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4 years ago

PLEASE HELP I'LL GIVE BRAINLIEST!!!! ASAP PLEASE

2 answers:

7 0

They traveled 5 miles (3 + 2 = 5)
you cannot calculate displacement without an origin point (how far away are they from it at the final destination?) because distance does not equal displacement

Gnesinka [82]4 years ago

3 0

Answer: 6

Explanation:

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Design your own plan for a scientific experiment. You do not have to conduct your experiment. Choose a topic that you are intere

vredina [299]

Answer:

Hello there. I know its too late, but you could use one of your scientific experiments you did over the past few years.

You could make an solinoid.

Hope this helps! Lol

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3 years ago

A car veers off course and runs straight into a brick wall. This is an example

ValentinkaMS [17]

Short time large force

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3 years ago

The vine has a fixed end, so Tarzan's path is circular. So the vine must not only support Tarzan's weight (or some component of

Minchanka [31]

Answer: hello your question lacks some data attached below is the missing data

answer : T - mg = ma $_{c}$

Explanation:

Given that the vine has a fixed end and Tarzan's path is circular

At Tarzans lowest point the point can be expressed as shown below.

It can be expressed as : T - mg = ma $_{c}$

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

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

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

7 0

3 years ago

Most ionic bonds form when elements from __________.

I am Lyosha [343]

A bonds with elements from b

5 0

4 years ago

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