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Setler [38]
3 years ago
9

A sprinter starts from rest and runs with constant acceleration to a top speed of 12m/s in 4.0 seconds.

Physics
1 answer:
bagirrra123 [75]3 years ago
4 0

Answer:

  a =  3,0 m/s²

Explanation:

En este ejercicio se pide calcular la aceleracion del cuerpo, usemos las ecuaciones de cinematica en una dimensión.

          v= v₀ + a t

como el corredor parte del reposo si velocidad inicial es cero

           v =  at

           a = v/t

calculemos

          a = 12 /4,0

          a =  3,0 m/s²

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8. Fig. 4.1 shows a heavy ball B of weight W suspended from a fixed beam by two ropes P and Q.
mart [117]

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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1.) I’m bored, so I decide to play catch by myself. I throw a 1.5kg ball in the air at 25 m/s. How long do I have to wait to cat
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A small block with mass 0.0400 kg slides in a vertical circle of radius 0.600 m on the inside of a circular track. During one of
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Answer:

Explanation:

Given that

The mass of the body is 0.04kg

M=0.04kg

The radius of the paths is 0.6m

r=0.6m

The normal force exerted at A is 3.9N

Fa=3.9N

The normal force exerted at B is 0.69N

Fb=0.69N

Then work done by friction from point A to B will be the change in K.E

W=∆K.E+P.E

So we need to know the velocity at both point A and B

Then since the centripetal force is given as

Ft=mv²/r

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v²=3.9/0.0667

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Now at point B

Fb=mv²/r

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P.E=0.471J

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W=K.Eb-K.Ea

K.E is given as 1/2mv²

W=1/2m(Vb²-Va²) +P.E

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W=0.5×0.04×(-48.1541) +0.471

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work was done on the block by friction during the motion of the block from point A to point B is 0.49J.

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