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lbvjy [14]
3 years ago
12

I need help please someone !!!!! Would appreciate it

Physics
1 answer:
babymother [125]3 years ago
5 0

Answer:

Yes, it would make it back up.

Explanation:

If it has 100,000 Joules of gravitational potential energy at the top of the hill, by the time the cart gets to the bottom, it will become PE = 0, KE = 90,000 since 10% of 100,000 is 10,000. The cart only requires 80,000J to climb back up so it should easily do so.

I didn't quite understand if the 10% energy loss is total, or every time it goes up or down, but it isn't a problem because 10% of 90,000 is 9,000, which means it would have 81,000J of energy on the way back up IF it loses energy due to friction on the way back up also.

The only physical law you need to prove this is the Law of Conservation of Energy: no energy is lost, only transformed; 10% of the energy becomes heat, the rest remains mechanical energy, which is the reason why the reasoning above works.

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2. Three blocks, A,B and C of mass 2kg. 3kg. 5kg respectively kept side by side with one another are accelerated at 2m/s2 across
gulaghasi [49]

Answer:

Total mass of combination = 2+3+5 = 10kg.

Acceleration produced = 2m/s^2

hence force =( total mass × acceleration)= (2×10)= 20 N.

Net force on 3kg block = acceleration × mass = (2 × 2 )= 4 N

applied force on 2 kg block = 20N

Force between 2 kg and 3 kg block = (20-4) = 16N. ans

Net force on 3 kg block = 3 × 2 =6N.

Applied force on 3 kg block due to 2 kg block = 16N.

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answers:-

(a) 20 N

(b) 16N

(c) 10 N

4 0
2 years ago
8. Fig. 4.1 shows a heavy ball B of weight W suspended from a fixed beam by two ropes P and Q.
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Answer:

The resultant tension of the two ropes is approximately 42.4 N

The length of the line representing the resultant tension is approximately 8.48 cm

Please find included  with the answer the scale drawing created with Microsoft Word

Explanation:

The given parameters are;

The tension in rope P, T_P = 30 N

The tension in rope Q, T_Q = 30 N

The angle the rope, 'P', makes with the horizontal = 45°

The angle the rope, 'Q', makes with the horizontal = 45°

The scale factor of the scale diagram, S.F. = 5.0 N/cm

By the resolution of forces at equilibrium, we have;

The sum of the vertical forces, \Sigma F_y = T_P_y + T_Q_y + W = 0

∴ W = -(T_P_y + T_Q_y)

W = -(30 × sin(45°) + 30 × sin(45°)) = -42.4264068712

The weight of the heavy ball, W ≈ 42.4 N acting downwards

The sum of the horizontal forces, \Sigma F_x = T_P_x + T_Q_x  = 0

The length of the resultant force, W = W/(S.F.) ≈ 42.4 N/(5.0 N/cm) = 8.48 cm

The drawing of the vectors using the scale factor of 5.0 N/cm is created using Microsoft Word is included

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C.) Sled Team C 28 kg moving at 12m/s

I'm pretty sure.

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