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1 year ago

The speed of sound in air is 340 m/s. What is the wavelength of a soundwave that has a frequency of 502 Hz?

Physics

1 answer:

Mama L [17]1 year ago

5 0

The formula for calculating velocity of wave is expressed as

v = fλ

where

v = velocity of the wave

λ = wavelength

f = frequency of the wave

From the information given,

v = 340

f = 502

Substituting these values into the formula, we have

340 = 502λ

Dividing both sides of the equation by 502,

340/502 = 502λ/502

λ = 0.68 m

The wavelength is 0.68 m

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0.017s

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3 years ago

A 4 kilogram box is at rest on a frictionless floor. A net force of 12 newtons acts on it. What is the acceleration of the box?

Rashid [163]

Answer:

Explanation:

Step one:

given

Mass m= 4kg

Force F= 12N

Required

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3 years ago

A brave but inadequate rugby player is being pushed backward by an opposing player who is exerting a force of 800 N on him. The

uysha [10]

Answer:

f = 692 N

Explanation:

given data:

f =800N

a =1.2 m s^{2}

m= 90 kg

from newton's second law

net force $F_{net} =\sum F = F_1 +F_2 +..... = ma$

therefore we have from above equation $F_{net} = F -f = ma$

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3 years ago

A ball is thrown toward a cliff of height h with a speed of 33 m/s and an angle of 60∘ above horizontal. It lands on the edge of

Veseljchak [2.6K]

Answer:

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(b). The maximum height of the ball is 41.67 m

(c). The ball's impact speed is 16.52 m/s.

Explanation:

Given that,

Speed = 33 m/s

Angle = 60°

Time = 3.0 sec

(a). We need to calculate the height of the cliff

Using equation of motion

$h=ut-\dfrac{1}{2}gt^2$

$h=(u\sin60)\times t-\dfrac{1}{2}gt^2$

Put the value into the formula

$h=33\times\sin60\times3.0-\dfrac{1}{2}\times9.8\times(3.0)^2$

$h=41.6\ m$

(b). We need to calculate the maximum height of the ball

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$h_{max}=\dfrac{(u\sin\theta)^2}{2g}$

Put the value into the formula

$h=\dfrac{(33\sin60)^2}{2\times 9.8}$

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(c). We need to calculate the vertical component of velocity of ball

Using equation of motion

$v=u-gt$

$v=u\sin\theta-gt$

Put the value into the formula

$v_{y}=33\times\sin 60-9.8\times3.0$

$v_{y}=-0.82\ m/s$

We need to calculate the horizontal component of velocity of ball

Using formula of velocity

$v_{x}=u\cos\theta$

Put the value into the formula

$v_{x}=33\times\cos60$

$v_{x}=16.5\ m/s$

We need to calculate the ball's impact speed

Using formula of velocity

$v=\sqrt{v_{x}^2+v_{y}^2}$

Put the value into the formula

$v=\sqrt{(16.5)^2+(-0.82)^2}$

$v=16.52\ m/s$

Hence, (a). The height of the cliff is 41.67 m.

(b). The maximum height of the ball is 41.67 m

(c). The ball's impact speed is 16.52 m/s.

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3 years ago

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