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WARRIOR [948]
2 years ago
14

A 0.50-kg red cart is moving rightward with a speed of 50 cm/s when it collides with a 0.50-kg blue cart that is initially at re

st. After the collision, the blue cart begins moving rightward with a speed of 40 cm/s. The red cart is still moving rightward but has slowed down to a speed of 10 cm/s. Enter the momentum values (in kg•cm/s) of each individual cart and of the system of two carts before and after the collision. Also indicate the change in momentum of each cart.
Physics
1 answer:
9966 [12]2 years ago
5 0

Explanation:

Step one:

given data

mass of red cart m1=0.5kg

initial velocity u1 of red cart=50cm/s

mass of blue cart m2=0.5kg

initial velocity u2 of blue cart= 0cm/s

final velocity v2 of blue cart= 40cm/s

final velocity v1 of red cart= 10cm/s

Step two:

Momentum values red cart before and after the collision.

Before collision

P1=m1u1

P1=0.5*50

P1=25kg•cm/s

After collision

P2=m1v1

P2=0.5*10

P2=5kg•cm/s

Change in momentum= P1-P2=25-5= 20kg•cm/s

Momentum values blue cart before and after the collision.

Before collision

P1=m1u1

P1=0.5*0

P1=0 kg•cm/s

After collision

P2=m1v1

P2=0.5*40

P2=20kg•cm/s

Change in momentum= P1-P2=0-20= -20kg•cm/s

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FromTheMoon [43]

Good morning dear...

Have a beautiful and joyful day ahead.

6 0
2 years ago
Jack and Jill have made up since the previous HW assignment, and are now playing on a 10 meter seesaw. Jill is sitting on one en
Airida [17]

Answer: 3 m.

Explanation:

Neglecting the mass of the seesaw, in order the seesaw to be balanced, the sum of the torques created by  gravity acting on both children  must be 0.

As we are asked to locate Jack at some distance from the fulcrum, we can take torques regarding the fulcrum, which is located at just in the middle of the length of the seesaw.

If we choose the counterclockwise direction as positive, we can write the torque equation as follows (assuming that Jill sits at the left end of the seesaw):

mJill* 5m -mJack* d = 0

60 kg*5 m -100 kg* d =0

Solving for d:

d = 3 m.

6 0
3 years ago
A particle moves along a straight line. Its position at any instant is given by x = 32t− 38t^3/3 where x is in metre and t in se
Rudik [331]

Answer:

The acceleration of the object is -69.78 m/s²

Explanation:

Given;

postion of the particle:

x = 32t - 38\frac{t^3}{3} \\\\

The velocity of the particle is calculated as the change in the position of the  particle with time;

v = \frac{dx}{dt} = 32 - 38t^2\\\\when \ the \ particle \ is \ at \ rest, \ v = 0\\\\32-38t^2 = 0\\\\38t^2 = 32\\\\t^2 = \frac{32}{38} \\\\t = \sqrt{\frac{32}{38} } \\\\t = 0.918 \ s

Acceleration is the change in velocity with time;

a = \frac{dv}{dt} = -76t\\\\recall , \ t = 0.918 \ s\\\\a = -76(0.918)\\\\a = -69.78 \ m/s^2

4 0
2 years ago
Mechanical energy is the sum of<br> energy and potential energy.
MAVERICK [17]

Answer:

True, the total amount of mechanical energy is merely the sum of the potential energy and the kinetic energy. This sum is simply referred to as the total mechanical energy.

Explanation: Hope it helps you:))))

have a good day

7 0
1 year ago
Why might it be necessary to ignore some of the data points just before and just after the collision?
krek1111 [17]
We have no idea. We need to examine the experimental set-up. You've given us no information, except that there may have been some sort of collision.
7 0
3 years ago
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