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Eduardwww [97]
3 years ago
8

What is pulling force? Give any two examples.,

Physics
1 answer:
Dima020 [189]3 years ago
8 0

Answer:

Push or Pull Forces - example

When you push against a wall the force that you exert is an example of a push force. When you pull a trolley car the force that you exert is an example of pull force.

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If the distance between two masses is tripled, the gravitational force between changes by a factor of:_______
adell [148]

Answer:

option A

Explanation:

given,

distance between two masses is doubled

new distance, r' = 3 r

using gravitational force equation

F = \dfrac{GMm}{r^2}............(1)

new gravitational force

F' = \dfrac{GMm}{r'^2}

now from the given condition

F' = \dfrac{GMm}{(3r)^2}

F' = \dfrac{GMm}{9r^2}

F' = \dfrac{1}{9}\dfrac{GMm}{r^2}

now, from equation (1)

F' = \dfrac{1}{9}F

\dfrac{F'}{F} = \dfrac{1}{9}

now, the change in gravitational force factor is equal to \dfrac{1}{9}

Hence, the correct answer is option A

3 0
3 years ago
What is the purpose of a kink in a thermometer?
AnnZ [28]

Answer:

A 'kink' in the glass tube which breaks the mercury as it contracts, storing the highest temperature reading. The glass tube is shaped like a lens to magnify the thin mercury thread. Shaking the thermometer resets the mercury back into the bulb.

3 0
2 years ago
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How can the process of climate change best be described?
seropon [69]
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3 0
3 years ago
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What is a safe following distance between your automobile and the vehicle in front of you?
Colt1911 [192]

Answer:

Many drivers follow the “three-second rule.” In other words, you should keep three seconds worth of space between your car and the car in front of you in order to maintain a safe following distance.

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4 0
2 years ago
A satellite is in a circular orbit 8200 km above the Earth’s surface; i.e., it moves on a circular path under the influence of n
andreyandreev [35.5K]

Answer:

5.23km/s

Explanation:

Given

Radius of Earth = 6.37 * 10^6 m

Altitude of Satellite = 8200km = 8200 * 10³m = 8.2 * 10^6 m

Gravity Acceleration on Satellite Altitude = 1.87965m/s²

For a satellite to remain in circular orbit, then it means the acceleration of gravity must be exact as the centripetal acceleration.

Centripetal Acceleration = V²/R

So, Acceleration of Gravity (A)= Centripetal Acceleration = V²/R

Make V the subject of formula

A = V²/R

V² = AR

V = √AR

Where R = (radius of earth) + (altitude of satellite)

R = 6.37 * 10^6 + 8.2 * 10^6

R = 14.57 * 10^6m

A = 1.87965m/s²

V = √(1.87965 * 14.57x10^6)

V = √27386500.5

V = 5233.211299001789

V = 5233.2113 m/s ------- Approximated

V = 5.23km/s

7 0
3 years ago
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