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3 years ago

An object that completes 20 vibrations in 10 seconds has a frequency of

Physics

1 answer:

nika2105 [10]3 years ago

7 0

Answer:

The object has frequency of 2 Hz and time period of 0.5 s.

Explanation:

Frequency is defined as number of oscillation per second ie back and forth swings done in single second.

Here it is given that the object oscillates 20 times in 10 seconds.

So f = $\frac{20}{10}$ = 2Hz

The time period is defined as time taken by the object to complete one full oscillation.

T = $\frac{1}{f}$

T= $\frac{1}{2}$ =0.5 s

Thus the object has frequency of 2 Hz and time period of 0.5 s.

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If all of the forces acting on an object are balanced, then:

kondaur [170]

A. the direction the object is moving in will not change.

B. the acceleration of the object will be 0 m/s2

Explanation:

We can answer this problem by using Newton's second law of motion:

$\sum F = ma$

where

$\sum F$ is the net force acting on an object

m is the mass of the object

a is its acceleration

In this problem, all the forces acting on the object are balanced, therefore the net force is zero:

$\sum F=0$

which means that also the acceleration is zero:

$a=0$

Acceleration is equal to the rate of change of velocity: therefore, zero acceleration means that the velocity of the object does not change. We can now analyze the given statements:

A. the direction the object is moving in will not change. --> TRUE, because the velocity is not changing.

B. the acceleration of the object will be 0 m/s2 --> TRUE, as we stated above

c. the object will not be in motion. --> FALSE: we just know that its velocity is constant, but it can be different from zero

D. the velocity of the object will be 0 m/s. --> FALSE, for the same reason stated in C

Learn more about Newton's second law of motion:

brainly.com/question/3820012

#LearnwithBrainly

3 0

3 years ago

a load of 800 newton is lifted by an effort of 200 Newton. if the load is placed at a distance of 10 cm from the fulcrum. what w

nataly862011 [7]

Answer:

40 cm

Explanation:

We are given that

Load=800 N

Effort=200 N

Load distance=10 cm

We have to find the effort distance.

We know that

$load\times load\;distance=Effort\times effort\;distance$

Using the formula

$800\times 10=200\times effort\;distance$

Effort distance= $\frac{800\times 10}{200}$

Effort distance= $\frac{8000}{200}$

Effort distance=40 cm

Hence, the effort distance will be 40 cm.

7 0

3 years ago

What’s the answer to this

ladessa [460]

Choices 1, 2, and 4 . . . . . Yes

Choices 3 and 5 . . . . . No

6 0

4 years ago

Applying the Law of Conservation of Energy. If a car was released down the track from a height what happens to the potential ene

erastova [34]

Answer:

According to the law of conservation of energy, energy cannot be created or destroyed, although it can be changed from one form to another. KE + PE = constant. A simple example involves a stationary car at the top of a hill. As the car coasts down the hill, it moves faster and so it’s kinetic energy increases and it’s potential energy decreases. On the way back up the hill, the car converts kinetic energy to potential energy. In the absence of friction, the car should end up at the same height as it started.

This law had to be combined with the law of conservation of mass when it was determined that mass can be inter-converted with energy.

One can also imagine the energy transformation in a pendulum. When the ball is at the top of its swing, all of the pendulum’s energy is potential energy. When the ball is at the bottom of its swing, all of the pendulum’s energy is kinetic energy. The total energy of the ball stays the same but is continuously exchanged between kinetic and potential forms

4 0

3 years ago

Definition of AM and Fm Wave? in your own word

vfiekz [6]

Answer:

In AM broadcasting, the amplitude of the carrier wave is modulated to encode the original sound. In FM broadcasting, the frequency of the carrier wave is modulated to encode the sound. A radio receiver extracts the original program sound from the modulated radio signal and reproduces the sound in a loudspeaker.

8 0

3 years ago