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Marianna [84]
2 years ago
6

10. A girl pulls a wagon along a level path for a distance of 44 m. The handle of

Physics
1 answer:
Rina8888 [55]2 years ago
4 0

The work done on the wagon is 3549 J

Explanation:

The work done by a force when moving an object is given by

W=Fd cos \theta

where :

F is the magnitude of the force

d is the displacement

\theta is the angle between the direction of the force and of the displacement

In this problem we have the following data:

F = 87 N is the magnitude of the force

d = 44 m is the displacement of the wagon

\theta=22^{\circ} is the angle between the direction of the force and the displacement

Substituting, we find the work done

W=(87)(44)(cos 22^{\circ})=3549 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

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A long-distance swimmer is able to swim through still water at 4.0 km/h. She wishes to try to swim from Port Angeles, Washington
Roman55 [17]

Let \theta be the direction the swimmer must swim relative to east. Then her velocity relative to the water is

\vec v_{S/W}=\left(4.0\dfrac{\rm km}{\rm h}\right)(\cos\theta\,\vec\imath+\sin\theta\,\vec\jmath)

The current has velocity vector (relative to the Earth)

\vec v_{W/E}=\left(3.0\dfrac{\rm km}{\rm h}\right)\,\vec\imath

The swimmer's resultant velocity (her velocity relative to the Earth) is then

\vec v_{S/E}=\vec v_{S/W}+\vec v_{W/E}

\vec v_{S/E}=\left(\left(4.0\dfrac{\rm km}{\rm h}\right)\cos\theta+3.0\dfrac{\rm km}{\rm h}\right)\,\vec\imath+\left(4.0\dfrac{\rm km}{\rm h}\right)\sin\theta\,\vec\jmath

We want the resultant vector to be pointing straight north, which means its horizontal component must be 0:

\left(4.0\dfrac{\rm km}{\rm h}\right)\cos\theta+3.0\dfrac{\rm km}{\rm h}=0\implies\cos\theta=-\dfrac{3.0}{4.0}\implies\theta\approx138.59^\circ

which is approximately 41º west of north.

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2 years ago
Describe the energy transformations that occur from the time a skydiver jumps out of a plane until landing on the ground.
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A car initially traveling at 24 m/s slams on the brakes and moves forward 196 m before coming to a complete halt. What was the m
Marrrta [24]

Answer:

-1.47 m/s^2

Explanation:

We can use the following SUVAT equation to solve the problem:

v^2 - u^2 = 2ad

where

v = 0 is the final velocity of the car

u = 24 m/s is the initial velocity

a is the acceleration

d = 196 m is the displacement of the car before coming to a stop

Solving the equation for a, we find the acceleration:

a=\frac{v^2-u^2}{2d}=\frac{0-(24)^2}{2(196)}=-1.47 m/s^2

4 0
2 years ago
Part A
irina [24]

Answer:

v' = -18 m/s

Explanation:

  • Assuming no external forces acting during the collision, total momentum must be conserved, as follows:

       p_{o} = p_{f} (1)

  • The initial momentum can be expressed as follows (taking as positive the initial direction of the ball):

       m_{b} * v_{b} -M_{c}*V_{c}  = m_{b} * 18 m/s + (-M_{c}* 20 m/s)  (2)

  • The final momentum can be expressed as follows (since we know that v'b is opposite to the initial vb):

        -(m_{b} * v'_{b}) + M_{c}*V'_{c} (3)

  • If we assume that Mc >> mb, we can assume that the car doesn't change its speed at all as a result of the collision, so we can replace V'c by Vc in (3).
  • So, we can write again (3) as follows:

       -(m_{b} * v'_{b}) +(- M_{c}*V_{c}) = -(m_{b} * v'_{b})  + (-M_{c} * 20 m/s)  (4)

  • Replacing (2) and (4) in (1), we get:

       m_{b} * 18 m/s + (-M_{c}* 20 m/s) = -(m_{b} * v'_{b})  + (-M_{c} * 20 m/s)  (5)

  • Simplifying, and rearranging, we can solve for v'b, as follows:
  • v'_{b} = -18 m/s (6), which is reasonable, because everything happens as if the ball had hit a wall, and the ball simply had  inverted its speed after the collision.
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