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Kruka [31]
1 year ago
8

1. Block A with a mass of 1 kg and block B with a mass of 2 kg lie on a smooth table. You wish to apply a horizontal force to ea

ch ball to give them the same horizontal acceleration (and you may safely ignore friction). Which of the following best describes the relative magnitude of the forces you must apply?
a) The magnitude of the force on block A must be twice as much as the force on block B.
b) The magnitude of the force on block B must be twice as much as the force on block A.
c) Both forces must have the same magnitude.
d) More information is needed to determine the relative magnitudes of the forces.

2. Block A has a mass of 1 kg and block B has a mass of 2 kg. Which of following best describes the relative magnitudes of the gravitational force (due to the Earth) on each block?
a)The gravitational force on block A is twice as much as the gravitational force on block B.
b)The gravitational force on block B is twice as much as the gravitational force on block A.
c)Both gravitational forces have the same magnitude.
d)More information is needed to determine the relative magnitudes of the forces.

3. You hold block A with a mass of 1 kg in one hand and block B with a mass of 2 kg in the other. You release them both from the same height above the ground at the same time. (Air resistance can safely be ignored as they fall.) Which of the following describes and explains the motion of the two objects after you release them?
a)Block A hits the ground before the block B. The same force of gravity acts on both objects, but because block A has a lower mass, it will have a larger acceleration.
b)Block B hits the ground before the block A. The larger force on block B causes it to have a greater acceleration than block A.
c)The two blocks hit the ground at the same time, because they both accelerate at the same rate while falling. The greater gravitational force on block B is compensated by its greater mass, giving it the same acceleration.
d)The two blocks hit the ground at the same time, because they both move at the same constant speed while falling. The forces on the blocks are irrelevant once they start falling.
Physics
1 answer:
goblinko [34]1 year ago
7 0

The gravitational force on block B is twice as much as the gravitational force on block A.

<h3>What is Force?</h3>

Force is the product of mass and acceleration. If the mass of A is 1Kg and the mass of B is 2Kg, In order to give the masses the same horizontal acceleration, the magnitude of the force on block B must be twice as much as the force on block A.

Given that the mass of A is 1Kg and the mass of B is 2Kg, the gravitational force on block B is twice as much as the gravitational force on block A.

If I hold block A with a mass of 1 kg in one hand and block B with a mass of 2 kg in the other  and release both from the same height above the ground at the same time, the two blocks hit the ground at the same time, because they both accelerate at the same rate while falling. The greater gravitational force on block B is compensated by its greater mass, giving it the same acceleration.

Learn more about force:brainly.com/question/12528243

#SPJ1

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An arrow is launched upward with an initial speed of 100 meters per second (m/s). The equations above describe the constant-acce
ankoles [38]

Answer:

d=510.2m

t=10.2s

Explanation:

The formulas for accelerated motion are:

v=v_0+at\\x=x_0+v_0t+\frac{at^2}{2}

From them we can get v^2=v_0^2+2ad.

We have:

v-v_0=at\\t=\frac{v-v_0}{a}

And substitute:

x=x_0+v_0(\frac{v-v_0}{a})+\frac{a}{2}(\frac{v-v_0}{a})^2\\x-x_0=\frac{v_0(v-v_0)}{a}+\frac{(v-v_0)^2}{2a}

We multiply both sides by 2a, and continue:

2a(x-x_0)=2v_0(v-v_0)+(v-v_0)^2=2v_0v-2v_0^2+v^2+v_0^2-2vv_0=v^2-v_0^2

Being d the displacement x-x_0, we have v^2=v_0^2+2ad

For our exercise, we will write this as:

d=\frac{v^2-v_0^2}{2a}

And taking upwards direction positive and imposing final velocity 0m/s (for maximum height), we have:

d=\frac{-v_0^2}{2a}=\frac{-(100m/s)^2}{2(-9,8m/s^2)}=510.2m

For the time we use:

t=\frac{v-v_0}{a}=\frac{-v_0}{a}=\frac{-(100m/s)}{(-9.8m/s^2)}=10.2s

6 0
3 years ago
A motorcycle is moving at a constant velocity of 15 meters/second. Then it starts to accelerate and reaches a velocity of 24 met
andrew11 [14]

Answer: 3 m/s^2

Explanation:

The acceleration of the motorcycle is given by

a=\frac{v-u}{t}

where

v=24 m/s is the final velocity of the motorcycle

u=15 m/s is the initial velocity

t=3 s is the time taken

Substituting these numbers into the equation, we find

a=\frac{24 m/s-15 m/s}{3 s}=\frac{9 m/s}{3s}=3 m/s^2

4 0
3 years ago
Read 2 more answers
please help with questions
kaheart [24]

Answer:

Please selection one of the following

3 0
2 years ago
40 POINTS!!! + BRAINLIEST!!!
Sergeu [11.5K]

Answer:

I think it is better if you read and shortly write my explanation

Explanation:

simple pendulum with no friction, mechanical energy is conserved. Total mechanical energy is a combination of kinetic energy and gravitational potential energy. As the pendulum swings back and forth, there is a constant exchange between kinetic energy and gravitational potential energy.

8 0
3 years ago
1. Calculate the height of tree, 250 m away that produces
gtnhenbr [62]

Answer:

Explanation:

1.5/30 = x/250

x = 12.5 m

5 0
2 years ago
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