All categories

3 years ago

A wave is traveling at 241 m/s. It has a wavelength of 30.0 m. What is its frequency?

Physics

1 answer:

Feliz [49]3 years ago

3 0

Answer:

A) 8.03Hz

Explanation:

f = V/λ

Where wavelength( λ )= 30m

Speed (V) =241m/S

f= 241/30=8.03Hz

You might be interested in

Absorbance measurements in the range of a = 0.3-2 are considered the most accurate. why would absorbance measurements of 0.05 an

zhenek [66]

Please provide the choices to select the possible choices.

6 0

3 years ago

What is the best conclusion for the events taking place on a molecular level in the images seen here?

slega [8]

Answer:

Explanation:

if usa test prep

8 0

3 years ago

In a game of angry birds you launch a bird with an angle of 53 degrees to horizontal. Unfortunatly, its not a good shot and the

Alisiya [41]

Answer:

The maximum height covered is 3.25 m.

The horizontal distance covered is 9.81 m.

The total time in the air is 1.63 seconds.

Explanation:

The launch speed, $u_0= 10 m/s$ .

Angle of launch with the horizontal, $\theta = 53 ^{\circ}$

So, the vertical component of the initial velocity,

$u_0\sin\theta=10 \sin 53 ^{\circ}\cdots(i)$ .

The horizontal component of the initial velocity,

$u_0\cos\theta=10 \cos 53 ^{\circ}$

Let, t be the time of flight, to the horizontal distance covered

$D=10 \cos (53 ^{\circ})t\cdots(ii)$ .

Not, applying the equation of motion in the vertical direction.

$s= ut +\frac 1 2 at^2$

Where s is the displacement in time t, u is the initial velocity and a is the acceleration.

In this case, $u =10 \sin 53 ^{\circ}$ (from equation (i), s=0 (as the final height is same as the launch height) and $a = -9.81 m/s^2$ (negative sign is due to the downward direction).

$\Rightarrow 0 = 10 (\sin 53 ^{\circ})t-\frac 1 2 (9.81)t^2$

$\Rightarrow t= \frac {2\times 10 (\sin 53 ^{\circ})}{9.81}=1.63$ seconds.

So, the total time in the air is 1.63 seconds.

From equation (i),

Total horizontal distance covered is

$D=10 \cos (53 ^{\circ})\times 1.63 = 9.81 m$ .

Now, for the maximum height, H, applying the equation of motion as

$v^2=u^2+2as$

Here, v is the final velocity and v=0 (at the maximum height), and h=H.

So, $0^2=(10 \sin 53 ^{\circ})^2-2(9.81)H$

$\Rightarrow H = \frac {(10 \sin 53 ^{\circ})^2}{2\times 9.81}$

$\Rightarrow H = 3.25 m$ .

Hence, the maximum height covered is 3.25 m.

8 0

3 years ago

Why is the solar system commonly represented as a scale model?

topjm [15]

The actual size of the Solar system is too big to show without making a much smaller model. If someone wants to see the orientation of the planets a model has to be made so we can see it without flying out too space.

6 0

3 years ago

A mass m = 550 g is hung from a spring with spring constant k = 2.8 N/m and set into oscillation at time t = 0. A second, identi

riadik2000 [5.3K]

Answer:

The second system must be set in motion $t=0.70s$ seconds later

Explanation:

The oscillation time, T, for a mass, m, attached to spring with Hooke's constant, k, is:

$T=2\pi\sqrt(\frac{m}{k} )$

One oscillation takes T secondes, and that is equivalent to a 2π phase. Then, a difference phase of π/2=2π/4, is equivalent to a time t=T/4.

If the phase difference π/2 of the second system relative to the first oscillator. The second system must be set in motion $t=\frac{\pi}{2}\sqrt(\frac{m}{k})=\frac{\pi}{2}\sqrt(\frac{0.55}{2.8}= 0.70s)$ seconds later

6 0

3 years ago