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3 years ago

Example of energy transfer from potential to kinetic and back

1 answer:

5 0

Swimmer and Divers. The Potential energy is transferred into Kinetic energy, and allows the diver to submerge into the water. The Kinetic energy then allows the diver to submerge and dive into the water. Potential energy however, is needed to allow the diver to get back out of the water after diving to get up and go and dive again, and then the Kinetic energy is transferred back to Potential energy to repeat the process.

Hope :) -Emilie Xo this is right and it helps! Xo

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Are u god tell me u have to tell me are u a god

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They tell me I'm a god, I'm lost in the facade

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Makin' so much bread, I don't care that they're just being fake

They tell me they're below me, I act like I'm above

The people blend together but I would be lost without their love

Can you heal me? Have I gained too much?

When you become untouchable, you're unable to touch

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It hurts me just to think and I don't do pain

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If I fall, I think I'll fly

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3 years ago

Magnetic and electric fields lab report guide

Digiron [165]

Answer:

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Explanation:

4 0

3 years ago

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A string is wrapped several times around the rim of a small hoop with radius 8.00 cm and mass 0.180 kg. The free end of the stri

Dmitrij [34]

Answer:

(a). The angular speed of the rotating is 33.8 rad/s.

(b). The speed of its center is 2.7 m/s.

Explanation:

Given that,

Radius = 8.00 cm

Mass = 0.180 kg

Height = 75.0 m

We need to calculate the angular speed of the rotating

Using conservation of energy

$\dfrac{1}{2}I\omega_{1}^2+\dfrac{1}{2}mv_{1}^{2}+mgh_{1}=\dfrac{1}{2}I\omega_{2}^2+\dfrac{1}{2}mv_{2}^{2}+mgh_{2}$

Here, initial velocity and angular velocity are equal to zero.

$mgh_{1}=\dfrac{1}{2}I\omega_{2}^2+\dfrac{1}{2}mv_{2}^{2}+mgh_{2}$

$mg(h_{1}-h_{2})=\dfrac{1}{2}I\omega_{2}^2+\dfrac{1}{2}mv_{2}^{2}$

$mgH=\dfrac{1}{2}mr^2\omega_{2}^2+\dfrac{1}{2}m(r\omega_{2})^2$

Here, $H = h_{1}-h_{2}$

$gH=r^2\omega_{2}^2$

$\omega_{2}^2=\dfrac{gH}{r^2}$

$\omega_{2}=\sqrt{\dfrac{9.8\times75.0\times10^{-2}}{(8.00\times10^{-2})^2}}$

$\omega_{2}=33.8\ rad/s$

The angular speed of the rotating is 33.8 rad/s.

(b). We need to calculate the speed of its center

Using formula of speed

$v=r\omega$

Put the value into the formula

$v=8.00\times10^{-2}\times33.8$

$v=2.7\ m/s$

Hence, (a). The angular speed of the rotating is 33.8 rad/s.

(b). The speed of its center is 2.7 m/s.

3 0

3 years ago

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If a 0.5 kg pair of running shoes would weigh 11.55 Newton’s on Jupiter, what is the strength of gravity there?

OverLord2011 [107]

The weight of an object is given by
$F=mg$
where m is the mass of the object, while g is the strength of the gravity (which corresponds to the gravitational acceleration of the planet).

In our problem, the shoes have a mass of m=0.5 kg, and their weight is F=11.55 N. So, we can re-arrange the previous formula to find the value of g:
$g= \frac{F}{m}= \frac{11.55 N}{0.5 kg}=23.1 N kg^{-1}$
and this is the strength of the gravity on Jupiter's surface.

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3 years ago

The BEST description of a light-year is A) the speed of light in a vacuum. B) the age of a one-year-old star. C) the distance li

Deffense [45]

C is the answer
D is impossible
A and B are false

Hope this helps!

Please crown;)

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3 years ago