 s) + (1/2)(-9.8 m/s^2)(7.14 s)^2
d = 247.5 m
Therefore, the peak height of the football is 247.5 meters.
Explain how a balloon is able to keep its shape?
Answers
Answer:
It depends on how they are made.
Explanation:
Rubber balloons are not always spherical in shape. When filled with air, the inflation forms a balance between the balloon material, including its shape and thickness and its elasticity and the pressure of the air. That’s what determines its shape. Although a sphere is often the shape, it could be tubular, offset, and most other shapes.
it the frequency of a wave is increased it’s
and
will decrease but its
will stay the same.
Answers
Answer:
Wavelength will decrease but frequency will stay the same.
Explanation:
The higher the frequency, the shorter the wavelength. The relationship between wavelength and frequency is called an inverse relationship, because as the frequency increases, the wavelength decreases.
However, the frequency usually remains the same because it is like a driven oscillation and maintains the frequency of the original source. If v changes and f remains the same, then the wavelength λ λ must change. Since v = f λ v = f λ , the higher the speed of a sound, the greater its wavelength for a given frequency.
Student A has a mass of 95 kg and student B has a mass of 82 kg. If their centers of mass are
0.6 meters apart, what will be the gravitational force each student exerts on the other?
Round your answer to the nearest whole number.
Answers
Answer:
0 Newtons, to the nearest whole number
Explanation:
Using the formula attached to this answer,
F = (6.67 x 10-¹¹ • 95 • 82) / 0.6²
F = 5.19593 x 10-⁷ / 0.36
F = 1.4433 x 10-⁶ N
F = 0.000001443 N
To the nearest whole number
F = 0 N
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If you lift one load up one story, how much more work do you do lifting one load up
three stories?
Answers
Answer:
Explanation:
alot