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3 years ago

An object that travels around another object in space is called a(n)

Physics

2 answers:

HACTEHA [7]3 years ago

3 0

Satellite. think of the moon.

Hope this helps!
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user100 [1]3 years ago

3 0

It's definitely D, satellite

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At a particular instant, a proton at the origin has velocity < 5e4, -2e4, 0> m/s. You need to calculate the magnetic field

vesna_86 [32]

Answer:

$9.7\times 10^{-5} T$

Explanation:

Velocity = $5\times 10^4i-2\times 10^4j$

r= $0.03i+0.05j$

r= $\mid r\mid=\sqrt{(0.03)^2+(0.05)^2}=0.058$

v= $\mid V\mid=\sqrt{(5\times 10^4)^2+(-2\times 10^{4})^2}=5.39\times 10^{2}$

We know that

$B=\frac{mv}{qr}$

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

Mass of proton= $1.67\times 10^{-27} kg$

Using the formula

$B=\frac{1.67\times 10^{-27}\times 5.39\times 10^2}{1.6\times 10^{-19}\times 0.058}$

$B=9.7\times 10^{-5} T$

3 0

3 years ago

Using the image provided, which moon phase is next?

ella [17]

Answer:

Waning Crescent

5 0

3 years ago

Read 2 more answers

Is sand a dependent or independent variable

anastassius [24]

Answer:

dependent variable

Explanation: Because sand changes when water touches it therefore it would be considered a dependent variable

6 0

3 years ago

Read 2 more answers

Your friend is trying to construct a clock for a craft show and asks you for some advice. She has decided to construct the clock

Anastasy [175]

Answer:

The period of the pendulum is $T = 1.68 \ sec$

Explanation:

The diagram illustrating this setup is shown on the first uploaded image

From the question we are told that

The length of the rod is $L = 80 \ cm$

The diameter of the ring is $d = 10 \ cm$

The distance of the hole from the one end $D = 15cm$

From the diagram we see that point A is the center of the brass ring

So the length from the axis of rotation is mathematically evaluated as

$AP = 80 + 10 -5 -15$

$AP = 70 \ cm = \frac{70}{100} = 0.7 \ m$

Now the period of the pendulum is mathematically represented as

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

$T = 2 \pi \sqrt{\frac{0.7}{9.8 } }$

$T = 1.68 \ sec$

4 0

3 years ago

A student is studying simple harmonic motion of a spring. She conducts an experiment where she measures the amplitude and period

myrzilka [38]

Answer:

5.9*10^{-4}m

Explanation:

to find the uncertainty of the displacement it is necessary to compute the uncertainty for the angular frequency:

$\omega=\frac{2\pi}{T}=\frac{2\pi}{0.40s}=15.707rad/s\\\\\frac{d \omega}{\omega}=\frac{dT}{T}\\\\d\omega=\omega \frac{dT}{T}=(15.707rad/s)\frac{0.020s}{0.40s}=0.785rad/s$

then, you can calculate the uncertainty in angular displacement:

$\theta=\omega t\\\\\frac{d\theta}{\theta}=\sqrt{(\frac{d\omega}{\omega})^2+(\frac{dt}{t})^2}\\\\d\theta=\theta\sqrt{(\frac{d\omega}{\omega})^2+(\frac{dt}{t})^2}=0.0422$

finally, by using:

$y=Acos(\omega t)\\\\dy=dAcos(\omega t)d(\omega t)=(dA)cos(\theta)d\theta=(0.002m)cos(0.785)(0.0422)\\\\dy=5.9*10^{-4}m$

7 0

3 years ago