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Illusion [34]
3 years ago
9

Why is newton's gravitational force called universal gravitational force​

Physics
2 answers:
Lelechka [254]3 years ago
6 0
Newton’s law of gravity is considered “universal” because it is believed to be applicable to the entire Universe (to a good approximation). Gravity does not just pull an apple from a tree, but also pulls the Moon to the Earth, the Earth to the Sun, and so on. Gravity is a fundamental force of nature.
Rasek [7]3 years ago
5 0

Answer:

is called so because it is applicable on all bodies having mass, and the bodies will be governed by the same law, that is newton's law of gravitation. Thus, as it is applicable universally, it is called as universal law

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Explain how objects in motion have kinetic energy, use examples
Ann [662]
Kinetic energy is energy that a body possesses by virtue of being in motion, there for if an object is moving, it has kinetic energy.
Example; A roller coaster sitting on top of hill has potential energy. When it starts to move and is going down the hill, it has kinetic energy. :)
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3 years ago
What type of electromagnetic radiation is being used in the picture?
pashok25 [27]

Answer: b i think

Explanation:

6 0
3 years ago
If a circle was flattened by pushing down on it, it would most likely form into the shape of ----------- which has ----------- f
murzikaleks [220]
The question is looking for "ellipse" and "two" to fill in the blanks.
8 0
4 years ago
Read 2 more answers
The velocity of a particle moving along the x-axis varies with time according to v(t) = A + Bt−1 , where A = 2 m/s, B = 0.25 m,
kondaur [170]

Answer:

a= -2\ m/s^2

a=-12.5\ m/s^2

x=2.17 m

x=8.4 m

Explanation:

Given that

v=A+Bt^{-1}

v=2+0.25t^{-1}

To find acceleration :

we know that

a=\dfrac{dv}{dt}

\dfrac{dv}{dt}=0-0.5t^{-2}

a=-0.5t^{-2}

Acceleration at t= 2 s

a=-0.5\times 2^{-2}

a= -2\ m/s^2

Acceleration at t= 5 s

a=-0.5\times 5^{-2}

a=-12.5\ m/s^2

We know that

v=\dfrac{dx}{dt}

dx=\left(2+\dfrac{1}{4t}\right)dt

Position at t= 2 s:

\int_{0}^{x}dx=\int_{1}^{2} \left(2+0.25\dfrac{1}{t}\right)dt

x=\left [2t+0.25\ lnt \right ]_{1}^{2}

x=2+0.25 ln2

x=2.17 m

Position at t= 5 s:

\int_{0}^{x}dx=\int_{1}^{5} \left(2+0.25\dfrac{1}{t}\right)dt

x=\left [2t+0.25\ lnt \right ]_{1}^{5}

x=8+0.25 ln5

x=8.4 m

4 0
3 years ago
A hopper jumps straight up to a height of 1.1 m. With what velocity did it leave the floor?
Amanda [17]

Answer:

4.64m/s

Explanation:

We can use the formula [ v = √2gh ] to solve for this problem. We know that g is constant acceleration (9.8), and h is height (1.1).

v = √2(9.8)(1.1)

v ≈ 4.64m/s

Best of Luck!

4 0
2 years ago
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