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Nadusha1986 [10]
3 years ago
12

A pendulum of length L is suspended from the ceiling of an elevator. When the elevator is at rest the period of the pendulum is

T. How would the period of the pendulum change if the supporting chain were to break, putting the elevator into freefall?
Physics
1 answer:
Juliette [100K]3 years ago
6 0

Answer:

Explanation:

When the pendulum falls freely the net acceleration due to gravity is zero.

As we know that the time period of simple pendulum is inversely proportional to the square root of acceleration due to gravity, thus the time period becomes infinity.

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How does gravity affect the motions of Earth and the Moon?
RoseWind [281]

Answer:

The Sun's gravitational pull keeps our planet orbiting the Sun. The motion of the Moon is affected by the gravity of the Sun and Earth. Moon's gravity pulls on the Earth and makes the tides rise and fall.

8 0
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Which of the following is true of our current knowledge of electrons?
frosja888 [35]
They have a negative charge and rotate around the nucleus
8 0
3 years ago
A lightning bolt transfers 6.0 coulombs of charge from a cloud to the ground in 2.0 x 10-3 second. what is the average current d
AlladinOne [14]
The current is defined as the amount of charge transferred through a certain point in a certain time interval:
I= \frac{Q}{\Delta t}
where
I is the current
Q is the charge
\Delta t is the time interval

For the lightning bolt in our problem, Q=6.0 C and \Delta t= 2.0 \cdot 10^{-3}s, so the average current during the event is
I= \frac{Q}{\Delta t} = \frac{6.0 C}{2.0 \cdot 10^{-3} s}=3000 A
4 0
4 years ago
A particle with charge 3.01 µC on the negative x axis and a second particle with charge 6.02 µC on the positive x axis are each
ra1l [238]

Answer:

The third particle should be at 0.0743 m from the origin on the negative x-axis.

Explanation:

Let's assume that the third charge is on the negative x-axis. So we have:

E_{1}+E_{3}-E_{2}=0

We know that the electric field is:

E=k\frac{q}{r^{2}}

Where:

  • k is the Coulomb constant
  • q is the charge
  • r is the distance from the charge to the point

So, we have:

k\frac{q_{1}}{r_{1}^{2}}+k\frac{q_{3}}{r_{3}^{2}}-k\frac{q_{2}}{r_{2}^{2}}=0

Let's solve it for r(3).

\frac{3.01}{0.0429^{2}}+\frac{9.03}{r_{3}^{2}}-\frac{6.02}{0.0429^{2}}=0

r_{3}=0.0743\:  

Therefore, the third particle should be at 0.0743 m from the origin on the negative x-axis.

I hope it helps you!

 

3 0
3 years ago
Why does static friction have a maximum value​
Naddik [55]

Answer:

If you push horizontally with a small force, static friction establishes an equal and opposite force that keeps the book at rest. As you push harder, the static friction force increases to match the force. Eventually maximum static friction force is exceeded and the book moves.

Explanation:

3 0
3 years ago
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