Answer:
7.98 Newton Seconds
Explanation:
J = F * (Δt)
J = 1450 (5.5 * 10^-3)
J = 7.975
We only use three significant digits, so our answer becomes 9.98
Answer:
B. Increases.
Explanation:
![{ \bf{F = \frac{m {v}^{2} }{r} }} \\ { \bf{F \: \alpha \: {v}^{2} }}](https://tex.z-dn.net/?f=%7B%20%5Cbf%7BF%20%3D%20%20%5Cfrac%7Bm%20%7Bv%7D%5E%7B2%7D%20%7D%7Br%7D%20%7D%7D%20%5C%5C%20%7B%20%5Cbf%7BF%20%5C%3A%20%20%5Calpha%20%20%5C%3A%20%20%7Bv%7D%5E%7B2%7D%20%7D%7D)
Keeping mass, and radius constant, speed or velocity is directly proportioanal to centripetal force.
Answer:
The girl is doing work on the ball because the energy in her muscles changed, even though the ball is not displaced.
Explanation:
The complete question is...
Which of these correctly describes whether a girl holding a ball in the same position is doing work on the ball?
-The girl is doing work on the ball because the energy of the ball changed, even though it is not displaced.
-The girl is doing work on the ball because the energy in her muscles changed, even though the ball is not displaced.
-The girl is doing no work on the ball because the ball is not displaced.
-The girl is doing no work on the ball because she is exerting a net force on the ball.
Holding up a ball costs energy, which is used to counter the work that would have otherwise be done on the ball by gravity. Although no physical distance is moved, we should consider the fact that by holding the ball, the girls hand exerts physical force to hold the ball in place. Also, there is a potential gravitational work on the ball due to gravity, but the force exerted by the girls hand does an equivalent of this gravito-potential work in order to counter it and hold the ball in place. All these activities eventually lead to a change in energy in her hand muscle to show that energy is expended.
Force can be exerted into an object with out it moving, but if you were to move the object due to force it would be considered work. (yes, you can exert force without having the object moving)
Answer:
Right shoe
Explanation:
Let the mass and velocity of incoming puck be m and v respectively.
Momentum of the colliding puck will be mv
In case of first case , the momentum of puck becomes zero so change in momentum after collision with left shoe
= mv - 0 = mv
If time duration of collision be t
rate of change of momentum
= mv / t
This is the force exerted by puck on the left shoe .
Now let us consider collision with right shoe
momentum after collision with right shoe
- mv
change in momentum
= mv - ( - mv ) = 2mv
If time duration of collision be t
rate of change of momentum
= 2mv / t
This is the force exerted by puck on the right shoe .
Since the force on the right shoe is more , this shoe will have greater speed
after collision.