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Alexxandr [17]
2 years ago
5

What affect does distance have on an electric field?

Physics
1 answer:
nikklg [1K]2 years ago
8 0

Answer:

the answer is equal to yes

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Calculate the magnitude of the average gravitational force between earth and the moon
zepelin [54]
F = G mM / r^2, where 
<span>F = gravitational force between the earth and the moon, </span>
<span>G = Universal gravitational constant = 6.67 x 10^(-11) Nm^2/(kg)^2, </span>
<span>m = mass of the moon = 7.36 × 10^(22) kg </span>
<span>M = mass of the earth = 5.9742 × 10^(24) and </span>
<span>r = distance between the earth and the moon = 384,402 km </span>

<span>F </span>
<span>= 6.67 x 10^(-11) * (7.36 × 10^(22) * 5.9742 × 10^(24) / (384,402 )^2 </span>
<span>= 1.985 x 10^(26) N</span>
5 0
3 years ago
A student pushes a 2.85 kg cart causing it to accelerate at a rate of 4.9 m/s squared .What amount of force must the student hav
uysha [10]
In order to find the force (F), you would have to use the formula for it:
F=ma
where m is mass and a is acceleration.
In the problem, the mass is 2.85kg and the acceleration is 4.9m/s^2.
Therefore,
F=2.85kg(4.9m/s^2)
F=13.965kg(m/s^2)
Since N=kg(m/s^2)
F=13.965N
And because the problem requires that we use only 2 significant figures,
F=13N
Therefore, the student must exert 13N of force.

7 0
3 years ago
The velocity of an object includes its speed and____?
mr_godi [17]

it includes an objects speed and direction

6 0
3 years ago
Objects with masses of 135 kg and a 435 kg are separated by 0.500 m. Find the net gravitational force exerted by these objects o
IrinaVladis [17]

Answer:

The net gravitational force on the mass is 1.27\times 10^{-5}

Explanation:

We have by Newton's law of gravity the force of attraction between masses m_{1},m_{2}

F_{att}=G\frac{m_{1}m_{2}}{r^{2}}

Applying vales we get

Force of attraction between 135 kg mass and 38 kg mass is

F_{1}=6.67\times 10^{-11}\frac{135\times 38}{(0.25)^{2}}\\\\F_{1}=5.47\times 10^{-6}N

Force of attraction between 435 kg mass and 38 kg mass is

F_{2}=6.67\times 10^{-11}\frac{435\times 38}{(0.25)^{2}}\\\\F_{2}=1.76\times 10^{-5}N

Thus the net force on mass 38.0 kg is F_{2}-F_{1}=1.76\times 10^{-5}-5.47\times 10^{-6}\\\\F_{2}-F_{1}=1.27\times 10^{-5}

8 0
3 years ago
An automobile moving along a straight track changes its velocity from 40 m/s to 80 m/s in a distance of 200 m. What is the (cons
Nata [24]

Answer:

Dear Kaleb

Answer to your query is provided below

Acceleration of the vehicle is 12m/s^2

Explanation:

Explanation for the same is attached in image

8 0
3 years ago
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