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2 years ago

Anyone going to be my friend

Physics

1 answer:

Artemon [7]2 years ago

5 0

Explanation:

I'd love to but we cant talk right now cause its 12:22 am here and I'm gonna sleep now lol.

but let's follow each other.

who knows we might be able to help each other.

whaddya say?

have a good day ♡

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Commander Shepard, an N7 spectre for Earth, weighs 799 N on the Earth's surface. When she lands on Noveria, a distant planet in

Anastasy [175]

Answer:

Acceleration of gravity on Noveria = 4.4 m/s²

Explanation:

Commander Shepard, an N7 spectre for Earth, weighs 799 N on the Earth's surface.

We have weight, W = mg

Acceleration due to gravity, g = 9.81m/s²

799 = m x 9.81

Mass of Shepard, m = 81.45 kg

She lands on Noveria, a distant planet in our galaxy, she weighs 356 N.

We have weight, W = mg'

356 = 81.45 xg'

Acceleration of gravity on Noveria, g' = 4.4 m/s²

6 0

3 years ago

What do you think would happen to earth's tides if the moon was not there? the tides would....?

defon

Their are things call solar tides which are the effects of the sun But it would be alot calmer with out the moon.

8 0

3 years ago

A charge of 7.00 mC is placed at opposite corners corner of a square 0.100 m on a side and a charge of -7.00 mC is placed at oth

andrew-mc [135]

Answer:

4.03\times10^{7}N[/tex], 135°

Explanation:

charge, q = 7 mC = 0.007 C

charge, - q = - 7 mC = - 0.007 C

d = 0.1 m

Let the force on charge placed at C due to charge placed at D is FD.

$F_{D}=\frac{kq^{2}}{DC^{2}}$

$F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N$

The direction of FD is along C to D.

Let the force on charge placed at C due to charge placed at B is FB.

$F_{B}=\frac{kq^{2}}{BC^{2}}$

$F_{B}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N$

The direction of FB is along C to B.

Let the force on charge placed at C due to charge placed at A is FA.

$F_{A}=\frac{kq^{2}}{AC^{2}}$

$F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1 \times\sqrt{2} \times 0.1 \times\sqrt{2}}=2.205 \times 10^{7}N$

The direction of FA is along A to C.

The net force along +X axis

$F_{x}=F_{A}Cos45-F_{D}$

$F_{x}=2.205\times10^{7}Cos45-4.41\times10^{7}=-2.85\times10^{7}N$

The net force along +Y axis

$F_{y}=F_{B}-F_{A}Sin45$

$F_{y}=4.41\times10^{7}-2.205\times10^{7}Sin45=2.85\times10^{7}N$

The resultant force is given by

$F=\sqrt{F_{x}^{2}+F_{y}^{2}}=\sqrt{(-2.85\times10^{7})^{2}+(2.85\times10^{7})^{2}}$

$F = 4.03\times10^{7}N$

The angle from x axis is Ф

tan Ф = - 1

Ф = -45°

Angle from + X axis is 180° - 45° = 135°

5 0

2 years ago

A 5,400 W motor is used to do work. If the motor is used for 640 s, about how much work could it do?

Gemiola [76]

Answer:

3,500,000 J

Explanation:

WORK = POWER * TIME

WORK= 5400 * 640

=6456000 J = 3,500,000 J

4 0

3 years ago

Give an example of a reputable website

Leni [432]

Well, almost any website can be, but if you are meaning for money, then google. if you are meaning reliable, then look for something with .org, its history and stuff

6 0

2 years ago

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