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3 years ago

How high will a ball go if thrown at 10m/s up?

1 answer:

4 0

And physics and physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsvand physicsand physicsand physicsand physicsand physicsand physicsand physicsvvand physicsand physicsand physicsand physicsvvand physicsand physicsvand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physicsand physics

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________ is a force acting through distance.

Andreas93 [3]

High school???
No way
It's work.

8 0

4 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 $T_{e}=365 \text { days }$

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

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

7 0

3 years ago

What is the "atomic mass" of an atom?

brilliants [131]

Answer:

It consists of one proton and one electron. The atomic mass for an atom of hydrogen is one dalton, which is calculated from the lone proton.

Explanation:

6 0

4 years ago

a solution contains equal amounts of ammonia and ammonium chloride, and is buffered at a pH of 9.25. Which action will most stre

telo118 [61]

Doubling the amounts of both ammonia and ammonium chloride is the action that will most likely strengthen the buffering capacity of a solution that contains equal amounts of ammonia and ammonium chloride, and is buffered at a pH of 9.25.

7 0

3 years ago

The unit of length most suitable for measuring the thickness of a cell phone is a

Elis [28]

centimeter or millimeter.

Explanation:

The unit of length most suitable for measuring the thickness of a cell phone is centimeter or millimeter.

Length refers to how long a material is.

There are different units of length.
Smaller units that are microscopic are used to measure length dimensions that are very small
For example, nanometre is used to express microscopic length values.
Centimeters and millimeters are used for small dimensions that we can easily see.
Longer lengths can take the order of a yard, kilometer, miles e.t.c

Learn more:

Length brainly.com/question/13035995

#learnwithBrainly

8 0

4 years ago