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tensa zangetsu [6.8K]
3 years ago
7

Jenny and Betty are having a great time at Busch Gardens riding the Ubanga Banga bumper cars. Jenny, who is traveling southward

in her bumper car, aims her car toward Betty, who is traveling northward in her bumper car. The cars collide and briefly come to a stop.
What is the direction of the acceleration of Jenny's car during the collision?

Southward or Northward?
Physics
2 answers:
Alex777 [14]3 years ago
6 0

Jenny is traveling southward.  In order to stop, she needs a northward acceleration.


A better way to say it:

Jenny is traveling southward in her bumper car, so the direction of her velocity is south.  In order to reduce her velocity to zero, a velocity of equal magnitude but directed north must be added to it.  Then the change in velocity is positive northward, and the change in velocity per unit time is acceleration.

max2010maxim [7]3 years ago
6 0

Answer:

The direction of the acceleration of Jenny's car during the collision is Northward.

Explanation:

Newton's Third Law is also known as the Principle of action and reaction. This law indicates that “Every action has an equal reaction but in the opposite direction” This means that when a body exerts a force (action) on another, it responds with a force of equal magnitude although opposite direction (reaction). Then this law explains that forces always occur in the form of pairs: an action and a reaction.

This law applies in this case. Jenny, who travels south in her bumper car, points her car at Betty. When crashing and stopping, the force received and its acceleration went north due to the pair of action (in this case the trip south in the bumper car) and reaction (the car responds with a force of equal magnitude although opposite direction, the North).

Then, the direction of the acceleration of Jenny's car during the collision is Northward.

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A skier is pulled by a towrope up a frictionless ski slope that makes an angle of 12 degrees with the horizontal. The rope moves
MArishka [77]

Answer:

Explanation:

Given,

  • Work done by the rope 900 m/s.
  • Angle of inclination of the slope = \theta\ =\ 12^o
  • Initial speed of the skier = v = 1.0 m/s
  • Length of the inclined surface = d = 8.0 m

part (a)

The rope is doing the work against the gravity on the skier to uplift up to the inclined surface. Therefore the work done by the rope is equal to the work done on the skier due to the gravity

\therefore W_r\ =\ W_g\ =\ 900\ J

In both cases the height attained by the skier is equal. and the work done by gravity does not depend upon the speed of the skier.

part (b)

  • Initial speed of the skier = v = 1.0 m/s.

Rate of the work done by the rope is power of the rope.

Power\ =\ \dfrac{\Delta W}{\Delta t}\\\Rightarrow P\ =\ \dfrac{\Delta W}{\dfrac{d}{v}}\\\Rightarrow P\ =\ \dfrac{\Delta W\times v}{d}\\\Rightarrow P\ =\ \dfrac{900\times 1.0}{8.0}\\\Rightarrow P\ =\ 112.5\ Watt

Part (c)

  • Initial speed of the skier = v = 2.0 m/s.

Rate of the work done by the rope is power of the rope.

Power\ =\ \dfrac{\Delta W}{\Delta t}\\\Rightarrow P\ =\ \dfrac{\Delta W}{\dfrac{d}{v}}\\\Rightarrow P\ =\ \dfrac{\Delta W\times v}{d}\\\Rightarrow P\ =\ \dfrac{900\times 2.0}{8.0}\\\Rightarrow P\ =\ 225\ Watt

4 0
2 years ago
a crate is being lifted into a truck. if it is moved with a 2470n force and 3650 j of work is done , then how far is the crate b
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Answer:

The crate was being lifted by a height of 1.48 meters.

Explanation:

In an attempt o move a crate;

Force applied = 2470 N

Work done by the force = 3650 J

We know that the work done is defined as the force used to move an object to a distance.

Given the Force used and the work done by that Force, we need to find out the distance the crate was lifted to.

Work done is defined as:

Work = Force*distance covered in the direction of the force

3650 = 2470*distance

distance = 3650/2470

distance = 1.48 meters

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In that case, their momentum must be equal. 
So, m1v1 = m2v2
20 * 20 = 40 * v2
v2 = 400 / 40
v2 = 10

In short, Your Answer would be: 10 m/s

Hope this helps!
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