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DiKsa [7]
3 years ago
7

1. How would the motion of a pendulum change at high altitude like a high mountain top? How would the motion change under weight

less conditions?
Physics
1 answer:
gavmur [86]3 years ago
6 0

Answer:

A pendulum moves with the principle of small angles that show that the frequency and period of the pendulum are not dependent on the initial angular displacement of the mass it carries.

On a high mountain top, a reduced force of gravity is exerted on the mass, therefore, producing long periods of oscillation on the pendulum.

However, under weightless conditions where gravity is zero, the pendulum will not work as what keeps the mass in oscillation is gravity. If therefore gravity is absent there can be no oscillation.

Explanation:

On a high mountain top, a reduced force of gravity is exerted on the mass, therefore, producing long periods of oscillation on the pendulum.

However, under weightless conditions where gravity is zero, the pendulum will not work as what keeps the mass in oscillation is gravity. If therefore gravity is absent there can be no oscillation.

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On a horizontal surface is located
Ierofanga [76]

By Newton's second law, the net vertical force acting on the object is 0, so that

<em>n</em> - <em>w</em> = 0

where <em>n</em> = magnitude of the normal force of the surface pushing up on the object, and <em>w</em> = weight of the object. Hence <em>n</em> = <em>w</em> = <em>mg</em> = 196 N, where <em>m</em> = 20 kg and <em>g</em> = 9.80 m/s².

The force of static friction exerts up to 80 N on the object, since that's the minimum required force needed to get it moving, which means the coefficient of <u>static</u> friction <em>µ</em> is such that

80 N = <em>µ</em> (196 N)   →   <em>µ</em> = (80 N)/(196 N) ≈ 0.408

Moving at constant speed, there is a kinetic friction force of 40 N opposing the object's motion, so that the coefficient of <u>kinetic</u> friction <em>ν</em> is

40 N = <em>ν</em> (196 N)   →   <em>ν</em> = (40 N)/(196 N) ≈ 0.204

And so the closest answer is C.

(Note: <em>µ</em> and <em>ν</em> are the Greek letters mu and nu)

3 0
3 years ago
Is it proper to use an infinitely long cylinder model when finding the temperatures near the bottom or top surfaces of a cylinde
Gelneren [198K]

Answer:

No, it is not proper to use an infinitely long cylinder model when finding the temperatures near the bottom or top surfaces of a cylinder.

Explanation:

A cylinder is said to be infinitely long when is of a sufficient length. Also, when the diameter of the cylinder is relatively small compared to the length, it is called infinitely long cylinder.

Cylindrical rods can also be treated as infinitely long when dealing with heat transfers at locations far from the top or bottom surfaces. However, it not proper to treat the cylinder as being infinitely long when:

* When the diameter and length are comparable (i.e have the same measurement)

When finding the temperatures near the bottom or top of a cylinder, it is NOT PROPER TO USE AN INFINITELY LONG CYLINDER because heat transfer at those locations can be two-dimensional.

Therefore, the answer to the question is NO, since it is not proper to use an infinitely long cylinder when finding temperatures near the bottom or top of a cylinder.

8 0
3 years ago
A batter hits a 0.140-kg baseball that was approaching him at 19.5 m/s and, as a result, the ball leaves the bat at 44.8 m/s in
Arada [10]

Answer:

5295.3 N

Explanation:

According to law of momentum conservation, the change in momentum of the ball shall be from the momentum generated by the batter force

mv + P = mV

P = mV - mv = m(V - v)

Since the velocity of the ball before and after is in opposite direction, one of them is negative

P = 0.14(44.8 - (-19.5)) = 9 kg m/s

Hence the force exerted to generate such momentum within 1.7ms (0.0017s) is

F = P/t = 9/0.0017 = 5295.3 N

4 0
3 years ago
What is the force of gravity between you and a standard tennis ball if you are 1 meter away
faltersainse [42]

Answer:

WHY MUST I CRYYYYYYYY

Explanation:

3 0
3 years ago
An airplane starts from rest at the end of a runway and accelerates at a constant rate. In the first second, the airplane travel
Licemer1 [7]

Answer:

v=4.44\frac{m}{s}

Explanation:

Given that the airplane starts from the rest (this is initial velocity equals to zero)  and accelerates at a constant rate, position can be described like this: x=v_{0}t +\frac{1}{2} at^{2} where x is the position, t is the time a is the acceleration and v_{0} is initial velocity. In this way acceleration can be found. a=\frac{2(x-v_{0}t) }{t^{2} } =\frac{2(1.11m-0)}{1s^{2} } =2.22\frac{m}{s^{2} }.

Now we are able to found velocity at any time with the formula: v=v_{0} +at = 0\frac{m}{s} +(2.22\frac{m}{s^{2}}.2s)=4.44\frac{m}{s}

3 0
3 years ago
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