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3 years ago

Calculate a pendulum's frequency of oscillation (in Hz) if the pendulum completes one cycle in 0.5 s.

1 answer:

3 0

Time taken to complete one oscillation for a pendulum is Time Period, T = 0.5 s
Frequency of the pendulum oscillation = 1 / Time Period => f = 1 / T = 1 / 0.5
Frequency f = 2 Hz

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Define neuron and nerve<br>cell

san4es73 [151]

Answer: Hi!

A neuron is a basic working unit of the brain. Neurons are special cells designed to transfer information to other nerve, muscle, or gland cells. They are pretty cool - looking too! (A slightly irregular circular shape with branches reaching out from all sides.) A neuron is a nerve cell. Nerve cells are the way of communication in the nervous system.

Hope this helps!

3 0

2 years ago

Read 2 more answers

How much work does it take to move a 50 μC charge<br> against a 12 V potential difference?

lukranit [14]

<span>work =V*Q =12*50*10^-6

The total work done will be equal to

work = V.Q

which means

w= 12 . 50.10^-6
Hence,
w= 0.0006 J</span>

8 0

2 years ago

A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 4.444 N

Dmitriy789 [7]

Answer:

The value of third charge is 0.8μC.

Explanation:

Given that.

Magnitude of net force=4.444 N

According to figure,

Suppose, First charge = 2.4 μC

Second charge = 6.2 μC

Distance r₁ = 9.8 cm

Distance r₂ = 2.1 cm

We need to calculate the value of r

Using Pythagorean theorem

$r=\sqrt{(r_{1})^2+(r_{2})^2}$

Put the value into the formula

$r=\sqrt{(9.8)^2+(2.1)^2}$

$r=10.02\ cm$

We need to calculate the force

Using formula of force

$F_{12}=\dfrac{kq_{1}q_{2}}{(r)^2}$

Force F₁₂,

$F_{12}=\dfrac{9\times10^{9}\times2.4\times10^{-6}\times6.2\times10^{-6}}{(10.02\times10^{-2})^2}$

$F_{12}=13.33\ N$

$F_{21}=-13.33\ N$

Force F₂₃,

$F_{23}=\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(10.02)^2}$

We need to calculate the value of third charge

$F_{net}=F_{21}+F_{23}$

$4.444=-13.33+\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(5.01)^2}$

$q_{3}=\dfrac{(4.444+13.33)\times(5.01\times10^{-2})^2}{9\times10^{9}\times6.2\times10^{-6}}$

$q_{3}=7.99\times10^{-7}\ C$

$q_{3}=0.8\times10^{-6}\ C$

Hence, The value of third charge is 0.8μC.

4 0

2 years ago

Which statement best explains why an object

slavikrds [6]

D, Mercury as a weaker gravitational pull! Due to mercury being farther from the sun and it being a smaller planet it has a weaker pull

4 0

2 years ago

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If the bar magnet is flipped over and the south pole is brought near the hanging ball, the ball will be?

disa [49]

The ball may attracted to the magnet.

<h3>How can we understand that the hanging ball will be attracted to the magnet or not?</h3>

From the question, we understand that the ball is attracted by the north pole of the bar magnet, then the bar magnet flipped over and the south pole is brought near the hanging ball.
As we know, in this type of experiments of bar magnet most of the times the ball is made out of steel.
Steel is a magnetic material.
Magnetic materials gets attracted to the magnet at both the North and South pole.
This can be compared to how neutral objects also gets attracted to the positively and negatively charged rods through the Polarization force.

So, If the bar magnet is flipped over and the south pole is brought near the hanging ball, The ball will be attracted to the magnet.

Learn more about the bar magnet:

brainly.com/question/27943723

#SPJ4

7 0

1 year ago