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4 years ago

Two objects must be in contact for them to exert a force on each other. True False

Physics

1 answer:

Irina-Kira [14]4 years ago

5 0

Answer:

False

Explanation:

The given statement "Two objects must be in contact for them to exert a force on each other" is not true as there are many types of forces that doesn't require being in contact for exerting a force.

One such example is the gravitational force acting between two bodies. Gravitational force is the force of pull with which a body pulls another body without being in contact.

For two bodies of masses 'M' and 'm' separated at a distance of 'R', the gravitational force is given as:

$Force=\frac{GMm}{R^2}\\Where,G\to \textrm{Universal Gravitational constant}$

The gravitational force acts always act between bodies that have mass. The bodies are not in contact yet experience force.

Therefore, the given statement is false.

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SI is considered a consistent system because it what

NemiM [27]

Answer:

because it is a worldwide system....

Explanation:

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3 years ago

What will be the ratio between the initial and final kinetic energy of a car, if its velocity is tripled?

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Answer:b

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2 years ago

A car starts from rest and travels for 5.8 s with a uniform acceleration of 1.6 m/s² in the negative direction. What is the fina

elena-s [515]

Answer:

Final velocity of the car will be -9.28 m/sec

Explanation:

We have given that the car starts from the rest so initial velocity of the car u = 0 m /sec

Acceleration of the car $a=1.6m/sec^2$ in negative direction so acceleration will be $a=-1.6m/sec^2$

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v = u+at

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8 0

3 years ago

Suppose that a planet were discovered between the sun and Mercury, with a circular orbit of radius equal to 2/3 of the average o

Musya8 [376]

Explanation:

It is given that,

A planet were discovered between the sun and Mercury, with a circular orbit of radius equal to 2/3 of the average orbit radius of Mercury.

Mass of the Sun, $M=1.99\times 10^{30}\ kg$

Radius of Mercury's orbit, $r=5.79\times 10^{10}\ m$

Radius of discovered planet, $R=\dfrac{2}{3}r$

$R=\dfrac{2}{3}\times 5.79\times 10^{10}\ m=3.86\times 10^{10}\ m$

Let T is the orbital period of such a planet. Using Kepler's third law of planetary motion as :

$T^2\propto R^3$

$T^2=\dfrac{4\pi^2R^3}{GM}$

$T^2=\dfrac{4\pi^2\times (3.86\times 10^{10})^3}{6.67\times 10^{-11}\times 1.99\times 10^{30}}$

$T=\sqrt{1.71\times 10^{13}}$

T = 4135214.625 s

T = 47.86 days

So, the orbital period of such a planet is 47.86 days. Hence, this is the required solution.

6 0

3 years ago

WILL MARK BRAINLIEST!!!!!!!!!!!!!!!!!!

stepan [7]

*FRICTIONAL FORCE* in the opposite direction of the way Bobby is pushing.
Friction is a force which varies but it is always opposing the direction of motion.

*APPLIED FORCE* is the force that Bobby is pushing with.
An applied force is literally the force that is applied to an object.

*WEIGHT FORCE* is also called the force of gravity. It is straight downward.
It is the weight of the object multiplied by the force of gravity. If the TV weighed 100kg, acceleration is always 9.81 m/s^2, so the weight force would be 981 N.

*NORMAL FORCE* is the force which is holding the TV above ground. The ground supplies a force upward against the TV.
Normal force is just the force that prevents the TV from falling through the ground. We don't normally realize it in our everyday life, but the floor must hold everything up because gravity is always "pushing" against it.

3 0

3 years ago