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2 years ago

14

Which layer of the atom sphere contacts the surface of earth

1 answer:

Keith_Richards [23]2 years ago

4 0

The troposphere is the layer of the atmosphere which contacts the surface of the Earth. This layer is <span>0 to 12 km (0 to 7 miles).

Faith xoxo</span>

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describe the journey that light takes from the sun to your eye when you are looking at fish in a pond

Tcecarenko [31]

Rays of the light fast moves to the water and to the fish and then they are reflected

<u>Explanation:</u>

The movement of the light in the water and which to the fish taste of thence they are reflected by the upper surface of the water and reaches the eye, this results in The refraction in the lens of our eyes and it reaches the retina.
The retina consists of the photosensitive pigments that helps in the giving off the lights when the electrochemical neuronal activity reaches the brain of human beings

To know more,

this is the science that studies the physiographic units such as ..

.brainly.in/question/22161298

7 0

1 year ago

When you throw a ball, the work you do to accelerate it equals the kinetic energy the ball gains. If you do twice as much work w

aniked [119]

Answer:

No. Twice as much work will give the ball twice as much kinetic energy. But since KE is proportional to the speed squared, the speed will be $sqrt{2}$ times larger.

Explanation:

The work done on the ball is equal to the kinetic energy gained by the ball:

$W=K$

So when the work done doubles, the kinetic energy doubles as well:

$2W = 2 K$

However, the kinetic energy is given by

$K=\frac{1}{2}mv^2$

where

m is the mass of the ball

v is its speed

We see that the kinetic energy is proportional to the square of the speed, $v^2$ . We can rewrite the last equation as

$v=\sqrt{\frac{2K}{m}}$

which also means

$v=\sqrt{\frac{2W}{m}}$

If the work is doubled,

$W'=2W$

So the new speed is

$v'=\sqrt{\frac{2(2W)}{m}}=\sqrt{2}\sqrt{\frac{2W}{m}}=\sqrt{2} v$

So, the speed is $\sqrt{2}$ times larger.

5 0

3 years ago

Moving force of air flows through areas of high pressure to areas of low pressure

vazorg [7]

The answer is true i hope this helps

7 0

3 years ago

Read 2 more answers

Two loudspeakers emit sound waves along the x-axis. A listener in front of both speakers hears a maximum sound intensity when sp

yKpoI14uk [10]

Answer

given,

difference between the two consecutive maximum

λ = 0.870 - 0.540

λ = 0.33 m

speed of sound = 340 m/s

b) frequency of the sound

v = f x λ

340 = f x 0.33

$f =\dfrac{340}{0.33}$

f = 1030.3 Hz

a) phase difference

the expression of phase difference is given by

$\phi = \dfrac{2\pi}{\lambda}\ \delta$

$\delta = \Delta x - \lambda$

$\delta = 0.540 - 0.33$

$\delta = 0.21\ m$

now,

$\phi = \dfrac{2\pi}{\lambda}\ \times 0.21$

$\phi = \dfrac{2\pi}{0.33}\ \times 0.21$

$\phi = 3.99 rad$

8 0

2 years ago

The kinetic energy of a rotating body is generally written as K=12Iω2, where I is the moment of inertia. Find the moment of iner

stira [4]

Answer:

See explanation

Explanation:

We have a mass $m$ revolving around an axis with an angular speed $\omega$ , the distance from the axis is $r$ . We are given:

$\omega = 10 [rad/s]\\r=0.5 [m]\\m=13[Kg]$

and also the formula which states that the kinetic rotational energy of a body is:

$K =\frac{1}{2}I\omega^2$ .

Now we use the kinetic energy formula

$K =\frac{1}{2}mv^2$

where $v$ is the tangential velocity of the particle. Tangential velocity is related to angular velocity by:

$v=\omega r$

After replacing in the previous equation we get:

$K =\frac{1}{2}m(\omega r)^2$

now we have the following:

$K =\frac{1}{2}m(\omega r)^2 =\frac{1}{2}Iw^2$

therefore:

$mr^2=I$

then the moment of inertia will be:

$I = 13*(0.5)^2=3.25 [Kg*m^2]$

3 0

3 years ago