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mart [117]
3 years ago
10

Point charge 3.5 μC is located at x = 0, y = 0.30 m, point charge -3.5 μC is located at x = 0 y = -0.30 m. What are (a)the magni

tude and (b)direction of the total electric force that these charges exert on a third point charge Q = 4.5 μC at x = 0.40 m, y = 0?
Physics
1 answer:
IgorLugansk [536]3 years ago
3 0
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Answer:

air resistance, gravitational force

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A 10n falling object encounters 4n of air resistance. what is the net force on the object?
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It would be 6n down.
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100 POINTS WILL GIVE BRAINIEST TO BEST ANSWER!!!!!!!!!!!!!<br><br> see below
gogolik [260]

Answer:

What’s your question

Explanation:

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3 0
3 years ago
A 1000 kg satellite and a 2000 kg satellite follow exactly the same orbit around the earth. What is the ratio F1/F2 of the gravi
Tamiku [17]

Answer:

the <em>ratio F1/F2 = 1/2</em>

the <em>ratio a1/a2 = 1</em>

Explanation:

The force that both satellites experience is:

F1 = G M_e m1 / r²       and

F2 = G M_e m2 / r²

where

  • m1 is the mass of satellite 1
  • m2 is the mass of satellite 2
  • r is the orbital radius
  • M_e is the mass of Earth

Therefore,

F1/F2 = [G M_e m1 / r²] / [G M_e m2 / r²]

F1/F2 = [G M_e m1 / r²] × [r² / G M_e m2]

F1/F2 = m1/m2

F1/F2 = 1000/2000

<em>F1/F2 = 1/2</em>

The other force that the two satellites experience is the centripetal force. Therefore,

F1c = m1 v² / r    and

F2c = m2 v² / r

where

  • m1 is the mass of satellite 1
  • m2 is the mass of satellite 2
  • v is the orbital velocity
  • r is the orbital velocity

Thus,

a1 = v² / r ⇒ v² = r a1    and

a2 = v² / r ⇒ v² = r a2

Therefore,

F1c = m1 a1 r / r = m1 a1

F2c = m2 a2 r / r = m2 a2

In order for the satellites to stay in orbit, the gravitational force must equal the centripetal force. Thus,

F1 = F1c

G M_e m1 / r² = m1 a1

a1 = G M_e / r²

also

a2 = G M_e / r²

Thus,

a1/a2 = [G M_e / r²] / [G M_e / r²]

<em>a1/a2 = 1</em>

4 0
3 years ago
to start an avalanche on a mountain slope, an artillery shell is fired with an initial velocity of 290 m/s at 53.0° above the ho
kap26 [50]
So this is easy to calculate when you split the velocity into x and y components. The x component is going to equal cos(53) * 290 and the y component is going to equal sin(53)*290.

The x location therefore is 290*cos(53)*35 = 6108.4m

The y location needs to factor in the downwards acceleration of gravity too, which is 9.81m/s^2. We need the equation dist. = V initial*time + 0.5*acceleration*time^2.

This gives us d=290*sin(53)*35 + (0.5*-9.81*35^2)=2097.5m

So your (x,y) coordinates equals (6108.4, 2097.5)
5 0
3 years ago
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