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2 years ago

Snowboarder Jump—Energy and Momentum

Physics

1 answer:

soldier1979 [14.2K]2 years ago

4 0

Answer:

THE ANSWER TERMS ARE DEFINED BLOW:-

Explanation:

MOMENTUM- IT IS THE ABILITY TO INCREASE OR DEVELOP CONSTANT FORCE.

KINETIC ENERGY:- IT IS THE ENERGY THAT A PRTICLE POSSES WHEN IT IS ACTUALLY IN MOTION.

POTENTIAL ENERGY:- IT IS THE ENERGY THAT A PARTICLE POSSES WHEN IT ACTUALLY IS IN RESTING STATE.

IN THIS ACIVITY THE SNOWBOARDER IS IN THE MOTION STATE THEREFORE HE POSSES KINETIC ENERGY AND TO MAINTAIN THAT KINEITC ENERG FOR A PERIOD OF TIME,MOMENTUM PLAYS IT'S ROLE.

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A and B, move toward one another. Object A has twice the mass and half the speed of object B. Which of the following describes t

BARSIC [14]

Answer:D

Explanation:

Given

mass of A is twice the mass of B half the velocity of B

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2 years ago

While playing basketball in PE class, Logan lost his balance after making a lay-up and colliding with the padded wall behind the

olga2289 [7]

Answer:

a.) F = 3515 N

b.) F = 140600 N

Explanation: given that the

Mass M = 74kg

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Time t = 0.16 s

Force F = change in momentum ÷ time

F = (74×7.6)/0.16

F = 3515 N

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2 years ago

A giant wall clock with diameter d rests vertically on the floor. The minute hand sticks out from the face of the clock, and its

Katyanochek1 [597]

Answer:

$d_{x}(t)=(D/2)cos(\frac{\pi}{30}*t)$

Explanation:

We can try writing the equation of the horizontal component of the length of the minute hand in terms of distance and the angle, that depends of time in this particular case.

The x-component of the length of the minute hand is:

$d_{x}(t)=dcos(\theta (t))$ (1)

d is the length of the minute hand (d=D/2)
D is the diameter of the clock
t is the time (min)

Now, using the angular kinematic equations we can express the angle in term of angular velocity and time. As we know, the minute hand moves with a constant angular velocity, so we can use this equation:

$\theta (t)=\omega *t$ (2)

Also we know, that the minute hand moves 90 degrees or π/2 rad in 15 min, so using the definition of angular velocity, we have:

$\omega=\frac{\Delta \theta}{\Delta t}=\frac{\theta_{f}-\theta_{i}}{t_{f}-t{i}}=\frac{\pi/2-0}{15-0}=\frac{\pi}{30}$

Now, let's put this value on (2)

$\theta (t)=\frac{\pi}{30}*t$

Finally the length x(t) of the shadow of the minute hand as a function of time t, will be:

$d_{x}(t)=(D/2)cos(\frac{\pi}{30}*t)$

I hope it helps you!

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