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VladimirAG [237]

3 years ago

15

Wind gusts create ripples on the ocean that have a wavelength of 2.23 cm and propagate at 4.12 m/s. What is their frequency (in

Hz)?

1 answer:

stellarik [79]3 years ago

8 0

Answer:

The frequency of the wave is 184.75 Hz.

Explanation:

Given;

wavelength of the wind, λ = 2.23 cm = 0.0223 m

speed of the wind wave, v = 4.12 m/s

the frequency of the wave = ?

The frequency of the wave is calculated as;

F = V / λ

Where;

F is the frequency of the wave

Substitute the given values and solve for F

F = (4.12) / (0.0223)

F = 184.75 Hz.

Therefore, the frequency of the wave is 184.75 Hz.

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At what wavelength would a star radiate the greatest amount of energy if the star has a surface temperature of 60,000 K?

kompoz [17]

Answer:

$\lambda=4.81\times 10^{-8}\ m$

Explanation:

We have,

The surface temperature of the star is 60,000 K

It is required to find the wavelength of a star that radiated greatest amount of energy. Wein's displacement law gives the relation between wavelength and temperature such that :

$\lambda T=2.89\times 10^{-3}$

Here,

$\lambda$ = wavelength

$\lambda=\dfrac{2.89\times 10^{-3}}{60000}\\\\\lambda=4.81\times 10^{-8}\ m$

So, the wavelength of the star is $4.81\times 10^{-8}\ m$ .

7 0

3 years ago

A soccer player takes a corner kick, lofting a stationary ball 30.0° above the horizon at 18.0 m/s. If the soccer ball has a mas

Radda [10]

Answer:

change in momentum, $\Delta p=7.65 \,kg.m.s^{-1}$

$\Delta p_x= 6.6251 \,kg.m.s^{-1}$

$\Delta p_y= 3.825 \,kg.m.s^{-1}$

Average Force, $F=144.3396\,N$

$F_x=125.0018\,N$

$F_y=72.1698\,N$

Explanation:

Given:

angle of kicking from the horizon, $\theta= 30^{\circ}$

velocity of the ball after being kicked, $v=18 m.s^{-1}$

mass of the ball, $m=0.425\, kg$

time of application of force, $t=5.3\times 10^{-2}\,s$

We know, since body is starting from the rest

$\Delta p=m.v$ .....................(1)

$\Delta p=0.425\times 18$

$\Delta p=7.65 \,kg.m.s^{-1}$

Now the components:

$\Delta p_x= 7.65\times cos 30^{\circ}$

$\Delta p_x= 6.6251 \,kg.m.s^{-1}$

similarly

$\Delta p_y= 7.65\times sin 30^{\circ}$

$\Delta p_y= 3.825 \,kg.m.s^{-1}$

also, impulse

$I=F\times t$ .........................(2)

where F is the force applied for t time.

Then from eq. (1) & (2)

$F\times t=m.v$

$F\times 5.3\times 10^{-2}= 7.65$

$F=144.3396\,N$

Now, the components

$F_x=144.3396\times cos 30^{\circ}$

$F_x=125.0018\,N$

&

$F_y=144.3396\times sin 30^{\circ}$

$F_y=72.1698\,N$

6 0

3 years ago

Plants need to....................synthesize

koban [17]

Answer:

nitrogen

Explanation:

because I also had this in exam and I was correct

5 0

3 years ago

An object is placed in front of a convex mirror with a radius of curvature of magnitude 10 cm. The mirror produces an image that

jek_recluse [69]

Answer:

u = - 20 cm

m = $\frac{1}{5}$

Given:

Radius of curvature, R = 10 cm

image distance, v = 4 cm

Solution:

Focal length of the convex mirror, f:

f = $\frac{R}{2} = \frac{10}{2} = 5 cm$

Using Lens' maker formula:

$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$

Substitute the given values in the above formula:

$\frac{1}{5} = \frac{1}{u} + \frac{1}{4}$

$\frac{1}{u} = \frac{1}{5} - \frac{1}{4}$

u = - 20 cm

where

u = object distance

Now, magnification is the ratio of image distance to the object distance:

magnification, m = $\frac{|v|}{|u|}$

magnification, m = $\frac{|4|}{|-20|}$

m = $\frac{4}{20}$

m = $\frac{1}{5}$

4 0

3 years ago

When you are noting the results of your decision, you will ?

Nutka1998 [239]

Judge a source's reliability

5 0

3 years ago

Read 2 more answers