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3 years ago

A wave has a velocity of 24 m/s and a period of 3.0 s. Calculate the wavelength of the wave.

Physics

1 answer:

Katyanochek1 [597]3 years ago

5 0

Velocity (unit:m/s) of the wave is given with the formula:

v=f∧,

where f is the frequency which tells us how many waves are passing a point per second (unit: Hz) and ∧ is the wavelength, which tells us the length of those waves in metres (unit:m)

f=1/T , where T is the period of the wave.

In our case: f=1/3

∧=v/f=24m/s/1/3=24*3=72m

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3 years ago

A small block of mass 20.0 grams is moving to the right on a horizontal frictionless surface with a speed of 0.68 m/s. The block

Usimov [2.4K]

Answer:

a) $v'=-0.227\ m.s^{-1}$

b) $v=1.36\ m.s^{-1}$

Explanation:

Given:

mass of the lighter block, $m'=0.02\kg$

velocity of the lighter block, $u'=0.68\ m.s^{-1}$

mass of the heavier block, $m=0.04\ kg$

velocity of the heavier block, $u=0\ m.s^{-1}$

Using conservation of linear momentum:

$m'.u'+m.u=m'.v'+m.v$

where:

$v'=$ final velocity of the lighter block

$v=$ final velocity of the heavier block

$m'.u'=m'.v'+m.v$

$m'(u'-v')=m.v$ ........................(1)

Since kinetic energy is conserved in elastic collision:

$\frac{1}{2}m'.u'^2=\frac{1}{2}m'.v'^2+\frac{1}{2}m.v^2$

$m'(u'^2-v'^2)=m.v^2$

$m'(u'-v')(u'+v')= m.v^2$

divide the above equation by eq. (1)

$v=u'+v'$ .............................(2)

now we substitute the value of v from eq. (2) in eq. (1)

$m'(u'-v')=m(u'+v')$

$\frac{m'+m}{m'-m} =\frac{u'}{v'}$

$\frac{0.02+0.04}{0.02-0.04} =\frac{0.68}{v'}$

$v'=-0.227\ m.s^{-1}$ (negative sign denotes that the direction is towards left)

now we substitute the value of v' from eq. (2) in eq. (1)

$m'(u'-v+u')=m.v$

$2m'.u'=(m-m')v$

$2\times 0.02\times 0.68=(0.04-0.02)\times v$

$v=1.36\ m.s^{-1}$

6 0

3 years ago

A coin is dropped in a 15.0 m deep well.

labwork [276]

Answer:

t = 1.75

t = 0.04

Explanation:

For part 1 we want to use a kenamatic equation with constant acceleration:

X = 1/2*a*t^2

isolate time

t = sqrt(2X / a)

Plugin known variables. Acceleration is the force of gravity which is 9.8 m/s^2

t = sqrt(2*15m / 9.8m/s^2)

t = 1.75 s

The speed of sound travels at a constant speed therefore we don't need acceleration and can use the equation:

v = d / t

isolate time

t = d / v

plug in known variables

t = 15m / 340m/s

t = 0.04 s

7 0

2 years ago

A student observes that it is hard to hear music underwater in a pool. They state that the sound is always muffled. They

s344n2d4d5 [400]

Answer:

FALSE

Explanation:

The answer is false.

The speed of the sound in water is faster when compared to the speed of sound in air. This is because, the particles in air is loosely packed and are far from each other as compared to water or liquid.

The water particles are close to each other than air particles, so water particles are able to transmit the vibrations of the sound faster than the air particles.

Therefore sound waves travels faster in water than in air.

5 0

3 years ago

Please help!! This is a final and I need a good grade

Ket [755]

Diagram 4 is the correct answer.

6 0

3 years ago