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3 years ago

9

Friction _moving things and turns motion energy into _

1 answer:

SashulF [63]3 years ago

8 0

Answer:

slowsdown, heat

Explanation:

Hope it helps!!

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What is the pitch of a sound wave?

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That's the impression you get from the sound, that you call 'high' or 'low', determined by the FREQUENCY of the wave.

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3 years ago

Unpolarized light with intensity S is incident on a series of polarizing sheets. The first sheet has its transmission axis orien

jeka94

Answer:

Explanation:

Given

Initial Intensity of light is S

when an un-polarized light is Passed through a Polarizer then its intensity reduced to half.

When it is passed through a second Polarizer with its transmission axis $\theat =45^{\circ}$

$S_1=S_0\cos ^2\theta$

here $S_0=\frac{S}{2}$

$S_1=\frac{S}{2}\times \frac{1}{(\sqrt{2})^2}$

$S_1=\frac{S}{4}$

When it is passed through third Polarizer with its axis $90^{\circ}$ to first but $\theta =45^{\circ}$ to second thus $S_2$

$S_2=S_0\cos ^2\theta$

$S_2=\frac{S}{4}\times \frac{1}{2}$

$S_2=\frac{S}{8}$

When middle sheet is absent then Final Intensity will be zero

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3 years ago

An electric fence displays a warning sign about the voltage and amperage of the fence. How would amperage and voltage affect the

Lorico [155]

V = volts
<span>I = amps </span>
<span>P = rate or energy transfer (power) </span>
<span>and </span>
<span>P = V * I</span>

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3 years ago

Two planets P1 and P2 orbit around a star S in circular orbits with speeds v1 = 40.2 km/s, and v2 = 56.0 km/s respectively. If t

Readme [11.4K]

Answer: $3.66(10)^{33}kg$

Explanation:

We are told both planets describe a circular orbit around the star S. So, let's approach this problem begining with the angular velocity $\omega$ of the planet P1 with a period $T=750years=2.36(10)^{10}s$ :

$\omega=\frac{2\pi}{T}=\frac{V_{1}}{R}$ (1)

Where:

$V_{1}=40.2km/s=40200m/s$ is the velocity of planet P1

$R$ is the radius of the orbit of planet P1

Finding $R$ :

$R=\frac{V_{1}}{2\pi}T$ (2)

$R=\frac{40200m/s}{2\pi}2.36(10)^{10}s$ (3)

$R=1.5132(10)^{14}m$ (4)

On the other hand, we know the gravitational force $F$ between the star S with mass $M$ and the planet P1 with mass $m$ is:

$F=G\frac{Mm}{R^{2}}$ (5)

Where $G$ is the Gravitational Constant and its value is $6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$

In addition, the centripetal force $F_{c}$ exerted on the planet is:

$F_{c}=\frac{m{V_{1}}^{2}}{R^{2}}$ (6)

Assuming this system is in equilibrium:

$F=F_{c}$ (7)

Substituting (5) and (6) in (7):

$G\frac{Mm}{R^{2}}=\frac{m{V_{1}}^{2}}{R^{2}}$ (8)

Finding $M$ :

$M=\frac{V^{2}R}{G}$ (9)

$M=\frac{(40200m/s)^{2}(1.5132(10)^{14}m)}{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}}$ (10)

Finally:

$M=3.66(10)^{33}kg$ (11) This is the mass of the star S

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4 years ago

What is the mass of a soda can on a triple beam balance

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It depends on the kind of soda can

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3 years ago