The centripetal acceleration of this ball is equal to 12 [tex]m/s^2[/tex]
Given the following data:
Diameter = 1.5 mSpeed, V = 3 m/s.Mass = 0.25 kgRadius = [tex]\frac{Diameter}{2} = \frac{1.5}{2} = 0.75 \;meters[/tex]
To find the centripetal acceleration of this ball:
The acceleration of an object along a circular track is referred to as centripetal acceleration.
Mathematically, the centripetal acceleration of an object is given by the formula:
[tex]A_c = \frac{V^2}{r}[/tex]
Where:
Ac is the centripetal acceleration.r is the radius of the circular track.V is the velocity of an object.
Substituting the given parameters into the formula, we have;
[tex]A_c = \frac{3^2}{0.75}\\\\A_c = \frac{9}{0.75}\\\\A_c = \frac{9}{0.75}[/tex]
Centripetal acceleration = 12 [tex]m/s^2[/tex]
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which materials are good for constructing bridges and tall buildings
Answer:
Cement, iron, and steel all work great!
A 0.035-kg bullet is fired vertically at 214 m/s into a 0.15-kg baseball that is initially at rest. How high does the combined bullet and baseball rise after the collision, assuming the bullet embeds itself in the ball
Answer:
Explanation:
conservation of momentum during the collision
0.035(214) + 0.15(0) = 0.185v
v = 40.486 m/s
The kinetic energy after impact will convert to gravity potential energy
(ignoring air resistance)
mgh = ½mv²
h = v²/2g
h = 40.486² / (2(9.8))
h = 83.6303...
h = 84 m
if the Periodic time of an oscillating object Triples then its frequency will?
Answer:
it would decrease
Explanation:
f=1/T
Please Help
A projectile fired over level ground has an initial total velocity of 41.3 m/s. It is in the air for 5.1 s. What is the x-component of the projectile's initial velocity?
Answer:
Explanation:
In the vertical analysis assuming launch from ground level.
0 = 0 + (41.3sinθ)(5.1) + ½(-9.8)5.1²
(41.3sinθ)(5.1) = ½(9.8)5.1²
(41.3sinθ) = ½(9.8)5.1
sinθ = ½(9.8)5.1/41.3
sinθ = 0.60508...
θ = 37.235°
vx = 41.3cos37.235
vx = 32.881452...
vx = 32.9 m/s
Air is pumped into the tyre to inflate it.
This increases the temperature and the pressure of the air in the tyre.
Use ideas about molecules to explain why the air pressure in the tyre increases. *
Objects 1 and 2 attract each other with a gravitational force of 34 units. If the distance separating objects 1 and 2 is changed to one-third the original value, then the new gravitational force will be ____ units.
Answer:
F12 = G M1 M2 / R12^2
F12' = G M1 M2 / R12'^2
F12' / F12 = R12'^2 / R12^2 = (1/3)^2
F12' = 1/9 F12
The new force is 1/9 the of the old force
HELP PLS!!
A 3 kg mass is raised a distance of 14 m above the earth by a vertical force of 93 N.
a
The gain in potential energy of the mass, to 3 significant figures, is:
Hi there!
We know that:
U (Gravitational Potential Energy) = mgh
Where:
g = acceleration due to gravity (m/s²)
m = mass (kg)
h = height/displacement (m)
Plug in the values:
U = 3 × 9.8 × 14 = 412 J
Which statement best describes how light behaves with liquids, gases, and solids?
A. Light is unable to travel through liquids but travels easily through solids and some gases.
B. Light is unable to travel through gases but does travel through liquids and solids.
C. Light travels easily through liquids and gases, as well as through some solids like
glass.
D. Light travels easily through solids but is unable to travel through liquids and gases.
(20 points!)
Answer:
C number is write i think
A rocket ship has several engines and thrusters. While the Solid Rocket Booster (SRB) and main engines only work together during the first 2 minutes of flight, the main engines operate for a total of 8.5 minutes after the launch. Once the SRBs are released, the main engines alone accelerate the rocket from about 1341 m/s to 7600 m/s.
What is the acceleration of the SRB and main engine during the first 2.0 minutes of flight?
A. 52 m/s2
B. 13 m/s2
C. 9.8 m/s2
D. 11 m/s2
The acceleration of the SRB and main engine during the first 2.0 minutes of flight is 52.16 m/s².
The given parameters;
initial velocity of the engine, u = 1341 m/sfinal velocity of the engine, v = 7600 m/stime of motion, t = 2 minutes = 2 x 60 s = 120 sThe acceleration of the SRB and main engine is calculated as follows;
[tex]a = \frac{\Delta v}{\Delta t } \\\\a = \frac{7600 - 1341}{2 \times 60 s} \\\\a = 52.16 \ m/s^2[/tex]
Thus, the acceleration of the SRB and main engine during the first 2.0 minutes of flight is 52.16 m/s².
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Two objects are a distance of 1.7 x 103 meters apart. One object has a mass of 3 x 107 kg and the other has a mass of 6 x 108. Determine the gravitational force between the objects.
Answer: You need to use Newton's law for the equation --->
Explanation: G × M × m / separation. Thats how youll get your answer !!
A student stomps sternly on a super-sized stomp rocket. They notice that the rocket lands 54
m away in a time of 5.5 s. Find the magnitude of the total initial velocity of the rocket.
The magnitude of the total initial velocity of the rocket is determined as 9.82 m/s.
What is the total initial velocity of the rocket?
The magnitude of the total initial velocity of the rocket is calculated as follows;
V = D/T
where;
D is the distanceT is time of motionV = (54)/(5.5)
V = 9.82 m/s
Thus, the magnitude of the total initial velocity of the rocket is determined as 9.82 m/s.
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Amy uses 20 N of force to push a lawnmower 10 meters. How much work does she do?
Answer:
200J
Explanation:
work done = force x distance
= 20 x 10
= 200J
What is the relationship between balancing equations and the law of conservation of matter
What happens to the iron in the coilgun if the electricity in the coil was turned on
Upton Chuck is riding the Giant Drop at Great America. If Upton free falls for 2.60 seconds, what will be his final velocity and how far will he fall?
Answer:
d = -33.1 m and Vf = -25.5 m/s
Explanation:
Given:a = -9.8 m
t = 2.6 s
Vᵢ = 0 m/s
To Find:d = ?
Vf = ?
Now,
d = Vᵢ × t + 0.5 × a × t²
d = (0 m/s) × (2.60 s) + 0.5 × (-9.8 m/s²) × (2.60 s)²
d = -33.1 m (- indicates direction)
Vf = Vᵢ + a × t
Vf = 0 + (-9.8 m/s²) × (2.60 s)
Vf = -25.5 m/s (- indicates direction)
Thus, d = -33.1 m and Vf = -25.5 m/s
-TheUnknownScientist 72
A branch falls from a tree. How fast is the branch moving after 0.28 seconds?
A. 2.7 m/s
B. 1.3 m/s
C. 4.4 m/s
D. 3.1 m/s
Answer:
A. 2.7 m/s
Explanation:
Answer:
[tex]\boxed {\boxed {\sf A. \ 2.7 \ m/s}}[/tex]
Explanation:
We want to find how fast a branch is falling after 0.28 seconds.
Essentially, we want to find its final velocity at exactly 0.28 seconds. We will use the following kinematic equation:
[tex]v_f= v_i+at[/tex]
The branch fell from the tree, so it initially started at rest or 0 meters per second. The branch is in free fall, so its acceleration is due to gravity, or 9.8 meters per second squared. It falls for 0.28 seconds.
[tex]v_i[/tex]= 0 m/s a= 9.8 m/s²t= 0.28 sSubstitute the values into the formula.
[tex]v_f= 0 \ m/s + (9.8 \ m/s^2)(0.28 \ s)[/tex]
Multiply the numbers in parentheses.
[tex]v_f= 0 \ m/s +(9.8 \ m/s/s * 0.28 \ s )[/tex]
[tex]v_f= 0 \ m/s +2.744 \ m/s[/tex]
Add.
[tex]v_f= 2.744 \ m/s[/tex]
The original measurement of time has 2 significant figures, so our answer must have the same. For the number we found, that is the tenth place. The 4 in the hundredth place tells us to leave the 7.
[tex]v_f \approx 2.7 \ m/s[/tex]
The branch is moving at a velocity of approximately 2.7 meters per second.
The elevation at the base of a ski hill is 350 m above sea level. A ski lift raises a skier (total mass=72 kg, including equipment) to the top of the hill. If the skier's gravitational potential energy relative to the base of the hill is now 9.2 x 105 J, what is the elevation at the top of the hill?
The elevation at the top of the hill is 1,653.85 m.
The given parameters;
initial height of the skier, h₁ = 350 mlet the final height of the skier at the hill top, = h₂total mass, m = 72 kggravitational potential energy of the skier, P.E = 9.2 x 10⁵ JThe elevation at the top of the hill is calculated as follows;
[tex]P.E = mg\Delta h\\\\P.E = mg(h_2 -h_1)\\\\h_2 -h_1 = \frac{P.E}{mg} \\\\h_2 = \frac{P.E}{mg} + h_1\\\\h_2 = \frac{9.2 \times 10^5 }{72 \times 9.8} \ + \ 350 \ m\\\\h_2 = 1,653.85 \ m[/tex]
Thus, the elevation at the top of the hill is 1,653.85 m.
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At which type of boundary is new oceanic crust created?
A. a convergent plate boundary
B. a divergent plate boundary
C. a subduction plate boundary
D. a transform plate boundary
Answer:
c.
Explanation:
If the two plates that meet at a convergent plate boundary both are of oceanic crust, the older, denser plate will subduct beneath the less dense plate. The older plate subducts into a trench, resulting in earthquakes. Melting of mantle material creates volcanoes at the subduction zone.
When two oceanic plates converge, the denser plate will end up sinking below the less dense plate, leading to the formation of an oceanic subduction zone. Old, dense crust tends to be subducted back into the earth. An example of a subduction zone formed from a convergent boundary is the Chile-Peru trench….
Answer:
a divergent plate boundary
A car starts from rest and accelerates uniformly over a time of 5.21 seconds for a distance of 110 m. Determine the acceleration of the car.
Answer:
The acceleration of the car is 8.10 m/s²
Explanation:
Given:d = 110 m
t = 5.21 s
vi = 0 m/s
To Find:a = ?
d = Vᵢ × t + 0.5 × a × t²
110 m = (0 m/s) × (5.21 s) + 0.5 × (a) × (5.21 s)²
110 m = (13.57 s²) × a
a = (110 m)/(13.57 s²)
a = 8.10 m/s²
Thus, The acceleration of the car is 8.10 m/s²
-TheUnknownScientist 72
Saturn's mass is 5.68 x 1024 kg and its radius is 6.03 x 107 m. A. Calculate the gravitational field strength at Saturn's surface. (2 marks) B. Calculate the force of gravity at Saturn's surface on an object with a mass of 50 kg.
Hi there!
A.
We can calculate the gravitational field strength using the following equation:
[tex]g = \frac{Gm_p}{r^2}[/tex]
G = Gravitational Constant
mp = mass of planet (kg)
r = radius (m)
Plug in the given values:
[tex]g = \frac{(6.67*10^{-11})*(5.68*10^{24})}{(6.03*10^7)^2} = \boxed{0.104 N/kg}[/tex]
B.
The force can be calculated using:
[tex]F_g = \frac{Gm_1m_2}{r^2}[/tex]
Plug in the values:
[tex]F_g = \frac{(6.67*10^{-11})(5.68*10^{24})(50)}{(6.04*10^7)^2} = \boxed{5.209N}[/tex]
Answer:
[tex]\boxed {\boxed {\sf g=0.104 \ N/kg \ and \ F_g= 5.2 \ N }}[/tex]
Explanation:
A. Gravitational Field Strength
The gravitational field strength can be calculated using the following formula:
[tex]g= \frac{Gm}{r^2}[/tex]
G, or the universal gravitational constant, is 6.67 × 10⁻¹¹ N*m²/kg². The mass of Saturn is 5.68 × 10²⁴ kilograms. The radius of Saturn is 6.03×10⁷ meters.
Substitute these values into the formula.
[tex]g= \frac{ (6.67 \times 10^{-11} \ N*m^2/kg^2) (5.68 \times 10^{24} \ kg)}{(6.03 \times 10^{7} \ m )^2}[/tex]
Multiply the numerator and square the denominator.
[tex]g= \frac{ 3.78856 \times 10^{14} \ N *m^2/kg }{3.63609 \times 10^{15} \ m^2}[/tex]
Divide.
[tex]g= 0.1041932405 \ N/kg[/tex]
The original measurements of mass and radius have 3 significant figures, so our answer must have the same. For the number we found, that is the thousandth place. The 1 in the ten-thousandth place tells us to leave the 4 in the thousandth place.
[tex]\boxed {g \approx 0.104 \ N/kg}[/tex]
B. Force of Gravity
The force of gravity is calculated using the following formula:
[tex]F_g= mg[/tex]
The mass of the object is 50 kilograms. We just calculated the gravitational field strength, which is 0.104 Newtons per kilogram. Substitute these values into the formula.
[tex]F_g= (50 \ kg)(0.104 \ N/kg)[/tex]
Multiply. The units of kilograms cancel.
[tex]\boxed {F_g=5.20 \ N}[/tex]
A stone is dropped from the edge of a roof, and hits the ground with a velocity of -150 feet per second. How high (in feet) is the roof
Answer:
how long does it take? we need it to answer ure question
Explanation:
cause we don't know how many feet until we know how long it was falling
If a stone is dropped from the edge of a roof and hits the ground with a velocity of -150 feet per second, then the height of the roof would have been 1148 feet.
What are the three equations of motion?There are three equations of motion given by Newton
v = u + at
S = ut + 1/2 × a × t²
v² - u² = 2 × a × s
As given in the problem, a penny is dropped from a building that is 95 m tall, the initial velocity of the penny is zero, and the acceleration acting is due to the acceleration due to gravity,
By using the second equation of the motion for the vertical motion ,
v² = ( 2 × g ×h )
150² = 2 × 9.8 × h
h = 22500 / 19.6
= 1148 feet
Thus, the height of the roof would have been 1148 feet.
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A gold doubloon 6.1 cm in diameter and 2.0mm thick is dropped over the side of a Pirate Ship. When it comes to rest on the ocean floor at a depth of 770m how much has its volume changed
The volume of a material is the total amount of matter that it can contain. The volume of the given coin has been determined to be 5.85 x [tex]10^{-6}[/tex] [tex]m^{3}[/tex]. Since the gold doubloon do not absorb water, then its volume remains constant at the ocean floor.
The volume of the gold doubloon can be determined by;
volume = [tex]\pi r^{2}[/tex] + h
where r is the radius of the coin and h is its thickness.
Such that; diameter = 6.1 cm (61 mm) and h = 2.0 mm
r = [tex]\frac{diameter}{2}[/tex]
= [tex]\frac{61}{2}[/tex]
r = 30.5 mm
Thus,
volume of the coin = [tex]\frac{22}{7}[/tex] x [tex](30.5)^{2}[/tex] x 2
= 5847.2857
Therefore, the volume of the gold doubloon is 5847.3 [tex]mm^{3}[/tex]. This can also be expressed as 5.85 x [tex]10^{-6}[/tex] [tex]m^{3}[/tex].
Since the gold doubloon is not miscible with water, thus its volume at a depth of 770 m at the ocean floor is the same as its initial volume.
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what is kinetic friction ? what causes it ? what does it generate ?
Answer:
What is kinetic friction?
Kinetic friction is defined as a force that acts between moving surfaces. A body moving on the surface experiences a force in the opposite direction of its movement.
What causes it?
When the mass is not moving, the object experiences static friction. The friction increases as the applied force increases until the block moves. After the block moves, it experiences kinetic friction, which is less than the maximum static friction.
What does it generate?
When surfaces in contact move relative to each other, the friction between the two surfaces converts kinetic energy into thermal energy (that is, it converts work to heat). This property can have dramatic consequences, as illustrated by the use of friction created by rubbing pieces of wood together to start a fire.
A wheel has a radius of 0.40 m and is mounted on frictionless bearings. A block is suspended from a rope that is wound on the wheel and attached to it (see figure). The wheel is released from rest and the block descends 1.5 m in 2.00 s without any slipping of the rope. The tension in the rope during the descent of the block is 20 N. What is the moment of inertia of the wheel?
The moment of inertia of the wheel is 4.27 kg.m²
The kinematics equation explains the variables associated and related of motion.
From the information given, applying the kinematic equation of motion to determine the acceleration of the block, we have:
[tex]\mathbf{y = ut + \dfrac{1}{2}at^2}[/tex]
[tex]\mathbf{y = (0)t + \dfrac{1}{2}at^2}[/tex]
[tex]\mathbf{y = \dfrac{1}{2}at^2}[/tex]
Making acceleration (a) the subject, we have:
[tex]\mathbf{a = \dfrac{2y}{t^2}}[/tex]
where;
y = 1.5 mt = 2.0 s[tex]\mathbf{a = \dfrac{2\times 1.5 }{2.0^2}}[/tex]
a = 0.75 m/s²
The angular acceleration of the wheel can be estimated by the formula:
[tex]\mathbf{\alpha = \dfrac{a}{r}}[/tex]
[tex]\mathbf{\alpha = \dfrac{0.75 \ m/s^2}{0.40 \ m}}[/tex]
[tex]\mathbf{\alpha = 1.875 \ rad/s^2}[/tex]
Finally, the torque acting on the wheel is:
[tex]\mathbf{\tau = I \alpha}[/tex]
[tex]\mathbf{Tr = I \alpha}[/tex]
where;
T = tensionr = radiusI = moment of inertia∝ = angular acceleration∴
[tex]\mathbf{I =\dfrac{T\times r}{\alpha} }[/tex]
[tex]\mathbf{I =\dfrac{20 \ N\times 0.40 \ m}{1.875 \ rad/s^2} }[/tex]
I = 4.27 kg.m²
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What happens when the object is placed at F? Explain
your answer.
Answer:
Sample Response: No image will be formed because the rays will not converge to or diverge from a common point.
Explanation:
Who is Albert Einstein?
Answer:
Albert Einstein WAS a very well known Theoretical physicist
The USA claims he did not ever get his hands directly on an atomic bomb but in fact, other country textbooks like in Germany say he did.
Explanation:
A fun fact is that he was hired by the United States to make the Atomic bomb. Albert Einstein was a german yet many believed him to be a Smart American, they were wrong.
What is the difference between real and apparent weightlessness?
Answer:
In space we feel weightlessness because the earth's gravity has less effect on us. The Earth's gravitational attraction at those altitudes is only about 11% less than it is at the Earth's surface. If you had a ladder that could reach as high as the shuttle's orbit, your weight would be 11% less at the top.
Explanation:
Hope this helps:)
7) Germanium, element 32 on the Periodic Table, is shown here. If a proton is added to the nucleus of germanium, what outcome(s) would occur? Select ALL That apply.
A) The atom would increase in mass but would remain germanium.
B) The atom would become arsenic and have different properties.
The atom would remain germanium, but it would have a positive charge.
D) The atom would increase in mass and have different elemental properties.
E) The atom would expel a neutron to maintain a constant mass and chemical properties.
Addition of a proton to germanium will convert it to arsenic (element 33) having different properties.
The atomic number of an atom is the number of protons in the nucleus of the atom. The atomic number serves as the identity of an atom. If the atomic number is changed by adding or removing protons, the identity of that atom changes.
Hence, when a proton is added to germanium, the atom would become arsenic (element 33) and have different properties.
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Answer: Its B and D
I had the same question on usatestprep
which of the following statements might be used to defend the Act of 1848
A 10kg crate sits at rest on a rough flat surface. Astudent decides to pull the crate by attaching a rope at a 37 degree angle. Although the student pulls the rope with a force of 600 newtons, the coefficient of kinetic friction is large and has the force the student applies remains constant, how much time after he begins pulling the crate will it take before the crate has traveled a distance of 1.0 meter?
Answer:
Explanation:
Normal force of the surface on the box will be
N = mg - Fsinθ
Ν = 10(9.8) - 600sin37
N = -263
As normal force cannot be less than zero, the applied force lifts the crate off the surface.
Now it's just a matter of finding the acceleration
In the horizontal direction, the acceleration is
a = F/m
a = (600cos37) / 10
a = 47.9181... m/s²
the crate weight is mg = 10(9.8) = 98 N.
In the vertical direction the acceleration is
a = ((600sin37 - 98) / 10)
a = 26.3089... m/s²
total acceleration is
a = √(47.9181² + 26.3089²)
a = 54.6653... m/s²
s = ½at²
t = √(2s/a)
t = √(2(1.0)/54.6653)
t = 0.19127...
t = 0.19 s