To determine the object which could give the greatest impact we will apply the concept of momentum. The object that has the highest momentum will be the object that will impact the strongest. Our values are
Mass of Object A

Velocity of object A

Mass of object B

Velocity of object B

The general formula for momentum is the product between mass and velocity, then

For each object we have then,


Since the momentum of object A is greater than that of object B, then object A will make you feel force upon impact.
The answer would be center of mass, B
Nothing happens. The frequency is determined at the source,
and it doesn't change along the way.
Answer:
The fraction of its energy that it radiates every second is
.
Explanation:
Suppose Electromagnetic radiation is emitted by accelerating charges. The rate at which energy is emitted from an accelerating charge that has charge q and acceleration a is given by

Given that,
Kinetic energy = 6.2 MeV
Radius = 0.500 m
We need to calculate the acceleration
Using formula of acceleration

Put the value into the formula

Put the value into the formula


We need to calculate the rate at which it emits energy because of its acceleration is

Put the value into the formula


The energy in ev/s


We need to calculate the fraction of its energy that it radiates every second


Hence, The fraction of its energy that it radiates every second is
.
The velocity of the canoe is 1.7 m/s.
<h3>What is momentum?</h3>
Momentum in physics is the products of mass and velocity. Now we have to find momentum with the formula; p = mv
a) Initial momentum = (15)8 m/s + 135 = 255 Kgms-1
b) Since momentum is conserved, the total momentum after throwing the anchor is still 255 Kgms-1
c) The final velocity of the boat is obtained from;
255 Kgms-1 = (15Kg + 135 Kg) v
v = 255 Kgms-1/(15Kg + 135 Kg)
v = 1.7 m/s
Learn more about momentum: brainly.com/question/904448