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andrew-mc [135]
3 years ago
10

According to Newton's law of universal gravitation, which of these objects would have the strongest gravitational force?

Physics
1 answer:
bixtya [17]3 years ago
4 0
A golf ball less resistance-smaller object
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A common carnival ride, called a gravitron, is a large r = 11 m cylinder in which people stand against the outer wall. the cylin
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Answer:

The minimum speed is 14.53 m/s.

Explanation:

Given that,

r = 11 m

Friction coefficient = 0.51

Suppose we need to find the minimum speed, that the cylinder must make a person move at to ensure they will stick to the wall

When frictional force becomes equal to or greater than the weight of person

Then, he sticks to the wall

We need to calculate the minimum speed

Using formula for speed

f_{s}=\mu N\geq mg

Where, N =\dfrac{mv^2}{r}

\mu\times\dfrac{mv^2}{r}\gep mg

v^2\geq\dfrac{gr}{\mu}

Put the value into the formula

v=\sqrt{\dfrac{9.8\times11}{0.51}}

v=14.53\ m/s

Hence, The minimum speed is 14.53 m/s.

8 0
3 years ago
The ends of a massless rope are attached to two stationary objects (e.g., two trees or two cars) so that the rope makes a straig
julia-pushkina [17]

Answer: They are all true

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d. The forces exerted on the two objects by the rope must be in the direction of the rope.

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5 0
4 years ago
What is process that involves the collection of information and ideas supported by belief or opinion for science?
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The answer is pseudoscience
4 0
3 years ago
A wildebeest runs with an average speed of 4.0\,\dfrac{\text m}{\text s}4.0 s m ​ 4, point, 0, start fraction, start text, m, en
devlian [24]

Answer:

60m

Explanation:

4 0
3 years ago
Suppose that in a lightning flash the potential difference between a cloud and the ground is 1.0*109 V and the quantity of charg
nata0808 [166]

Answer:

a) U_{e} = 3 \times 10^{10}\,J, b) v \approx 7745.967\,\frac{m}{s}

Explanation:

a) The potential energy is:

U_{e} = Q \cdot \Delta V

U_{e} = (30\,C)\cdot (1.0\times 10^{9}\,V)

U_{e} = 3 \times 10^{10}\,J

b) Maximum final speed:

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The final speed is:

v=\sqrt{\frac{2\cdot (3 \times 10^{10}\,J)}{1000\,kg} }

v \approx 7745.967\,\frac{m}{s}

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