3 years ago

Ocean waves are hitting a beach at a rate of 3.2 Hz. What is the period of the waves? (2 decimal places)

Physics

1 answer:

insens350 [35]3 years ago

7 0

Answer:

0.31s

Explanation:

Given parameters:

Frequency of the ocean waves = 3.2Hz

Unknown:

Period of the waves = ?

Solution:

The period of a wave is the inverse of the frequency of a wave;

Period = $\frac{1}{Frequency}$

Now insert the parameters and solve;

Period = $\frac{1}{3.2}$ = 0.31s

You might be interested in

A police officer is using her radar to check speeds. She is moving south at 50 mph. You are moving

Luda [366]

Answer:

the relative speed should be 40mph

Explanation:

8 0

3 years ago

Using the free-body diagram, calculate the net force

yuradex [85]

Answer:yes it is

Explanation:

5 0

3 years ago

Read 2 more answers

Which chemical reaction absorbs energy

kondaur [170]

Answer:

endothermic reactions

8 0

3 years ago

Read 2 more answers

Racing greyhounds are capable of rounding corners at very high speeds. A typical greyhound track has turns that are 45 m diamete

Readme [11.4K]

Answer:

In m/s^2:

a=11.3778 m/s^2

In units of g:

a=1.161 g

Explanation:

Since the racing greyhounds are capable of rounding corners at very high speed so we are going use the following formula of acceleration for circular paths.

$a=\frac{v^2}{r}$

where:

v is the speed

r is the radius

Now,

$a=\frac{16^2}{45/2}\\ a=11.3778 m/s^2$

In g units:

$a=\frac{11.3778\ g}{9.8}\\ a=1.161\ g$

7 0

3 years ago

A professor sits at rest on a stool that can rotate without friction. The rotational inertia of the professor-stool system is 4.

Anestetic [448]

Answer:

$\omega=0.37 [rad/s]$

Explanation:

We can use the conservation of the angular momentum.

$L=mvR$

$I\omega=mvR$

Now the Inertia is I(professor_stool) plus mR², that is the momentum inertia of a hoop about central axis.

So we will have:

$(I_{proffesor - stool}+mR^{2})\omega=mvR$

Now, we just need to solve it for ω.

$\omega=\frac{mvR}{I_{proffesor-stool}+mR^{2}}$

$\omega=\frac{1.5*2.7*0.4}{4.1+1.5*0.4^{2}}$

$\omega=0.37 [rad/s]$

I hope it helps you!

5 0

3 years ago