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3 years ago

11

you throw a rock out of a 7th story window. you time that it takes 3.7 seconds to hit the ground, and measure that it hit the gr

ound 115 m from the base of the building

1 answer:

seropon [69]3 years ago

7 0

First of all, I doubt it ... you threw the rock out of your hand
at least 69.5 miles per hour ? ? Maybe more, depending on
the angle.

And that's all I have to say about this one, since you haven't
asked any question.

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Maria designs a test to see if lemon trees that receive more water produce larger lemons.

denis-greek [22]

Do you know the answer

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3 years ago

What is the measurement 111.009 mm rounded off to four significant digits?

dedylja [7]

111.0 because 111.009 rounds off to 111.01, thus rounding again off to 111.0

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3 years ago

One recently discovered extrasolar planet, or exoplanet, orbits a star whose mass is 0.70 times the mass of our sun. This planet

Stels [109]

0.078 times the orbital radius r of the earth around our sun is the exoplanet's orbital radius around its sun.

Answer: Option B

<u>Explanation:</u>

Given that planet is revolving around the earth so from the statement of centrifugal force, we know that any

$\frac{G M m}{r^{2}}=m \omega^{2} r$

The orbit’s period is given by,

$T=\sqrt{\frac{2 \pi}{\omega r^{2}}}=\sqrt{\frac{r^{3}}{G M}}$

Where,

$T_{e}$ = Earth’s period

$T_{p}$ = planet’s period

$M_{s}$ = sun’s mass

$r_{e}$ = earth’s radius

Now,

$T_{e}=\sqrt{\frac{r_{e}^{3}}{G M_{s}}}$

As, planet mass is equal to 0.7 times the sun mass, so

$T_{p}=\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}$

Taking the ratios of both equation, we get,

$\frac{T_{e}}{T_{p}}=\frac{\sqrt{\frac{r_{e}^{3}}{G M_{s}}}}{\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}}$

$\frac{T_{e}}{T_{p}}=\sqrt{\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}}$

$\left(\frac{T_{e}}{T_{p}}\right)^{2}=\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}$

$\left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}=\frac{r_{e}^{3}}{r_{p}^{3}}$

$\frac{r_{e}}{r_{p}}=\left(\left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}$

Given $T_{p}=9.5 \text { days }$ and $T_{e}=365 \text { days }$

$\frac{r_{e}}{r_{p}}=\left(\left(\frac{365}{9.5}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=\left(\frac{133225}{90.25} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=(2108.82)^{\frac{1}{3}}$

$r_{p}=\left(\frac{1}{(2108.82)^{\frac{1}{3}}}\right) r_{e}=\left(\frac{1}{12.82}\right) r_{e}=0.078 r_{e}$

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2 years ago

Wave C has an amplitude of 1 and wave D has an amplitude of 3 as shown below. What will happen when the crests of these two wave

Leya [2.2K]

Answer:
Amplitude 4

When the two waves collide, because they are both crests there is constructive interference. This mean that at the point of collision the amplitude increases.

You add both the amplitudes
1+3=4

6 0

2 years ago

A body is piloted at a point. A force of 10 N is applied at a distance of 30 cm from the pivot. Find the amount of force about t

bekas [8.4K]

Question :-

A Body is Pivoted at a Point. A Force of 10 N is Applied at a Distance of 30 cm from the Pivot. Find the Amount of Force about the Pivot.

Answer :-

Amount of Force is 3 Nm .

Explanation :-

As per the provided information in the given question, The Force is given as 10 Newton . The Distance is given as 30 cm [ 0.3 m ] . And, we have been asked to calculate the Amount of Force .

For calculating the Force , we will use the Formula :-

$\bigstar \: \: \: \boxed{ \: \sf {Moment \: of \: Force \: = \: Force \: \times \: Distance} \: }$

Therefore , by Substituting the given values in the above Formula :-

$\dag \: \: \: \sf {Moment \: of \: Force \: = \: Force \: \times \: Distance}$

$\longmapsto \: \: \: \sf {Moment \: of \: Force \: = \: 10 \: \times \: 0.3}$

$\longmapsto \: \: \: \textbf {\textsf {Moment \: of \: Force \: = \: 3}}$

Hence :-

Amount of Force = 0.3 Nm .

$\underline {\rule {185pt}{4pt}}$

8 0

2 years ago