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lord [1]
3 years ago
9

The gravitational attraction between a 20 kg cannonball and a 0.002 kg

Physics
1 answer:
Naya [18.7K]3 years ago
3 0

Answer:

2.966\times 10^{-11}\ N

Explanation:

Given:

Mass of the cannonball (M) = 20 kg

Mass of the marble (m) = 0.002 kg

Distance between the cannonball and marble (d) = 0.30 m

Universal gravitational constant (G) = 6.674\times 10^{-11}\ m^3 kg^{-1} s^{-2}

Now, we know that, the gravitational force (F) acting between two bodies of masses (m) and (M) separated by a distance (d) is given as:

F=\dfrac{GMm}{d^2}

Plug in the given values and solve for 'F'. This gives,

F=\frac{(6.674\times 10^{-11}\ m^3 kg^{-1} s^{-2})\times (20\ kg)\times (0.002\ kg)}{(0.30\ m)^2}\\\\F=\frac{6.674\times 20\times 0.002\times 10^{-11}\ m^3 kg^{-1+2} s^{-2}}{0.09\ m^2}\\\\F=2.966\times 10^{-11}\ kg\cdot m\cdot s^{-2}\\\\F=2.966\times 10^{-11}\ N.........(1\ N = 1\ kg\cdot m\cdot s^{-2})

The same force is experienced by both cannonball and marble.

Therefore, the gravitational  force of the marble is 2.966\times 10^{-11}\ N

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In a double‑slit interference experiment, the wavelength is lambda=487 nm , the slit separation is d=0.200 mm , and the screen i
pantera1 [17]

Answer:

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Therefore,

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3 years ago
A baseball pitcher throws the ball in a motion where there is rotation of the forearm about the elbow joint as well as other mov
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Answer: 330.88 J

Explanation:

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w = 17.1 / 0.47

w = 36.38 rad/s

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KE(rot) = 1/2Iw²

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