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Colt1911 [192]
3 years ago
14

Jerome plays middle linebacker for South's varsity football team. In a game against cross-town rival North, he delivered a hit t

o North's 82-kg running back, changing his eastward velocity of 5.6 m/s into a westward velocity of 2.5 m/s.
Physics
1 answer:
anygoal [31]3 years ago
5 0

Explanation:

We have,

Mass of the running back, m = 82 kg

Initial speed of the running back, u = +5.6 m/s

Final speed of the running back, v = -2.5 m/s

It is required to find the momentum change of the running back. For this we will subtract final momentum to the initial momentum. So,

\Delta p=m(v-u)\\\\\Delta p=82\times (-2.5-5.6)\\\\\Delta p=-664.2\ kg-m/s

So, the magnitude of the change in momentum of the running back is 664.2 m/s in west direction.

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A hockey puck that has a mass of 0.170 kg travels with a speed of 42.0 m/s.
Rina8888 [55]

Answer:

Momentum = 7.14 Kgm/s

Explanation:

Given the following data;

Mass = 0.170 kg

Velocity = 42m/s

To find the momentum;

Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.

Mathematically, momentum is given by the formula;

Momentum = mass * velocity

Substituting into the equation, we have;

Momentum = 0.170 * 42

Momentum = 7.14kgm/s

Therefore, the momentum of the honey puck is 7.14 Kgm/s.

3 0
3 years ago
Problem 21-40a:
Greeley [361]

Answer:

Part a)

\frac{F_e}{F_g} = 2.74 \times 10^{12}

Part b)

q = 3.37 \times 10^{-4} C

Explanation:

As we know that electric force on electric charge is given as

F = qE

here we have

q = 1.6 \times 10^{-19}C

E = 153 N/C

now force is given as

F = (1.6 \times 10^{-19})(153) = 2.45 \times 10^{-17} N

Gravitational force on electric charge near surface of earth is given as

F_g = mg

F_g = (9.1 \times 10^{-31})(9.81) = 8.93 \times 10^[-30} N

now the ratio of two forces is given as

\frac{F_e}{F_g} = \frac{2.45 \times 10^{-17}}{8.93 \times 10^{-30}}

\frac{F_e}{F_g} = 2.74 \times 10^{12}

Part b)

Now the ball is balanced by the electric force and the force of gravity on it

so here we have

F_g = qE

mg = qE

(5.25 \times 10^{-3})(9.81) = q(153)

here we have

q = 3.37 \times 10^{-4} C

8 0
4 years ago
If the frequency of a wave that is traveling through a medium is doubled, the _
Sonja [21]
If the wavelength is doubled, frequency is halved. The wavelength and frequency of a wave are inversely proportional to each other and both are directly proportional to the velocity of the wave.
7 0
3 years ago
I’ve looked though the text book and looked it up but I dunno the answer
Feliz [49]
C. Surface.
Hope it helps!
3 0
3 years ago
What aspect of bouncing balls relates to matter
Lynna [10]

Answer: An aspect of the event of various types of balls bouncing off the same floor, being matter is that all the balls consist of matter. They all occupy space and have a form of energy when moved by a force, such as a person. And for energy, like I just said, when they bounce they create energy as they bounce up and down, so if the ball were to hit some other object, it would have an impact on the still object.

The combination of the material properties of a ball (surface textures, actual materials, amount of air, hardness/ softness, and so on) affects the height of its bounce.

Hope this helps.......... Stay safe and have a Merry Christmas!!!!!!!!!! :D

Explanation:

8 0
3 years ago
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