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2 years ago

11

If the amplitude of a sound wave is increased by a factor of four,how does the energy carry by the sound wave in each time inter

val change?

1 answer:

OlgaM077 [116]2 years ago

6 0

Answer:

The energy will be increased by a factor 16

Explanation:

Mathematically, we have it that the energy carried by the sound wave is directly proportional to the square of the amplitude

So we have it that;

E = k * A^2

where E is the wave energy

K is the constant value

A is the amplitude value

So;

If E1 = k * (A1)^2

E2 = k * (A2)^2

But A2 = 4A1

E2 = k * (4A1)^2

E2 = k * 16(A1)^2

Divide E2 by E1

E2/E1 = k/k * 16(A1)^2/(A1)^2

E2/E1 = 16

E2 = 16 * E1

so simply, the energy will be increased by a factor of 16

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What occurs when the moon blocks the view of the sun

Sever21 [200]

Answer:

solar eclipse

Explanation:

because at that time, the moon completely covers the path and casts its shadow on earth because it is present between sun and earth's path. so, solar eclipse occurs.

hope it helps :)

8 0

3 years ago

A golfer is on the edge of a 12.5 m bluff overlooking the 18th hole which is located 60 m from the base of the bluff. She launch

Lina20 [59]

Answer:

The ball impact velocity i.e(velocity right before landing) is 6.359 m/s

Explanation:

This problem is related to parabolic motion and can be solved by the following equations:

$x=V_{o}cos \theta t$ ----------------------(1)

$y=y_{o}+V_{o} sin \theta t - \frac{1}{2}gt^{2}$ ---------(2)

$V=V_{o}-gt$ ----------------------- (3)

Where:

x = m is the horizontal distance travelled by the golf ball

$V_{o}$ is the golf ball's initial velocity

$\theta=0\°$ is the angle (it was a horizontal shot)

t is the time

y is the final height of the ball

$y_{o}$ is the initial height of the ball

g is the acceleration due gravity

V is the final velocity of the ball

Step 1: finding t

Let use the equation(2)

$t=\sqrt{\frac{2 y_{o}}{g}}$

$t=\sqrt{\frac{2 (12.5 m)}{9.8 m/s^{2}}}$

$t=1.597$ s

Substituting (6) in (1):

$67.1 =V_{o} cos(0\°) 1.597$ -------------------(4)

Step 2: Finding $V_{o}$ :

From equation(4)

$67.1 =V_{o}(1) 1.597$

$V_0 = \frac{6.71}{1.597}$

$V_{o}=42.01$ m/s (8)

Substituting $V_{o}$ in (3):

$V=42.01 -(9.8)(1.597)$

v =42 .01 - 15.3566

V=26.359 m/s

5 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

Objects in space that are moving at a constant velocity in a straight line ___________.

nikdorinn [45]

The best answer is A) <span>keep moving at a constant velocity until some forces act on them

As the man you're probably tired of hearing about said:

"Every object persists in its state of rest or in uniform motion in a straight line unless a new force acts upon it"
This is Isaac Newton's 1st law of motion, or the law of inertia.

Put more simply, objects in motion tend to stay in motion, and tend the maintain the same velocity (direction and speed) and objects at rest tend to stay at rest. </span>

6 0

3 years ago

Read 2 more answers

How do you find the total magnification of a microscope?

Alla [95]

In order to find total magnification of a microscope, you need to multiply the power of eyepiece and objective lens.

Hope this helps!

6 0

3 years ago