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3 years ago

If da is a diameter of point z and m

1 answer:

3 0

Your question is totally unintelligible, and apparently incomplete
since there is no independent clause or something to match the
"If" at the beginning..

Other than that, it promises quite an intriguing challenge.

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Why are satellites placed into orbit at least 150 km above Earth’s surface?

dybincka [34]

Because its just enough to where its not out f the gravitational pull and not close enough to be pulled back to earth. Hope it helps<span />

8 0

2 years ago

Read 2 more answers

What speed would a proton need to achieve in order to circle Earth 1790 km above the magnetic equator, where the Earth's mag- ne

irinina [24]

Answer:

The velocity is 31.25 m/s and direction is toward west.

Explanation:

Given that,

Distance $h= 1790 km = 1.790\times10^{6}\ m$

Magnetic field $B=4\times10^{-8}\ T$

Mass of proton $m=1.673\times10^{-21}\ Kg$

Radius of earth $R =6.38\times10^{6}\ m$

Radius of orbit $r=R+h$

$r=6.38\times10^{6}+1.790\times10^{6}$

$r=8170000\ m$

We need to calculate the speed

Using formula of magnetic field

$Bvq=\dfrac{mv^2}{r}$

$v=\dfrac{Bqr}{m}$

Put the value into the formula

$v=\dfrac{4\times10^{-8}\times1.6\times10^{-19}\times8170000}{1.673\times10^{-21}}$

$v=31.25\ m/s$

Hence, The velocity is 31.25 m/s and direction is toward west.

6 0

2 years ago

The constant forces F1 = 8 + 29 + 32 N and F2 = 48 - 59 - 22 N act together on a particle during a displacement from the point A

steposvetlana [31]

Answer:

- 600 J

Explanation:

A (20, 15, 0 ) m

B (0, 0, 7) m

$\overrightarrow{F_{1}}=8\widehat{i}+29\widehat{j}+32\widehat{k}$

$\overrightarrow{F_{2}}=48\widehat{i}-59\widehat{j}-22\widehat{k}$

Net force

$\overrightarrow{F}=\overrightarrow{F_{1}}+\overrightarrow{F_{2}}$

$\overrightarrow{F}}=\left ( 8+48 \right )\widehat{i}+\left ( 29-59 \right )\widehat{j}+\left ( 32-22 \right )\widehat{k}$

$\overrightarrow{F}}=56\widehat{i}-30\widehat{j}+10\widehat{k}$

$\overrightarrow{S}=\overrightarrow{OB}-\overrightarrow{OA}$

$\overrightarrow{S}=\left ( 0-20 \right )\widehat{i}+\left ( 0-15 \right )\widehat{j}+\left ( 7-0 \right )\widehat{k}$

$\overrightarrow{S}=-20\widehat{i}-15\widehat{j}+7\widehat{k}$

Work done is defined as

$W = \overrightarrow{F}.\overrightarrow{S}$

$W = \left ( 56\widehat{i}-30\widehat{j}+10\widehat{k} \right ).\left (-20\widehat{i}-15\widehat{j}+7\widehat{k} \right )$

W = -1120 + 450 + 70

W = - 600 J

3 0

2 years ago

if a man hits a golf ball (0.2 kg) which accelerates at a rate of 20 m/s squared. what amount of force acted on the ball?

kkurt [141]

F = ma

We have mass = 0.2kg

and acceleration = 20 m/s^2

So..

F = (0.2)(20)

F = 4 N

6 0

2 years ago

Brainliest if correct Question 1 of 10

Art [367]

Answer:

Explanation:

hope this helps you

4 0

1 year ago