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ivolga24 [154]
3 years ago
6

Which statement about the reaction is correct?

Physics
1 answer:
balu736 [363]3 years ago
8 0

Answer:

I think D am not pretty show

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A 35 kg trunk is pushed 4.8 m at constant speed up a 30° incline by a constant horizontal force. The coefficient of kinetic fric
netineya [11]

Answer:

We need to separate the x- and y-components of the applied force. For simplicity, I will denote the direction along the inclined plane as x-direction, and the perpendicular direction as y-direction.

F_x = F\cos(30^\circ)\\F_y = F\sin(30^\circ)

Only the x-component of the applied horizontal force does work on the trunk.

But we need to find the magnitude of the force. We know that the trunk is moving with constant speed. So, the x-component of the applied force is equal to the x-component of the gravitational force plus the force of friction.

0 = F\cos(30^\circ) - mg\sin(30^\circ) - mg\cos(30^\circ)\mu\\0.86F = 35(9.8)(0.5) + 35(9.8)(0.86)(0.19)\\F =264.5N

F_x = 264.5\cos(30^\circ) = 227.5N\\F_y = 264.5\sin(30^\circ) = 132.25N

W_{F_x} = F_xd = 227.5\times 4.8 = 1092J

The work done by the weight of the trunk can be calculated similarly. Only the x-component of the weight does work on the trunk.

W_g = -mg\sin(30^\circ)d

Note that the direction of the weight force is opposite of the direction of the motion, so this force does negative work on the trunk.

W_g = -823J

The energy dissipated by the frictional force can be found as follows:

W_f = -mg\cos(30^\circ)\mu d = -35(9.8)(0.86)(0.19)(4.8) = -269J

Additionally, the sum of work done by the friction and weight is equal in magnitude to the work done by the applied force. This shows that our calculations are consistent.

In the second part of the question, the applied force is on the x-direction. We will follow a similar procedure but a different force.

0 = F - mg\sin(30^\circ) - mg\cos(30^\circ)\mu\\0 = F - 35(9.8)(0.5) - 35(9.8)(0.86)(0.19)\\0 = F - 171.5 - 56\\F = 227.5N

W_F = Fd = 227.5\times 4.8 = 1092J

W_g = -mg\sin(30^\circ)d = -35(9.8)(0.5)(4.8) = -823J

W_f = -mg\cos(30^\circ)\mu d = -35(9.8)(0.86)(0.19)(4.8) = -269J

Explanation:

As you can see above, the answers are the same, although the directions of the applied forces are different. The reason for this situation is that in the first part the y-component does no work.

8 0
4 years ago
You are driving on a slippery road, and have stopped at an intersection. when you pull away from the intersection, your vehicle
AlexFokin [52]

<u>Answer</u>

You should still ensure you are steering straight in the direction of skidding.

Take your foot off the brake unless you are about to hit something. steer in the direction of the skid to straighten your car.

5 0
3 years ago
Read 2 more answers
element x has two isotopes: x-27 and x-29. x-27 has an atomic mass of 26.975 and a relative abundance of 82.33%, and x-29 has an
meriva

Answer:

27.336

Explanation:

The atomic mass is the weighted average of the atomic masses of the isotopes.  Simply multiply each isotopes's atomic mass by its relative abundance, then sum the results.

m = 0.8233 × 26.975 + 0.1767 × 29.018

m = 27.336

4 0
3 years ago
initially at rest. The bullet is traveling 300 m/s when it hits the wood block and sticks inside it. Now the bullet and the wood
madreJ [45]
Needs mass to be reasonably solvable, sorry
6 0
4 years ago
Objects a and b each have a mass of 25 kilograms. object a has a velocity of 5.98 meters/second. object b is stationary. they un
mash [69]

The total kinetic energy of the system after collision is 223.5J

In elastic collision kinetic energy and momentum are conserved.

According to the question

               mass of object a = 25kg

               mass of object b =  25 kg

        initial velocity of a u1 = 5.98 m/s

        initial velocity  of b u2 = 0

so from momentum conservation-

       mau1 + mbu2 = (m1+m2)v

       25kg × 5.98m/s + 25×0 = (25+25)v

              v = 2.99 m/s

Now the total kinetic energy after the collision will be:

          final kinetic energy = 1/2 (m1+m2) v²

                                           = 1/2 (25+25)× (2.99)²

                                           = 223.5 J

    Thus, total kinetic energy of the system after collision is 223.5J

Learn more about elastic collision here:

  brainly.com/question/1808045

    #SPJ4

7 0
2 years ago
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