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PSYCHO15rus [73]
3 years ago
9

Please help ASAP. In what direction do you need to apply force to move an object vertically?

Physics
1 answer:
PilotLPTM [1.2K]3 years ago
5 0
The answer is up . tylrhscjwizn
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a 2100-kg car drives with a speed of 18 m/s onb a flat road around a curve that has a radius of curvature of 83m. The coefficien
Law Incorporation [45]

Answer:

<u>The magnitude of the friction force is 8197.60 N</u>

Explanation:

Using the definition of the centripetal force we have:

\Sigma F=ma_{c}=m\frac{v^{2}}{R}

Where:

  • m is the mass of the car
  • v is the speed
  • R is the radius of the curvature

Now, the force acting in the motion is just the friction force, so we have:

F_{f}=m\frac{v^{2}}{R}

F_{f}=2100\frac{18^{2}}{83}

F_{f}=8197.60 \: N

<u>Therefore the magnitude of the friction force is 8197.60 N</u>

I hope it helps you!

7 0
3 years ago
What is one way to lower gravitational potential energy?
il63 [147K]

Answer:

decrease the height

Explanation:

height is directly proportional to the g.p.e

4 0
2 years ago
Question 1 of 25
finlep [7]

Answer:

<em>2.753*10^-11N</em>

Explanation:

According to Newton's law of gravitation, the force between the masses is expressed as;

F = GMm/d²

M and m are the distances

d is the distance between the masses

Given

M = 3.71 x 10 kg

m = 1.88 x 10^4 kg

d = 1300m

G = 6.67 x 10-11 Nm²/kg

Substitute into the formula

F = 6.67 x 10-11* (3.71 x 10)*(1.88 x 10^4)/1300²

F = 46.52*10^(-6)/1.69 * 10^6

F = 27.53 * 10^{-6-6}

F = 27.53*10^{-12}

F = 2.753*10^-11

<em>Hence the gravitational force between the asteroid is 2.753*10^-11N</em>

<em></em>

6 0
2 years ago
Help please in science
slamgirl [31]
The amplitude is 0.8 cm. The wavelength is 2.0 cm.
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3 years ago
Particles q1, q2, and q3 are in a straight line. Particles q1 = -28.1 μC, q2 = +25.5 μC, and q3 = -47.9 μC. Particles q1 and q2
Arturiano [62]

Answer:-88.4

Explanation:

3 0
2 years ago
Read 2 more answers
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