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Jet001 [13]
1 year ago
7

A frictionless pendulum clock on the surface of the earth has a period of 1.00 s. On a distant planet, the length of the pendulu

m must be shortened slightly to have a period of 1.00 s. What is true about the acceleration due to gravity on the distant planet?AnswerWe cannot tell because we do not know the mass of the pendulum.The gravitational acceleration on the planet is slightly less than g.The gravitational acceleration on the planet is slightly greater than g.The gravitational acceleration on the planet is equal to g.
Physics
1 answer:
Sergio [31]1 year ago
7 0

Answer:

The gravitational acceleration on the planet is slightly less than g.

Explanation:

The period of a pendulum is given by:

T=2\pi \sqrt{\frac{L}{g}}

where

L is the length of the pendulum

g is the acceleration due to gravity

The formula can also be rewritten as

g=(\frac{2\pi}{T})^2 L (1)

In this problem, we have a pendulum which has a period of T=1.00 s on Earth. The length of the same pendulum must be shortened on the distant planet to have the same period of T'=1.00 s: this means that the length of the pendulum on the distant planet, L', is shorter than the length of the pendulum on Earth, L

L'

By looking at formula (1), we see that g (the gravitational acceleration) is directly proportional to L. therefore, if L is shortened on the distant planet, it means that also the value of g is lower than on Earth:

so, the correct answer is

The gravitational acceleration on the planet is slightly less than g.

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Lorico [155]

The force tending to lift the load (vertical force) is equal to <u>22.5N.</u>

Why?

Since the boy is pulling a load (150N) with a string inclined at an angle of 30° to the horizontal, the total force will have two components (horizontal and vertical component), but we need to consider the given information about the tension of the string which is equal to 105N.

We can calculate the vertical force using the following formula:

VerticalForce=Force*Sin(30\° )=(BoysForce-StringForce)*\frac{1}{2}\\\\VerticalForce=(150N-105N)*\frac{1}{2}=VerticalForce=45N*\frac{1}{2}=22.5N

Hence, we can see that <u>the force tending to lift the load</u> off the ground (vertical force) is equal to <u>22.5N.</u>

Have a nice day!

8 0
2 years ago
---&gt;Two aircraft P and Q are flying at the same speed. 300 m/s, The direction along which P is flying is at right angles to t
REY [17]

Answer:

The magnitude of the velocity of the aircraft P relative to aircraft Q is zero

Explanation:

The velocity of the two aircraft, P & Q, v = 300 m/s

The angle of the direction between them, Ф = 90°

The magnitude of the velocity of aircraft P relative to aircraft Q is given by the formula

                                  <em> V = v cos Ф </em>

Substituting the values in the above equation

                                   v = 300 x cos 90°

                                      = 300 x 0

                                      = 0

Since the aircraft are at right angles, the velocity of one aircraft relative to the other is zero.

5 0
2 years ago
A teacher uses the model that little invisible gremlins speed up or slow down objects and the direction they push gives the dire
Vlada [557]
Newtons second law.. <span>The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.</span>
4 0
2 years ago
Read 2 more answers
In the sport of parasailing, a person is attached to a rope being pulled by a boat while hanging from a parachute-like sail. A r
scoray [572]

Answer:

W = 506.75 N

Explanation:

tension = 2300 N

Rider is towed at a constant speed means there no net force acting on the rider.

hence taking all the horizontal force and vertical force in consideration.

net horizontal  force:

F cos 30° - T cos 19° = 0

F cos 30° = 2300 × cos 19°

F = 2511.12 N

net vertical force:

F sin 30° - T sin 19°- W = 0

W = F sin 30° - T sin 19°

W =  2511.12 sin 30° - 2300 sin 19°

W = 506.75 N

8 0
2 years ago
Wrapping paper is being unwrapped from a 5.0-cm radius tube, free to rotate on its axis. if it is pulled at the constant rate of
lisov135 [29]
So the equation for angular velocity is

Omega = 2(3.14)/T

Where T is the total period in which the cylinder completes one revolution.

In order to find T, the tangential velocity is

V = 2(3.14)r/T

When calculated, I got V = 3.14

When you enter that into the angular velocity equation, you should get 2m/s
5 0
2 years ago
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