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dimulka [17.4K]
3 years ago
8

Write the formula for the Newton’s Law of Gravitation.​please answer this question!!!

Physics
2 answers:
Stells [14]3 years ago
5 0

Answer:

F = G(m1m2)/R2.

Explanation:

belka [17]3 years ago
3 0

Newton's law of universal gravitation:

<em>Gravitational Force, in Newtons, between two objects = </em>

<em>(a constant)·(one mass)·(the other mass)/(distance between them)²</em>

<em>acting on EACH object, in the direction of the other object.</em>

If the masses are in kilograms and the distance is in meters, then the constant is 6.67 x 10⁻¹¹ m³/kg-sec² .

I may be wrong, but I don't think Newton had any number to use for the constant.  It had not been measured yet, the kilogram and the meter had not been invented yet, and there certainly was no unit called "a Newton" during his lifetime.

He might have been able to calculate the value of the constant by applying his law of gravity to the motion of one or two  planets. But he would have needed to know the mass of the sun and the planets he used, and I don't think those were known yet in Newton's time.  

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Which of the following is an example of kinetic energy?
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3 years ago
A 15-µF capacitor and a 25-µF capacitor are connected in parallel, and charged to a potential difference of 60 V. How much energ
Alborosie

Answer:

Energy stored, E = 0.072 J

Explanation:

Given that,

Capacitance, C_1=15\ \mu F

Capacitance, C_2=25\ \mu F

These two capacitor are connected in parallel, and charged to a potential difference of, V = 60 volts

We know that in parallel combination of capacitor, the equivalent capacitance is given by :

C=C_1+C_2\\\\C=(15+25)\ \mu F\\\\C=40\times 10^{-6}\ F

The energy stored in the capacitor is given by :

E=\dfrac{1}{2}CV^2\\\\E=\dfrac{1}{2}\times 40\times 10^{-6}\times (60)^2\\\\E=0.072\ J

So, the energy stored in the capacitor in this capacitor combination is 0.072 J.

4 0
3 years ago
In terms of matter and resources, Earth is essentially a(n) ________ system ; in terms of energy, Earth is a(n) ________ system.
Savatey [412]

Answer:

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Explanation:

6 0
3 years ago
A block (mass m1) lying on a frictionless inclined plane is connected to a mass (mass m2) by a massless cord passing over a pull
azamat

Answer:

a) System aceleration:

  • a=\frac{g(m_2-m_1sin(\theta))}{m_1+m_2}

b) Direction of movement:

The block m_1 down the plane when the acceleration is negative. This occur when:

m_2-m_1sin(\theta)

The block m_2 up the plane when the acceleration is positive. This occur when:

m_2-m_1sin(\theta)>0

Explanation:

For the block m_1 the move direction is parallel (||) to the plane

\sum F_{||}=m_1a=T-sin(\theta)mg  (1)

For the block m_2  the move direction is vertical (y)

\sum F_y=m_2a=m_2g-T  (2)

Both blocks are connected with the same cable, therefore, the tension on the cable and the acceleration is the same for both.

Solving the equation 2 for T:

T=m_2g-m_2a   (3)

replacing (3) in the equation (1)

m_2g-m_2a- m_1gsin(\theta)=m_1a  (4)

solving (4) for a:

a=\frac{g(m_2-m_1sin(\theta))}{m_1+m_2}

8 0
3 years ago
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