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larisa [96]
3 years ago
8

A pendulum has a 0.35\ \text{kg}0.35 kg0, point, 35, space, start text, k, g, end text mass oscillating at a small angle from a

light string of length 1.0\ \text{m}1.0 m1, point, 0, space, start text, m, end text. What is the frequency of oscillation?
Physics
1 answer:
expeople1 [14]3 years ago
4 0

Answer:

The frequency of oscillation of the simple pendulum is 0.49 Hz.

Explanation:

Given that,

Mass of the simple pendulum, m = 0.35 kg

Length of the string to which it is attached, l = 1 m

We need to find the frequency of oscillation. The frequency of oscillation of the simple pendulum is given by :

f=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}} \\\\f=\dfrac{1}{2\pi}\sqrt{\dfrac{9.8}{1}} \\\\f=0.49\ Hz

So, the frequency of oscillation of the simple pendulum is 0.49 Hz. Hence, this is the required solution.

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At a given instant an object has an angular velocity. It also has an angular acceleration due to torques that are present. There
katen-ka-za [31]

a) Constant

b) Constant

Explanation:

a)

We can answer this question by using the equivalent of Newton's second law of motion of rotational motion, which can be written as:

\tau_{net} = I \alpha (1)

where

\tau_{net} is the net torque acting on the object in rotation

I is the moment of inertia of the object

\alpha is the angular acceleration

The angular acceleration is the rate of change of the angular velocity, so it can be written as

\alpha = \frac{\Delta \omega}{\Delta t}

where

\Delta \omega is the change in angular velocity

\Delta t is the time interval

So we can rewrite eq.(1) as

\tau_{net}=I\frac{\Delta \omega}{\Delta t}

In this problem, we are told that at a given instant, the object has an angular acceleration due to the presence of torques, so there is a non-zero change in angular velocity.

Then, additional torques are applied, so that the net torque suddenly equal to zero, so:

\tau_{net}=0

From the previous equation, this implies that

\Delta \omega =0

Which means that the angular velocity at that instant does not change anymore.

b)

In this second case instead, all the torques are suddenly removed.

This also means that the net torque becomes zero as well:

\tau_{net}=0

Therefore, this means that

\Delta \omega =0

So also in this case, there is no change in angular velocity: this means that the angular velocity of the object will remain constant.

So cases (a) and (b) are basically the same situation, as the net torque is zero in both cases, so the object acts in the same way.

8 0
3 years ago
Which type of rays have a wavelength shorter than that of visible light?
aniked [119]

Answer:

c

Explanation:

wavelength shorter means energy is higher

the wavelength

radio waves>microwave>infrared rays>gamma rays

6 0
1 year ago
A swimming pool is 4.0 m in depth; a swimmer at this depth feels discomfort in the ear. Calculate the net force on a 0.50-cm-dia
Mashcka [7]

The net force on a 0.50-cm-diameter eardrum is mathematically given as

F= 0.76969 N

<h3>What is the net force on a 0.50-cm-diameter eardrum?</h3>

Generally, the equation for Pressure is  mathematically given as

P = ρgh

Therefore

P= 1000*9.8*4

P= 39200 Pa

Where

A= pi*(0.005/2)^2

Generally, the equation for Net force is  mathematically given as

F = PA

F= 39200 *( pi*(0.005/2)^2)

F= 0.76969 N

In conclusion, The net force is

F= 0.76969 N

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5 0
2 years ago
"A block of metal weighs 40 N in air and 30 N in water. What is the buoyant force on the block due to the water? The density of
Alja [10]

Answer:

buoyant force on the block due to the water= 10 N

Explanation:

We know that

buoyant force(F_B) on a block= weight of the block in air (actual weight) - weight of block in water.

Given:

A block of metal weighs 40 N in air and 30 N in water.

F_B =  40-30= 10 N

therefore,  buoyant force on the block due to the water= 10 N

6 0
3 years ago
Read 2 more answers
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