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3 years ago

Is loudness affected by change in frequency? Explain I will mark you the brainliest

Physics

1 answer:

zmey [24]3 years ago

4 0

Answer:

Explanation:

To increase or decrease loudness, you have to change the amplitude

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CAN SOMEONE PLEASE TELL ME WHAT THIS WHEEL IS CALLED.WILL GIVE BRAINLIEST

kolbaska11 [484]

Answer:

its a pie chart

Explanation:

.........................

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3 years ago

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One major difference between plant and animal cells is that plant cells are the only ones with ?

sineoko [7]

Plant cells have a cell wall.

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2 years ago

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Find the equivalent resistance of this

frosja888 [35]

Answer:

Re=160ohm

Explanation:

Step#1

Rt=R1+R2 ( because both are in series)

Rt=(100+220 ) ohm

Rt=320 ohm

Step#2

Rt and R3 are parallel so,

Re= (Rt× R3) ÷ (Rt+R3)

Re= (320×320)÷( 320+320)

Re = 102,400÷ 640

Re=160ohm

7 0

3 years ago

A "moving sidewalk" in an airport terminal moves at 1.0 m/s and is 35.0 m long. If a woman steps on at one end and walks at 1.5

pishuonlain [190]

Answer:

a.14 s

b.70 s

Explanation:

a.Let the sidewalk moving in positive x- direction.

Speed of sidewalk relative to ground= $v_s=1m/s$

Speed of women relative to sidewalk=v=1.5m/s

The speed of women relative to the ground

$v_w=v_s+v=1+1.5=2.5m/s$

Distance=35 m

Time= $\frac{distance}{speed}$

Using the formula

Time taken by women to reach the opposite end if she walks in the same direction the sidewalk is moving= $\frac{35}{v_w}=\frac{35}{2.5}=14s$

b.If she gets on at the end opposite the end in part (a)

Then, we take displacement negative.

Speed of sidewalk relative to ground= $v_s=1m/s$

Speed of women relative to sidewalk=v=-1.5 m/s

The speed of women relative to the ground= $v_w=v_s+v=1-1.5=-0.5m/s$

Time= $\frac{-35}{-0.5}=70 s$

Hence, the women takes 70 s to reach the opposite end if she walks in the opposite direction the sidewalk is moving.

3 0

3 years ago

A ball is dropped from rest from a height h above the ground. another ball is thrown vertically upwards from the ground at the i

Darya [45]

The position of the first ball is

$y_1=h-\dfrac g2t^2$

while the position of the second ball, thrown with initial velocity $v$ , is

$y_2=vt-\dfrac g2t^2$

The time it takes for the first ball to reach the halfway point satisfies

$\dfrac h2=h-\dfrac g2t^2$

$\implies\dfrac h2=\dfrac g2t^2$

$\implies t=\sqrt{\dfrac hg}$

We want the second ball to reach the same height at the same time, so that

$\dfrac h2=v\sqrt{\dfrac hg}-\dfrac g2\left(\sqrt{\dfrac hg}\right)^2$

$\implies h=2v\sqrt{\dfrac hg}-g\left(\dfrac hg\right)$

$\implies h=v\sqrt{\dfrac hg}$

$\implies v=\sqrt{hg}$

8 0

3 years ago