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3 years ago

7

a surface recieving sound is moved from it original position to a position three times farther away from the source of the sound

. The intensity of the recieved sound thus becomes

1 answer:

Delicious77 [7]3 years ago

7 0

Sound intensity = 1/(r^2)

That is Sound intensity is indirectly proportional to the distance. Therefore, sound becomes 9 times less intense.

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A child sits on a merry-go-round, 1.5 meters from the center. The merry-go-round is turning at a constant rate, and the child is

exis [7]

Answer:

5 .07 s .

Explanation:

The child will move on a circle of radius r

r = 1.5 m

Let the velocity of rotation = v

radial acceleration = v² / r

v² / r = 2.3

v² = 2.3 r = 2.3 x 1.5

= 3.45

v = 1.857 m /s

Time of revolution = 2π r / v

= 2 x 3.14 x 1.5 / 1.857

= 5 .07 s .

5 0

3 years ago

What is the escape velocity from a planet with a mass that is 10.8% Earths mass and a size that is 53% Earths radial size?

DENIUS [597]

The escape velocity from the planet is 5052 m/s

Explanation:

The formula to calculate the escape velocity from a planet is:

$v=\sqrt{\frac{2GM}{R}}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

M is the mass of the planet

R is the radius of the planet

For Earth,

$M_E=5.98\cdot 10^{24} kg$ is the mass

$R_E=6.37\cdot 10^6 m$ is the radius

Here we have a planet that:

- its mass is 10.8% of that of Earth, so

$M=0.108M_E$

- its radius is 53% of that of Earth, so

$R=0.53 R_E$

Substituting everything into the first equation, we find the escape velocity from this planet:

$v=\sqrt{\frac{2G(0.108M_E)}{0.53R_E}}=\sqrt{\frac{2(6.67\cdot 10^{-11})(0.108)(5.98\cdot 10^{24})}{(0.53)(6.37\cdot 10^6)}}=5052 m/s$

#LearnwithBrainly

4 0

3 years ago

A cylindrical fishing reel has a mass of 0.7 kg and a radius of 5.24 cm. A friction clutch in the reel exerts a restraining torq

Korolek [52]

Answer:

Explanation:

moment of inertia of reel = 1/2 mr² , m is mass and r is radius of the reel

= .5 x .7 x .0524²

= 96.1 x 10⁻⁵

If force required be F

F x .0524 - 1.88 is net torque acting on the reel

F x .0524 - 1.88 = moment of inertia x angular acceleration.

F x .0524 - 1.88 = 96.1 x 10⁻⁵ x 65.7

F x .0524 - 1.88 = .063

F x .0524 = .063 + 1.88

F = 37.08 N

7 0

3 years ago

A) A real object 4.0 cm high stands 30.0 cm in front of a converging lens of focal length 23 cm. Find the image distance, the im

vfiekz [6]

Explanation:

Given that,

Height of object = 4.0 cm

Distance of the object u= -30.0 cm

Focal length = 23 cm

We need to calculate the image distance

Using lens's formula

$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

Where, u = object distance

v = image distance

f = focal length

Put the value into the formula

$\dfrac{1}{v}-\dfrac{1}{-30}=\dfrac{1}{23}$

$\dfrac{1}{v}=\dfrac{1}{23}-\dfrac{1}{30}$

$\dfrac{1}{v}=\dfrac{7}{690}$

$v=98.57\ cm$

We need to calculate the height of the image

Using formula of height

$\dfrac{h'}{h}=\dfrac{-v}{u}$

Where, h' = height of image

h = height of object

$\dfrac{h'}{4}=\dfrac{-98.57}{-30}$

$h'=\dfrac{98.57\times4}{30}$

$h'=13.14\ cm$

The image is real and inverted.

(b). Now, object distance u = 13.0 cm

We need to calculate the image distance

Using lens's formula

$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

Put the value into the formula

$\dfrac{1}{v}=\dfrac{1}{23}-\dfrac{1}{13.0}$

$\dfrac{1}{v}=-\dfrac{10}{299}$

$v=-29.9\ cm$

We need to calculate the height of the image

$\dfrac{h'}{4}=\dfrac{-(-29.9)}{-13.0}$

$h=-\dfrac{29.9\times4}{13.0}$

$h=-9.2\ cm$

The image is virtual and erect.

Hence, This is required solution.

6 0

4 years ago

A block of cheese is pulled on by a string and slides to the right along a rough surface.

Ulleksa [173]

Answer:

Explanation:

KHAN ACADEMY

3 0

3 years ago