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alukav5142 [94]
3 years ago
8

A 3.0-kg cart is rolling across a frictionless, horizontal track toward a 1.3-kg cart that is held initially at rest. The carts

are loaded with strong magnets that cause them to attract one another. Thus, the speed of each cart increases. At a certain instant before the carts collide, the first cart's velocity is +4.6 m/s, and the second cart's velocity is −1.9 m/s.
(Indicate the direction with the sign of your answer.)

(a) What is the total momentum of the system of the two carts at this instant?
kg · m/s

(b) What was the velocity of the first cart when the second cart was still at rest?
m/s
Physics
1 answer:
Karolina [17]3 years ago
6 0

Answer:

Explanation:

a ) Momentum of first cart = mass x velocity

= 3 x 4.6 =+13.8 kg m /s

Momentum of second cart = 1.3 x - 1.9 = - 2.47 kg m /s

Total momentum = 13.8 - 2.47

= +11.33 kg m /s

b )

Let the velocity of first cart be v at the moment when second cart was at rest

total momentum = 3 x v + 0 = 3 v

Applying conservation of momentum law

3 v  = +11.33

v = +3.77 m /s

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meriva

Answer:

300 Nm ; 300 J

Explanation:

Given that:

Force (F) = 20 N

Distance (d) = 15 m

The kinetic energy (Workdone) = Force * Distance

Kinetic Energy = 20N * 15m

Kinetic Energy = 300Nm

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4 0
3 years ago
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SCALCET8 3.9.018.MI. A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the
Firlakuza [10]

Answer:

The length of his shadow is decreasing at a rate of 1.13 m/s

Explanation:

The ray of light hitting the ground forms a right angled triangle of height H, which is the height of the building and width, D which is the distance of the tip of the shadow from the building.

Also, the height of the man, h which is parallel to H forms a right-angled triangle of width, L which is the length of the shadow.

By similar triangles,

H/D = h/L

L = hD/H

Also, when the man is 4 m from the building, the length of his shadow is L = D - 4

So, D - 4 = hD/H

H(D - 4) = hD

H = hD/(D - 4)

Since h = 2 m and D = 12 m,

H = 2 m × 12 m/(12 m - 4 m)

H = 24 m²/8 m

H = 3 m

Since L = hD/H

and h and H are constant, differentiating L with respect to time, we have

dL/dt = d(hD/H)/dt

dL/dt = h(dD/dt)/H

Now dD/dt = velocity(speed) of man = -1.7 m/s ( negative since he is moving towards the building in the negative x - direction)

Since h = 2 m and H = 3 m,

dL/dt = h(dD/dt)/H

dL/dt = 2 m(-1.7 m/s)/3 m

dL/dt = -3.4/3 m/s

dL/dt = -1.13 m/s

So, the length of his shadow is decreasing at a rate of 1.13 m/s

5 0
3 years ago
Describe and explain how the movement and arrangement of the particles in a block of ice change as the ice melts to form liquid
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Answer:

Ice is water in solid phase, in this phase, the particles are very close together and relatively in fixed positions.

As the temperature starts to increase (thermal energy), also does the kinetic energy of the particles (so we have a change from thermal energy to kinetic energy), so they start to move "more", and the position of the particles starts to be less "fixed". There is a point where the particles have enough energy, and this point is where the phase of the water changes from solid to liquid phase (the fusion point). After this point the water can not hold his shape, and takes the shape of the container where it is.

4 0
3 years ago
A top-fuel dragster starts from rest and has a constant acceleration of 42.0 m/s2. What are (a) the final velocity of the dragst
disa [49]

Answer:

a)  Final velocity of the dragster at the end of 1.8 s = 75.6 m/s

b) Final velocity of the dragster at the end of 3.6 s = 151.2 m/s

c) The displacement of the dragster at the end of 1.8 s = 68.04 m

d) The displacement of the dragster at the end of 3.6 s = 272.16 m

Explanation:

a) We have equation of motion v = u + at

  Initial velocity, u =  0 m/s

 Acceleration , a = 42 m/s²

 Time = 1.8 s    

Substituting

  v = u + at

  v  = 0 + 42 x 1.8 = 75.6 m/s

Final velocity of the dragster at the end of 1.8 s = 75.6 m/s

b) We have equation of motion v = u + at

  Initial velocity, u =  0 m/s

 Acceleration , a = 42 m/s²

 Time = 3.6 s    

Substituting

  v = u + at

  v  = 0 + 42 x 3.6 = 75.6 m/s

Final velocity of the dragster at the end of 3.6 s = 151.2 m/s

c) We have equation of motion s= ut + 0.5 at²

  Initial velocity, u =  0 m/s

 Acceleration , a = 42 m/s²

 Time = 1.8 s    

Substituting

   s= ut + 0.5 at²

    s = 0 x 1.8 + 0.5 x 42 x 1.8²

    s = 68.04 m

The displacement of the dragster at the end of 1.8 s = 68.04 m

d) We have equation of motion s= ut + 0.5 at²

  Initial velocity, u =  0 m/s

 Acceleration , a = 42 m/s²

 Time = 3.6 s    

Substituting

   s= ut + 0.5 at²

    s = 0 x 3.6 + 0.5 x 42 x 3.6²

    s = 272.16 m

The displacement of the dragster at the end of 3.6 s = 272.16 m

3 0
3 years ago
The total kinetic energy of two cars is measured before and after they crash into each other. The total kinetic energy of the tw
jenyasd209 [6]

Answer:

Option B, Some of the cars' kinetic energy was converted to sound and heat energy.

Explanation:

In an elastic collision, no energy is lost during and after collision. Thus, it can be said that in an elastic collision both momentum and kinetic energy remains conserved.  

While in non-elastic collision, kinetic energy of the system is lost. However, the momentum of the system is conserved. Generally, during and after collision some of the kinetic energy is lost as thermal energy, sound energy etc.  

Hence, option B is correct

4 0
3 years ago
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