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Jlenok [28]
2 years ago
11

What is the force of friction acting on the crate? I am offering 30 points.

Physics
1 answer:
olganol [36]2 years ago
4 0
I think the answer is C


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Two electric charges, held a distance, dd, apart experience an electric force of magnitude, FF, between them. If one of the char
lorasvet [3.4K]

Answer:

F'=2F

Explanation:

The Coulomb's law states that the magnitude of the electrostatic force between two charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them:

F=\frac{kq_1q_2}{d^2}

In this case, we have q_1'=2q_1:

F'=\frac{kq'_1q_2}{d^2}\\F'=\frac{k(2q_1)q_2}{d^2}\\F'=2\frac{kq_1q_2}{d^2}\\F'=2F

3 0
3 years ago
A person kicks a rock off a cliff horizontally with a speed of 20 m/s. It takes 7.0 seconds to hit the ground.
EastWind [94]
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6 0
3 years ago
An object propelled upwards with an acceleration of 2.0 m / s ^ 2 is launched from rest. After 6 seconds the fuel runs out. Dete
dezoksy [38]

Answer:43.34 m

Explanation:

Given

acceleration(a)=2 m/s^2

Initial Velocity(u)=0 m/s

After 6 s fuel runs out

Velocity after 6 s

v=u+at

v=0+2\times 6=12 m/s

After this object will start moving under gravity

height reached in first 6 s

s=ut+\frac{at^2}{2}

s=0+\frac{2\times 6^2}{2}

s=36 m

After fuel run out distance traveled in upward direction is

v^2-u^2=2as_0

here v=0

u=12 m/s

a=9.8 m/s^2

0-12^2=2(-9.8)(s)

s_0=\frac{144}{2\times 9.8}=7.34 m

s+s_0=36+7.34=43.34 m

7 0
3 years ago
HOW MANY KILOMETERS IS EARTH FROM THE SUN?
maria [59]
Earth is 150 million kilometers away for the sun 

7 0
3 years ago
Read 2 more answers
A 9 newtwon charge is Felt on a (1x10^-4) C charge from another charge that is 3 meters away. What is the magnitude of that othe
Marina86 [1]

The magnitude of that other charge will be 9×10⁻⁵ C. The force on the charge is inverse of the distance.

<h3>What is Columb's law?</h3>

The force of attraction between two charges, according to Coulomb's law, is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

The magnitude of that other charge is found as;

\rm F = \frac{Kq_1q_2}{r^2} \\\\ \rm 9  = \frac{9 \times 10^9 \times 1\times 10^{-4}\times q_2}{(3)^2} \\\\ q_2=9\times 10^{-5} \ C

Hence, the magnitude of that other charge will be 9×10⁻⁵ C.

To learn more about Columb's law, refer to the link;

brainly.com/question/1616890

#SPJ1

6 0
2 years ago
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