Momentum = mass x velocity
Thus Option A is the correct answer
Momentum (dog) = 10 kg x (0.447 x 30) m/s
= 134.1 Kg m/s
Momentum ( bullet) = 0.02 kg x (0.447 x 800) m/s
= 7.152 Kg m/s
Momentum ( truck) = 0, as v = 0
tightrope has both low mass and low speed, thus its momentum will be low
Answer:
Required rate of return = 18.5 %
Explanation:
given,
rate of inflection = 4 %
risk free rate = 3 %
market risk premium = 5 %
firm has a beta = 2.30
rate of return has averaged 15.0% over the last 5 years
now,
Nominal risk free rate = risk free rate + inflation
= 3% + 4%
= 7%
Required rate of return = Nominal risk free rate + β (RPM)
= 7% + 2.3 x 5.0%
Required rate of return = 18.5 %
Answer:
c. The temperature at which a glass transforms from a solid to liquid.
Explanation:
The glass transition temperature is said to be a temperature range when a polymer structure transition from a glass or hardy(solid) material to a rubber like or viscous liquid material.
The glass transition temperature is an important property that is critical in product design.
Answer:
The child represented by a star on the outside path.
Explanation:
The given question is incomplete. The complete question is as follows.
In a nuclear physics experiment, a proton (mass 
kg, charge +e = 
C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed 
m/s. The proton comes momentarily to rest at a distance 
m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are 
m apart?
Explanation:
The given data is as follows.
Mass of proton = 
kg
Charge of proton = 
Speed of proton = 
Distance traveled = 
We will calculate the electric potential energy of the proton and the nucleus by conservation of energy as follows.
= 
where, 
U = 
Putting the given values into the above formula as follows.
U =
= 
= 
Therefore, we can conclude that the electric potential energy of the proton and nucleus is .
