Answers
Answer:
it is whatever the temperature is at 5(I cant seem to see it clearly)
Explanation:
Vaporisation takes place at all temperatures above room temperature ( from 20 to 100 degrees)
Related Questions
A 20-kg bicycle carrying a 50-kg girl is traveling with a kinetic energy of 2,290 J.
What is the velocity of the girl and bicycle?
8 m/s
16 m/s
64 m/s
4 m/s
Answers
Answer:
64 m/s
Explanation:
64m/s is the answer of question
64/m/s the answer
A +5.00 pC charge is located on a sheet of paper.
(a) Draw to scale the curves where the equipotential surfaces due to these charges intersect the paper. Show only the surfaces that have a potential (relative to infinity) of 1.00 V, 2.00 V, 3.00 V, 4.00 V, and 5.00 V.
(b) The surfaces are separated equally in potential. Are they also separated equally in distance?
(c) In words, describe the shape and orientation of the surfaces you just found.
Answers
Answer:
a) V = - x ( σ / 2ε₀)
c) parallel to the flat sheet of paper
Explanation:
a) For this exercise we use the relationship between the electric field and the electric potential
V = - ∫ E . dx (1)
for which we need the electric field of the sheet of paper, for this we use Gauss's law. Let us use as a Gaussian surface a cylinder with faces parallel to the sheet
Ф = ∫ E . dA = [tex]q_{int}[/tex] /ε₀
the electric field lines are perpendicular to the sheet, therefore they are parallel to the normal of the area, which reduces the scalar product to the algebraic product
E A = q_{int} /ε₀
area let's use the concept of density
σ = q_{int}/ A
q_{int} = σ A
E = σ /ε₀
as the leaf emits bonnet towards both sides, for only one side the field must be
E = σ / 2ε₀
we substitute in equation 1 and integrate
V = - σ x / 2ε₀
V = - x ( σ / 2ε₀)
if the area of the sheeta is 100 cm² = 10⁻² m²
V = - x (10⁻²/(2 8.85 10⁻¹²) = - x ( 5.6 10⁻¹⁰)
x = 1 cm V = -1 V
x = 2cm V = -2 V
This value is relative to the loaded sheet if we combine our reference system the values are inverted
V ’= V (inf) - V
x = 1 V = 5
x = 2 V = 4
x = 3 V = 3
These surfaces are perpendicular to the electric field lines, so they are parallel to the sheet.
In the attachment we can see a schematic representation of the equipotential surfaces
b) From the equation we can see that the equipotential surfaces are parallel to the sheet and equally spaced
c) parallel to the flat sheet of paper
A compass needle moves when it is near a wire with electric current. which of the following explains this phenomenon?
Answers
Answer:
D
Explanation:
Your chosen answer is correct .
Hope this helps :)
The electric current creates a magnetic field that explains this phenomenon. So, the correct option is D.
What is Magnetic field?A magnetic field is defined as the vector field that describes the magnetic effect on electric charges, electric currents and magnetic materials where a charge moving in a magnetic field experiences a force perpendicular to its velocity and magnetic field.
The magnetic field is explained as the region around a magnet that has a magnetic force where all magnets have north and south poles. Opposite poles attract each other, while like poles repel each other. The electric current creates a magnetic field that explains a compass needle moves when it is near a wire with electric current phenomenon.
Therefore, the correct option is D.
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write .00000023 in scientific notation
Answers
Answer: 2.3 x 10^-7
Explanation:
When you move the decimal to the right, you decrease the exponent. So lets count how many spots you have to move the decimal spot to get an answer between 1 and 10. I counted seven spots.
So:
[tex]2.3 *10^-^7[/tex]
2.3 * 10 °−7
The degree indicates that the number is that small thing that floats above
A 12,000-N car is raised using a hydraulic lift, which consists of a U-tube with arms of unequal areas, filled with oil with a density of 800 kg/m3 and capped at both ends with tight-fitting pistons. The wider arm of the U-tube has a radius of 18.0 cm and the narrower arm has a radius of 5.00 cm. The car rests on the piston on the wider arm of the U-tube. The pistons are initially at the same level. What is the force that must be applied to the smaller piston in order to lift the car after it has been raised 1.20 m
Answers
Answer:
The answer is below
Explanation:
Pascal's law states that pressure is exerted in all parts of a static fluid equally.
The area of the narrower arm with a radius of 0.05 m (5 cm) is given as:
Area of narrow arm = π(0.05)²
The area of the wider arm with a radius of 0.18 m (18 cm) is given as:
Area of narrow arm = π(0.18)²
The ratio if the wider arm area to the narrow arm area = π(0.18)² / π(0.05)² = 12.96
To move the 12000 N car, the amount of force needed = 12000/12.96 = 926 N
If a bird applies a 5 N upward force on a branch to lift the branch of ground to a
height of 24 m, how much work did the bird do?
Answers
What type of climate would be near cold ocean currents? PLEASE HELP QUICK!!
Answers
Answer:
hot humid with lots of rain.
Explanation:
ocean currents act as conveyer belts of warm and cold water sending heat to the polar regions and helping the tropical areas cool off, thus influencing both weather and climate. the tropics are particularly rainy because heat absorption , and thus ocean evaporation, is highest.
Mild with cooler weather than similar latitudes type of climate would be near cold ocean currents.
What is cold ocean currents?On the eastern side of ocean basins, lower latitudes or the tropics are where cold currents migrate in a direction toward the equator. They carry cold water into regions of the continent that have warm water, which are usually found on the west coastlines.
The Labrador Ocean Current and East Greenland Current flows are two noteworthy examples of cold currents.
From the perspectives of ecology and climate, ocean currents are also advantageous. For example, they are responsible for changing global temperatures.
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please help me with this physics problem
Answers
Answer:
It says the answer. (5 marks.)
Explanation:
The FitnessGram Pacer Test is a multistage aerobic capacity test that progressively gets more difficult as it continues. The 20 meter pacer test will begin in 30 seconds. Line up at the start. The running speed starts slowly, but gets faster each minute after you hear this signal. A single lap should be completed each time you hear this sound. Remember to run in a straight line, and run as long as possible.
list of energy changes when hydroelectricity city is made
Answers
Answer:
A turbine and generator produce the electricity
Explanation:
"A hydraulic turbine converts the energy of flowing water into mechanical energy. A hydroelectric generator converts this mechanical energy into electricity.
f F= {mango, apple, banana, orange)
Answers
Answer:
n(F) = 4Explanation:
Cardinality of a set is the number of elements in that set. Given the set.
F= {mango, apple, banana, orange), we are to determine the cardinality of the set i.e the amount of fruit present in the set. Cardinality of the set F is represented as n(F).
Since there are 4 different fruit in the given set F, hence the cardinality of the set F is n(F) = 4
What is the product of 0.0322 cm x 6.5 cm rounded to the correct number of significant figures?
Answers
Answer:
2.093 × 10⁻³ cm²
Explanation:
Given:
0.0322 cm × 6.5 cm
Find:
Product
Computation:
⇒ 0.0322 cm × 6.5 cm
⇒ 0.2093 cm²
2.093 × 10⁻³ cm²
You have purchased an inexpensive USB oscilloscope (which measures and displays voltage waveforms). You wish to determine if the oscilloscope has an error bias; in other words, you wish to determine if the errors made by the oscilloscope have a population mean that is not equal to zero. So you use a very accurate voltmeter to find the measurement errors for 13 different measurements made by your USB oscilloscope. A data file containing these measurements is HTMean1.csvPreview the document . Do a statistical analysis on this data to determine if the oscilloscope has an error bias.
Answers
Complete Question
The complete question is shown on the first uploaded image
Answer:
1
A
2
A
Explanation:
From the question the data given for Error (mV) is -15
-15.17
8.67
-13.74
-20.69
-6.96
-1.36
-2.96
-9.26
3.11
-14.12
6.39
-14.77
Generally
The null hypothesis is [tex]H_o : \mu = 0[/tex]
The alternative hypothesis is [tex]H_a : \mu \ne 0[/tex]
The sample size is n = 13
Here [tex]\mu[/tex] represents the true error bias (i.e population error bias)
Generally the sample error bias is mathematically represented as
[tex]\= x = \frac{ \sum x_i}{n}[/tex]
=> [tex]\= x = \frac{ -15.17 + 8.67 + (-13.74) + \cdots + (-14.77) }{13}[/tex]
[tex]\= x = -7.37[/tex]
Generally the standard deviation is mathematically represented as
[tex]\sigma = \sqrt{\frac{\sum (x_i - \= x )^2}{n} }[/tex]
=> [tex]\sigma = \sqrt{\frac{ (-15.17-( -7.37) )^2 + (8.67 -( -7.37) )^2 + \cdots + (-14.77 -( -7.37) )^2 }{13} }[/tex]
=> [tex]\sigma = \sqrt{ 119.385}[/tex]
=> [tex]\sigma = 10.926[/tex]
Generally the test statistics is mathematically represented as
[tex]t = \frac{\= x - \mu }{\frac{\sigma }{\sqrt{n} } }[/tex]
=> [tex]t = \frac{ -7.37 - 0 }{\frac{10.926}{\sqrt{13} } }[/tex]
=> [tex]t = -2.838[/tex]
Generally the p-value is mathematically represented as
[tex]p-value = 2 P(t < -2.432)[/tex]
From the z-table [tex]P(t < -2.432) = 0.0075 [/tex]
So [tex]p-value = 2* 0.0075 [/tex]
=> [tex]p-value = 0.015 [/tex]
So given that p-value is less than the [tex] \alpha = 0.05[/tex] then we reject the null hypothesis and conclude that the oscilloscope has an error bias
Change in Energy Quick Check
1 of 5
Which change occurs as an object falls toward the surface of Earth? (1 point)
The force of gravity on the object decreases.
The potential energy of the object decreases.
The acceleration due to gravity decreases.
The kinetic energy of the object decreases
IF SOMEONE GETS THE ANSWERS TO 1-5 of quick check pls post them here
Answers
Answer:
The force of gravity on the object decreases. (FALSE)
The potential energy of the object decreases. (TRUE)
The acceleration due to gravity decreases. (FALSE)
The kinetic energy of the object decrease (FALSE)
Explanation:
FORCE OF GRAVITY:
Force of gravity on the object is the weight of object, which depends upon the mass and value of acceleration due to gravity (W = mg). Since, the value mass is constant on Earth and acceleration due to gravity is also constant on earth.
Therefore, force of gravity remains same on the object. The statement is false.
POTENTIAL ENERGY:
Potential energy of object depends upon the mass, value of acceleration due to gravity, and change in height of object (P.E = mgΔh). Since, the value mass is constant on Earth and acceleration due to gravity increases as the object moves towards the surface of earth. But, the height of object is decreasing.
Therefore, potential energy of object decreases. The statement is true.
ACCELERATION DUE TO GRAVITY:
Acceleration due to gravity depends upon the altitude (gh = g[1 - 2h/Re]). Since, the height of object is decreasing.
Therefore, acceleration due to gravity increases. The statement is false.
But, this change is not significant.
KINETIC ENERGY:
Kinetic Energy of object, which depends upon the mass and velocity of the object (K.E = mv²/2). Since, the value mass is constant on Earth and velocity increases as the object moves towards the surface of earth.
Therefore, kinetic energy of the object also increases. The statement is false.
As the object falls towards the surface of the earth, the potential energy of the object decreases.
The earth is a large gravitational field. The potential energy of an object is dependent on its height. As a body falls through a height, its mass remains constant but its height decreases therefore its potential energy decreases also.
Hence, the true statement is that as the object falls towards the surface of the earth, the potential energy of the object decreases.
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Researchers have measured the acceleration of racing greyhounds as a function of their speed; a simplified version of their results is shown in (Figure 1). The acceleration at low speeds is constant and is limited by the fact that any greater acceleration would result in the dog pitching forward because of the force acting on its hind legs during its power stroke. At higher speeds, the dog's acceleration is limited by the maximum power its muscles can provide.
How far does the dog run until its speed reaches 4.0 m/s?
Answers
[tex]\Huge\boxed{\boxed{\dfrac{4}{5}\ \text{meters}}}[/tex]
Let's start by finding the time it takes for the dog to reach a velocity of [tex]4[/tex] m/s.
We can use the following equation, where [tex]v_i[/tex] is initial velocity, [tex]v_f[/tex] is final velocity, [tex]t[/tex] is time, and [tex]a[/tex] is acceleration.
[tex]v_f-v_i=at[/tex]
We're trying to solve for [tex]t[/tex] first, so divide both sides by [tex]a[/tex].
[tex]\dfrac{v_f-v_i}{a}=t[/tex]
Substitute in the known values.
[tex]\dfrac{4-0}{10}=t[/tex]
[tex]\dfrac{4}{10}=t[/tex]
[tex]\dfrac{2}{5}=t[/tex]
Now, we can use the following formula to find the distance.
[tex]s=v_it+\dfrac{1}{2}at^2[/tex]
Substitute in the known values.
[tex]s=0*\dfrac{2}{5}+\dfrac{1}{2}*10*(\frac{2}{5})^2[/tex]
Anything multiplied by [tex]0[/tex] is
[tex]s=\dfrac{1}{2}*10*(\frac{2}{5})^2[/tex]
Just simplify from there.
[tex]s=\dfrac{1}{2}*10*\dfrac{4}{25}[/tex]
[tex]s=5*\dfrac{4}{25}[/tex]
[tex]s=\dfrac{20}{25}[/tex]
[tex]s=\boxed{\dfrac{4}{5}}[/tex]
Hey there!
The hind legs of the dog create it to accelerate.
We know the mass of the dog is 36km (m).
We know that the acceleration of the dog it 10m/s² (a).
Find the average force with the formula [ f = ma ] where m = mass and a = acceleration.
f = 36*10
f = 360 newtons
We don't know the traveled distance but we do know that the starting speed of the dog was 0m/s and the ending speed was 4m/s.
We can use the formula [ vf² = vo² + 2ad ] where vf = final speed, vo = starting speed, a = acceleration, and d = distance.
We know all the variables except the distance, so we are going to solve for d.
(4)² = (0)² + 2(10)(d)
16 = 0 + 20d
16 = 20d
16/20 = 20d/20
0.8 = d
Therefore, the dog runs 0.8 meters until it reaches 4m/s.
Best of Luck!
what are the answers to the question
Answers
Answer: the link isnt loading
NEED HELP DUE AT 11:59!! A ball is thrown horizontally from the top of
a building 130 m high. The ball strikes the
ground 53 m horizontally from the point of
release.
What is the speed of the ball just before it
strikes the ground?
Answer in units of m/s.
Answers
Answer:
Since the ball was thrown horizontally, there was no vertical component in that force. and hence, the initial vertical velocity of the ball is 0 m/s and the initial horizontal velocity is r.
We are given:
initial velocity (u) = 0 m/s [vertical]
final velocity (v) = v m/s [vertical]
time taken to reach the ground (t) = t seconds
acceleration (a) = 10 m/s/s [vertical , due to gravity]
height from the ground (h) = 130 m
displacement (s) = 53 m [horizontal]
Solving for time taken:
From the third equation of motion:
s = ut + 1/2 at²
130 = (0)(t) + 1/2 * (10) * t²
130 = 5t²
t² = 26
t = √26 seconds or 5.1 seconds
Final Horizontal velocity of the ball
Since the horizontal velocity of the ball will remain constant:
the ball covered 53 m in 5.1 seconds [horizontally]
horizontal velocity of the ball = horizontal distance covered / time taken
Velocity of the ball = 53 / 5.1
Velocity of the ball = 10.4 m/s
Answer:
51.51519 m/s
Explanation:
Given: [tex]a_{x} =0[/tex] [tex]a_{y} -g[/tex] [tex]v_{yo} =0[/tex] [tex]x_{o} =0[/tex] [tex]x=53[/tex][tex]y_{o} =130[/tex]
X-direction | Y-direction
[tex]x=x_{o} +v_{xo}t[/tex] | [tex]y=y_{o} +v_{yo}t+\frac{1}{2}a_{y}t^2[/tex]
[tex]53=0v_{xo}(5.15078)[/tex] | [tex]0=130+\frac{1}{2}(-9.8)t^2[/tex]
[tex]53=v_{xo} (5.15078)[/tex] | [tex]-130=-4.9t^2[/tex]
[tex]\frac{53}{5.15078} =v_{xo}[/tex] | [tex]\sqrt{\frac{-130}{-4.9} }=\sqrt{t^2}[/tex]
[tex]10.2897=v_{xo}[/tex] | [tex]5.15078=t[/tex]
[tex]v=\sqrt{v_{y}^2+ v_{x}^2}[/tex] | [tex]v_{y}^2 =v_{yo}+2a_{y} d[/tex]
[tex]v=\sqrt{(50.27771)^2+(10.2897)^2}[/tex] | [tex]\sqrt{v_{y}^2} =\sqrt{2(-9.8)(0-130)}[/tex]
[tex]v=51.51519 m/s[/tex] | [tex]v_{y}=50.47771[/tex]
A block of mass 12.2 kg is sliding at an initial velocity of 6.65 m/s in the positive x-direction. The surface has a coefficient of kinetic friction of 0.253. (Indicate the direction with the signs of your answers.)
a.) What is the force of kinetic friction (in N) acting on the block?
(b What is the block's acceleration (in m/s2)?
(c) How far will it slide (in m) before coming to rest?
Answers
Explanation:
Given that,
Mass of the block, m = 12.2 kg
Initial velocity of the block, u = 6.65 m/s
The coefficient of kinetic friction, [tex]\mu_k=0.253[/tex]
(a)The force of kinetic friction is given by :
[tex]f=\mu_k mg[/tex]
mg is the normal force
So,
[tex]f=0.253\times 12.2\times 9.8\\\\f=30.24\ N[/tex]
(b) Net force acting on the block in the horizontal direction,
f = ma
a is the acceleration of the block
[tex]a=\dfrac{f}{m}\\\\a=\dfrac{30.24}{12.2}\\\\a=2.47\ m/s^2[/tex]
(c) Let d is the distance covered by the block before coming to the rest. Using third equation of motion as follows :
[tex]v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{-(6.65)^2}{2\times 2.47}\\\\d=-8.95\ m[/tex]
Hence, this is the required solution.
This question involves the concepts of the frictional force, Newton's second law of motion, and the equations of motion.
a) The force of kinetic friction acting on the block is "-30.28 N".
b) The acceleration of the block is "-2.48 m/s²".
c) The block will slide "8.9 m" before coming to rest.
a)
The kinetic friction force is given by the following formula:
[tex]F=\mu_k mg[/tex]
where,
F = frictional force = ?
[tex]\mu_k[/tex] = coefficient of kinetic friction = 0.253
m = mass of block = 12.2 kg
g = acceleration dueto gravity = 9.81 m/s²
Therefore,
[tex]F=(0.253)(12.2\ kg)(9.81\ m/s^2)[/tex]
F = - 30.28 N (negative sign due to resistive force)
b)
According to Newton's Second Law of Motion:
F = ma
[tex]a=\frac{F}{m} = \frac{-30.28\ N}{12.2\ kg}\\\\[/tex]
a = -2.48 m/s² (negative sign due to deceleration)
c)
Using the third equation of motion to calculate the distance:
[tex]2as=v_f^2-v_i^2[/tex]
where,
s = distance = ?
vf = final speed = 0 m/s
vi = initial speed = 6.65 m/s
Therefore,
[tex]2(-2.48\ m/s^2)s=(0\ m/s)^2-(6.65\ m/s)^2\\\\s=\frac{-(6.65\ m/s)^2}{2(-2.48\ m/s^2)}[/tex]
s = 8.9 m
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The attached picture shows Newton's Second Law of Motion.
25° 20' 42" S 131° 2' E
1. What is the place?
Answers
An airplane flies with a constant speed of 540 miles per hour. How far can it travel in 1 1/4 hours?
Answers
Answer:
Hey!
Your answer is 675 MILES!
Explanation:
Using the S = D x T formula...
We first need to convert any values... 1 1/4 hours--75 mins (1.25hrs)
So now we substitute into the formula!
D (distance)= 540 x 1.25 = 675 (Distance = Speed x Time)
Distance Travelled= 675 MILES!
I HOPE THIS HELPED YOU!why bananas are curved
Answers
they are bananas and bananas are curved because they grow curved
I need help with this answer
Answers
Answer:
Single Replacement
Explanation:
A boy kicked off a cliff and lands 151m away 45s later. What was the initial velocity? How tall is the cliff?
Answers
That’s the height of the cliff
The original sideways velocity = 151/45 = 3.35 m/s
3 Points
The half-value layer for lead for g rays from 60Co is 12.8 mm. An electrometer gave a count rate of 35 pulses per second when at a
certain distance from a source, with no lead between source and detector. If 6 mm of lead were then inserted between source and
detector, the count rate (in pulses per second) would be nearest to:
A. 16
OB. 8
OC. 20
D. 25
OE. 75
Answers
Answer:
D. 25
Explanation:
The number of pulses per second (p) for x mm of lead can be modeled by the exponential equation ...
p = 35(0.5^(x/12.8))
Then for x=6, the value of p is ...
p = 35·0.5^(6/12.8) ≈ 25.29
The expected count rate would be nearest 25 pps.
PLZ EXPLAIN AND I WILL GIVE YOU BRANILEST
How do two interacting objects exert equal and opposite forces on each other when they collide, even though they have different masses?
Answers
Answer:
Did a little research.
Explanation:
The Law of Action-Reaction:
Newton's third law of motion is naturally applied to collisions between two objects. In a collision between two objects, both objects experience forces that are equal in magnitude and opposite in direction. Such forces often cause one object to speed up (gain momentum) and the other object to slow down (lose momentum). According to Newton's third law, the forces on the two objects are equal in magnitude. While the forces are equal in magnitude and opposite in direction, the accelerations of the objects are not necessarily equal in magnitude. In accord with Newton's second law of motion, the acceleration of an object is dependent upon both force and mass. Thus, if the colliding objects have unequal mass, they will have unequal accelerations as a result of the contact force that results during the collision.Consider the collision between the club head and the golf ball in the sport of golf. When the club head of a moving golf club collides with a golf ball at rest upon a tee, the force experienced by the club head is equal to the force experienced by the golf ball. Most observers of this collision have difficulty with this concept because they perceive the high speed given to the ball as the result of the collision. They are not observing unequal forces upon the ball and club head, but rather unequal accelerations. Both club head and ball experience equal forces, yet the ball experiences a greater acceleration due to its smaller mass. In a collision, there is a force on both objects that causes an acceleration of both objects. The forces are equal in magnitude and opposite in direction, yet the least massive object receives the greatest acceleration.
Consider the collision between a moving seven ball and an eight ball that is at rest in the sport of table pool. When the seven ball collides with the eight ball, each ball experiences an equal force directed in opposite directions. The rightward moving seven ball experiences a leftward force that causes it to slow down; the eight ball experiences a rightward force that causes it to speed up. Since the two balls have equal masses, they will also experience equal accelerations. In a collision, there is a force on both objects that causes an acceleration of both objects; the forces are equal in magnitude and opposite in direction. For collisions between equal-mass objects, each object experiences the same acceleration.
How many sets of planets would you need to create the mass of the Sun?
Answers
Answer:
volume about 1.3 million Earths could fit inside the Sun the mass of the Sun is 1.989 X 10 to the 30 kg about 333000 times the mass of the Earth
Explanation:
step-by-step explanation hope this answer your question
4. A tankful of liquid has a volume
of 0.2m3. What is the volume in (a)
lities (b) cm3 (c)ml
Answers
Explanation:
sjaaqqjajkslxxjxn vbnvisowpsjsndncnmcnvngieoowpwisnxxnnccnnckfkdkdjcncjcmdmcmcmdmcmdmdmcmcmcdjqowieurisosksjsjsosoapqpskdj
please help me. i have 2 hours
Answers
the missing word is clockwise moment. I hope this helps good luck
A pile of bricks of mass M is being raised to the tenth floor of a building of height H = 4y above the ground by a crane that is on top of the building. During the first part of the lift, the crane lifts the bricks a vertical distance h1=3y in a time t1=4T. During the second part of the lift, the crane lifts the bricks a vertical distance h2=y in t2=T. Which of the following correctly relates the power P1 generated by the crane during the first part of the lift to the power P2 generated by the crane during the second part of the lift?
A. P2=4P1
B. P2=43P1
C. P2=P1
D. P2=34P1
E. P2=13P1
Answers
Complete Question
A pile of bricks of mass M is being raised to the tenth floor of a building of height H = 4y above the ground by a crane that is on top of the building. During the first part of the lift, the crane lifts the bricks a vertical distance h1=3y in a time t1=4T. During the second part of the lift, the crane lifts the bricks a vertical distance h2=y in t2=T. Which of the following correctly relates the power P1 generated by the crane during the first part of the lift to the power P2 generated by the crane during the second part of the lift?
[tex]A.\ \ P_2=4P_1[/tex]
[tex]B.\ \ P_2=\frac{4}{3} P1[/tex]
[tex]C.\ \ P_2=P_1[/tex]
[tex]D. \ \ P_2=\frac{3}{4} P_1[/tex]
[tex]E. \ \ \ P_2=\frac{1}{3} P_1[/tex]
Answer:
The correct option is B
Explanation:
From the question we are told that
The mass of the brick is M
The height height of the 10th floor is H = 4y
The height attained during the first part of the lift is [tex]h_1 = 3y[/tex]
The time taken is [tex]t_1 = 4T[/tex]
The height attained during the second part of the lift is [tex]h_2 = y[/tex]
The time taken is [tex]t_2 = T[/tex]
Generally the velocity of the crane during the first lift is mathematically represented as
[tex]v_1 = \frac{h_1}{t_1}[/tex]
=> [tex]v_1 = \frac{3y}{4T}[/tex]
Generally the velocity of the crane during the first lift is mathematically represented as
[tex]v_1 = \frac{h_2}{t_2}[/tex]
=> [tex]v_1 = \frac{y}{T}[/tex]
Generally the power generated during the first lift is
[tex]P_1 = F_1 * v_1[/tex]
Here [tex]F_1[/tex] force applied during the first lift which is mathematically represented as
[tex]F_1 = M * g[/tex] here g is the acceleration due to gravity
So
[tex]P_1 = Mg * \frac{3y}{4T}[/tex]
Generally the power generated during the second lift is
[tex]P_2 = F_2 * v_2[/tex]
Here [tex]F_2[/tex] force applied during the second lift which is mathematically represented as
[tex]F_2 = M * g[/tex] here g is the acceleration due to gravity
So
[tex]P_2 = Mg * \frac{y}{T}[/tex]
So the ratio of the first power to the second power is
[tex]\frac{P_1}{P_2} = \frac{Mg * \frac{3y}{4T}[}{Mg * \frac{y}{T}}[/tex]
=> [tex]\frac{P_1}{P_2} = \frac{3}{4}[/tex]
=> [tex]P_2 = \frac{4}{3} P_1[/tex]
If dx denotes the change in position of an object and dt denotes the corresponding time interval, then instantaneous velocity is given by:
Answers
Answer:
[tex]Velocity=\frac{dx}{dt}[/tex]
Explanation:
Remember that instantaneous velocity is just a measure to know the velocity that any object has at any point given in time, so we just need to know the distance it has travel, which would be the change in position, and the time it took that change in position to occurr, this means distance by time, so we just divide dx by dt and we have the solution for instantaneous velocity.
If a change in position as denoted by [tex]dx[/tex] and [tex]dt[/tex] change in time, the instantaneous velocity will be given by,
[tex]v = \dfrac {dx}{dt}[/tex]
What is Velocity?It can be defined by the change in position of the object over time. This is a vector quantity. Vector quantity is a quantity that has both magnitude and direction.Instantaneous velocity:The velocity of the object at a point of time is known as instantaneous velocity. Instantaneous velocity can be calculated by the ratio of change in position to the elapsed point of time.
[tex]v = \dfrac {dx}{dt}[/tex]
Where,
[tex]v[/tex] - instantaneous velocity
[tex]dt[/tex] - change in distance (position)
[tex]dt[/tex]- change in time
Therefore, if a change in position as denoted by [tex]dx[/tex] and [tex]dt[/tex]change in time, the instantaneous velocity will be given by,
[tex]v = \dfrac {dx}{dt}[/tex]
Learn more about Instantaneous velocity.
https://brainly.com/question/13372043
A 2.60-kg box is sliding across the horizontal floor of an elevator. The coefficient of kinetic friction between the box and the floor is 0.380. Determine the kinetic frictional force that acts on the box for each of the following cases. (a) The elevator is stationary. N (b) The elevator is accelerating upward with an acceleration whose magnitude is 1.20 m/s2. N (c) The elevator is accelerating downward with an acceleration whose magnitude is 1.20 m/s2. N Additional Materials
Answers
Answer:
a) F = 9.69 N
b) F = 10.88 N
c) F = 8.51 N
Explanation:
a) The kinetic frictional force when the elevator is stationary is the following:
[tex] F_{k} = \mu_{k}N = \mu(mg) [/tex]
Where:
F(k) is the kinetic frictional force
N is the normal force = mg
m: is the mass = 2.60 kg
g: is the gravity = 9.81 m/s²
μ(k) is the coefficient of kinetic friction = 0.380
[tex] F_{k} = \mu(mg) = 0.380*2.60 kg*9.81 m/s^{2} = 9.69 N [/tex]
b) When the elevator is accelerating upward with acceleration "a" equal to 1.20 m/s²
[tex] F_{k} = \mu_{k}N = \mu[m(g + a)] [/tex]
The normal is equal to mg plus ma because the elevator is accelerating upward
[tex] F_{k} = \mu m(g + a) = 0.380*2.60 kg(9.81 m/s^{2} + 1.20 m/s^{2}) = 10.88 N [/tex]
c) When the elevator is accelerating downward with a =1.20 m/s² we can find the kinetic frictional force similar to the previous case:
[tex] F_{k} = \mu m(g - a) = 0.380*2.60 kg(9.81 m/s^{2} - 1.20 m/s^{2}) = 8.51 N [/tex]
I hope it helps you!