All categories

2 years ago

What is the frequency of a wave with speed 3m/s and wavelength 6m?

Physics

1 answer:

Ugo [173]2 years ago

6 0

Answer:

f = 0.5 Hz

Explanation:

frequency = f

wave length = l

speed of wave = v

f = v/l → f = 3/6 = ½ = 0.5 Hz

You might be interested in

What is the farthest distance parallaxes can be used to measure star distances from Earth?

JulsSmile [24]

Using current technology, useful parallax measurements can only be found for stars up to about 340 light years (100 parsecs) away.

8 0

3 years ago

Energy from solar radiation may be ________ or taken in by a surface or an object.

tester [92]

Answer:

Absorbed

Explanation:

I hope this helps you

7 0

3 years ago

a car initially at 39.40 m/s accelerates at -3.00 m/s^2 how far has a car travel when it comes to a stop 9.35 seconds later

Kipish [7]

Answer:

499.523. meter

I hope it's helps you

6 0

3 years ago

A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate

Viefleur [7K]

Answer:

The rate of change of the area when the bottom of the ladder (denoted by $b$ ) is at 36 ft. from the wall is the following:

$\frac{dA}{dt}|_{b=36}=-571.2\, ft^2/s$

Explanation:

The Area of the triangle is given by $A=h\times b$ where $h=\sqrt{l^2-b^2}$ (by using the Pythagoras' Theorem) and $b$ is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

The area is then

$A=\sqrt{l^2-b^2}b$

The rate of change of the area is given by its time derivative

$\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)$

$\implies \frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\right)\cdot b+\frac{db}{dt}\cdot\sqrt{l^2-b^2}$

$\implies\frac{dA}{dt}=\frac{1}{2\sqrt{l^2-b^2}}\frac{d}{dt}(l^2-b^2)\cdot b+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Product rule

$\implies\frac{dA}{dt}=-\frac{1}{2\sqrt{l^2-b^2}}\cdot 2\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Chain rule

$\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$

$\implies\frac{dA}{dt}=\frac{db}{dt}\left(-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2+\sqrt{l^2-b^2}}\right)$

In here we can identify $b=36\, ft$ , $l=39$ and $\frac{db}{dt}=8\,ft/s$ .

The result is then

$\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s$

3 0

2 years ago

Can u explain cause i dont understand

galben [10]

Answer:a

Explanation:

7 0

3 years ago