1. An object moving with velocity 15 ms accelerates at 2 m/s for 5 seconds. Calculate the final velocity.

A 2500‐kg vehicle traveling at 25 m/s can be stopped by gently applying the breaks for 20 seconds. What is the average force sup

katen-ka-za [31]

Momentum = (mass) x (speed)

Change in momentum = (force) x (time)

The initial momentum is (mass) x (speed) = 2500x 25 = 62,500 kg-m/s.

Since you want to <u>stop</u> the vehicle, that number is also the required <em>change</em>
in momentum ... you want the vehicle to wind up with zero momentum.

62,500 = (force) x (time) = 20 x force

Divide each side by 20 :

force = 62,500 / 20 = <em>3,125 newtons </em>

3 0

3 years ago

Two spheres having masses M and 2M and radii R and 3R, respectively, are released from rest when the distance between their cent

Andrei [34K]

Answer:

$v_2 = \sqrt{\frac{GM}{3R}}$

$v_1 = 2\sqrt{\frac{GM}{3R}}$

Explanation:

As we know by energy conservation that change in gravitational potential energy of the system = change in kinetic energy of the two ball

So here we can say

$-\frac{GM(2M)}{12R} + 0 = -\frac{GM(2M)}{4R} + \frac{1}{2}Mv_1^2 + \frac{1}{2}(2M)v_2^2$

Also since there is no external force on the system of two masses so here total momentum of the two balls will remains conserved

$0 = Mv_1 + 2Mv_2$

$v_1 = -2v_2$

now we have

$\frac{GM^2}{2R} - \frac{GM^2}{6R} = \frac{1}{2}M(-2v_2)^2 + \frac{1}{2}(2M)v_2^2$

$\frac{GM^2}{3R} = Mv_2^2$

$v_2 = \sqrt{\frac{GM}{3R}}$

$v_1 = 2\sqrt{\frac{GM}{3R}}$

4 0

4 years ago

During a lab investigation, students added four 50 g masses to two boxes and arranged the boxes so that they were motionless on

IceJOKER [234]

When the resultant force is not equal to zero termed an unbalanced force. By procedures 4 and 5 students observe an unbalanced upward force on Box 1. Hence option 1 is right for the problem.

<h3>What is an unbalanced force?</h3>

The forces operating on a body are known as unbalanced forces when the resulting force exerted on it is not equal to zero.

Unbalanced forces acting on the body, causing it to modify its state of motion. To further grasp the nature of imbalanced forces.

<h3 />

The following reasons by which we can understand the unbalanced force caused by the box.

Due to these two reasons, books will move up.

By adding another mass to box 2. The box becomes lighter. As the box becomes lighter the gravity force acting on the box will be less due to which the box easily can move up.

By removing the two masses from box 1. Due to which other become heavier other becomes heavier pulling it down causing box 1 one to go up.

Hence by procedures 4 and 5 students observe an unbalanced upward force on Box 1. Hence option 1 is right for the problem.

To learn more about the unbalanced force refer to the link;

brainly.com/question/227461

3 0

3 years ago

A toy rocket is launched vertically from ground level (y = 0.00 m), at time t = 0.00 s. The rocket engine provides constant upwa

Ksju [112]

The toy rocket is launched vertically from ground level, at time t = 0.00 s. The rocket engine provides constant upward acceleration during the burn phase. At the instant of engine burnout, the rocket has risen to 72 m and acquired a velocity of 30 m/s. The rocket continues to rise in unpowered flight, reaches maximum height, and falls back to the ground with negligible air resistance.

The total energy of the rocket, which is a sum of its kinetic energy and potential energy, is constant.

At a height of 72 m with the rocket moving at 30 m/s, the total energy is m*9.8*72 + (1/2)*m*30^2 where m is the mass of the rocket.

At ground level, the total energy is 0*m*9.8 + (1/2)*m*v^2.

Equating the two gives: m*9.8*72 + (1/2)*m*30^2 = 0*m*9.8 + (1/2)*m*v^2

=> 9.8*72 + (1/2)*30^2 = (1/2)*v^2

=> v^2 = 11556/5

=> v = 48.07

<span>The velocity of the rocket when it impacts the ground is 48.07 m/s</span>

3 0

3 years ago

A golf club rotates 215 degrees and has a length (radius) equal to 29 inches. The time it took to swing the club was 0.8 seconds

vichka [17]

Answer:

The average linear velocity (inches/second) of the golf club is 136.01 inches/second

Explanation:

Given;

length of the club, L = 29 inches

rotation angle, θ = 215⁰

time of motion, t = 0.8 s

The angular speed of the club is calculated as follows;

$\omega = (\frac{\theta}{360} \times 2\pi, \ rad) \times \frac{1}{t} \\\\\omega = (\frac{215}{360} \times 2\pi, \ rad) \times \frac{1}{0.8 \ s} \\\\\omega = 4.69 \ rad/s$

The average linear velocity (inches/second) of the golf club is calculated as;

v = ωr

v = 4.69 rad/s x 29 inches

v = 136.01 inches/second

Therefore, the average linear velocity (inches/second) of the golf club is 136.01 inches/second

8 0

3 years ago

1. An object moving with velocity 15 ms accelerates at 2 m/s for 5 seconds. Calculate the final velocity.​

1. An object moving with velocity 15 ms accelerates at 2 m/s for 5 seconds. Calculate the final velocity.