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3 years ago

A satellite is held in orbit by a 2000- N gravitational

Physics

1 answer:

KengaRu [80]3 years ago

4 0

Answer:

The work done by gravity is zero.

Explanation:

Given that,

Gravitational force = 2000 N

Circumference = 80000 km

We need to calculate the work done by gravity

Using formula of work done

$W=F\cdot d\cos\theta$

Here, $\cos\theta$ = 0

Because, for a circular motion is always perpendicular to the force.

Where, F = force

d = distance

Put the value into the formula

$W=2000\times80000\times0$

$W=0$

Hence, The work done by gravity is zero.

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Compare and contrast the terms vaporizing and condensation.

JulsSmile [24]

The terms are both about changing states. Vaporizing is when you heat something up into a vapor; condensation is when you lower a vapors temperature to make it become a liquid state.

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3 years ago

A particle with a mass of 0.500 kg is attached to a horizontal spring with a force constant of 50.0 N/m. At the moment t = 0, th

svp [43]

a) $x(t)=2.0 sin (10 t) [m]$

The equation which gives the position of a simple harmonic oscillator is:

$x(t)= A sin (\omega t)$

where

A is the amplitude

$\omega=\sqrt{\frac{k}{m}}$ is the angular frequency, with k being the spring constant and m the mass

t is the time

Let's start by calculating the angular frequency:

$\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{50.0 N/m}{0.500 kg}}=10 rad/s$

The amplitude, A, can be found from the maximum velocity of the spring:

$v_{max}=\omega A\\A=\frac{v_{max}}{\omega}=\frac{20.0 m/s}{10 rad/s}=2 m$

So, the equation of motion is

$x(t)= 2.0 sin (10 t) [m]$

b) t=0.10 s, t=0.52 s

The potential energy is given by:

$U(x)=\frac{1}{2}kx^2$

While the kinetic energy is given by:

$K=\frac{1}{2}mv^2$

The velocity as a function of time t is:

$v(t)=v_{max} cos(\omega t)$

The problem asks as the time t at which U=3K, so we have:

$\frac{1}{2}kx^2 = \frac{3}{2}mv^2\\kx^2 = 3mv^2\\k (A sin (\omega t))^2 = 3m (\omega A cos(\omega t))^2\\(tan(\omega t))^2=\frac{3m\omega^2}{k}$

However, $\frac{m}{k}=\frac{1}{\omega^2}$ , so we have

$(tan(\omega t))^2=\frac{3\omega^2}{\omega^2}=3\\tan(\omega t)=\pm \sqrt{3}\\$

with two solutions:

$\omega t= \frac{\pi}{3}\\t=\frac{\pi}{3\omega}=\frac{\pi}{3(10 rad/s)}=0.10 s$

$\omega t= \frac{5\pi}{3}\\t=\frac{5\pi}{3\omega}=\frac{5\pi}{3(10 rad/s)}=0.52 s$

c) 3 seconds.

When x=0, the equation of motion is:

$0=A sin (\omega t)$

so, t=0.

When x=1.00 m, the equation of motion is:

$1=A sin(\omega t)\\sin(\omega t)=\frac{1}{A}=\frac{1}{2}\\\omega t= 30\\t=\frac{30}{\omega}=\frac{30}{10 rad/s}=3 s$

So, the time needed is 3 seconds.

d) 0.097 m

The period of the oscillator in this problem is:

$T=\frac{2\pi}{\omega}=\frac{2\pi}{10 rad/s}=0.628 s$

The period of a pendulum is:

$T=2 \pi \sqrt{\frac{L}{g}}$

where L is the length of the pendulum. By using T=0.628 s, we find

$L=\frac{T^2g}{(2\pi)^2}=\frac{(0.628 s)^2(9.8 m/s^2)}{(2\pi)^2}=0.097 m$

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3 years ago

Why is it harder to breathe on a

Law Incorporation [45]

I think it’s d but I’m not sure

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2 years ago

If we double the frequency of a system undergoing simple harmonic motion, which of the following statements about that system ar

AnnyKZ [126]

Answer:

a. The angular frequency is doubled.

e. The period is reduced to one-half of what it was.

Explanation:

Angular frequency is given as;

ω = 2πf

$\frac{\omega _1}{f_1} = \frac{\omega _2}{f_2}$

when the frequency is doubled

$\frac{\omega _1}{f_1} = \frac{\omega _2}{(2f_1)} \\\\\omega _1 = \frac{\omega _2}{2}\\\\\omega _2 = 2\omega _1$

Thus, the angular frequency will be doubled.

Amplitude in simple harmonic motion is the maximum displacement.

Frequency is related to period in simple harmonic motion as given in the equation below;

$f = \frac{1}{T} \\\\f_1T_1= f_2T_2\\\\T_2 = \frac{f_1T_1}{f_2}$

when the frequency is doubled;

$T_2 = \frac{f_1T_1}{2f_1} \\\\T_2 = \frac{T_1}{2}$

Thus, the period will be reduced to one-half of what it was.

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3 years ago

An instrument that measures wind speed is called a(n ________.

Oduvanchick [21]

The instrument that measures wind is called anemometer. Anemometer not only measures wind, but also, this measures the direction of the wind. And anemometer is also a common device used in a station weather. It is dervived from the greek work anemos which means wind.

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3 years ago