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3 years ago

Calculate the potential energy of a 20-kg sled at 40 meters

Physics

2 answers:

irakobra [83]3 years ago

5 0

Answer: 7840N

Explanation:

Given that

Potential energy = ?

Mass of sled = 20-kg

Distance = 40 meters

Acceleration due to gravity = 9.8m/s^2

Recall that potential energy is the energy possessed by a body at rest

i.e potential energy = mass m x acceleration due to gravity g x distance h

P.E = mgh

P.E = 20kg x 9.8m/s^2 x 40m

P.E = 7840N

Thus, the potential energy of the sled is 7840N

Natali [406]3 years ago

4 0

The above answer is correct however gravitational potential energy is measured in joules. So the final answer would be 7840 J (joules) not newton’s.

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Answer:

in the downward movement of the movement when the constant is lost

Explanation:

When the coin is on the piston it has a relationship given by

a = d²x / dt²

the piston position is

x = A cos wt

a = - A w² cos wt

the maximum acceleration is

a = - A w²

When the piston raises the acceleration of gravity and that of the piston go in the same direction, when the piston descends they relate it is contrary to gravity, therefore when the frequency increases, the point where the acceleration of the piston is greater than gravity arrives and the coin loses contact.

The point where you lose contact is

a = g

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In short, in the downward movement of the movement when the constant is lost

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Explanation:

It is given that,

Wavelength, $\lambda_1=500\ nm=5\times 10^{-7}\ m$

Wavelength, $\lambda_2=500.10\ nm=5.001\times 10^{-7}\ m$

We need to find the frequencies from corresponding wavelengths. The frequency of the light is given by :

$f=\dfrac{c}{\lambda}$

c is the speed of light

Frequency 1,

$f_1=\dfrac{c}{\lambda_1}$

$f_1=\dfrac{3\times 10^8\ m/s}{5\times 10^{-7}\ m}$

$f_1=6\times 10^{14}\ Hz$

Frequency 2,

$f_2=\dfrac{c}{\lambda_2}$

$f_1=\dfrac{3\times 10^8\ m/s}{5.001\times 10^{-7}\ m}$

$f_1=5.99\times 10^{14}\ Hz$

Hence, this is the required solution.

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Answer:

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Given that,

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$v=\dfrac{A\times\Delta P}{8\pi\eta\rho}$

Where, A = area

$\rho$ = density of oil

$\eta$ = viscosity

Put the value into the formula

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Answer:

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Answer:

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