Please help meee please please

The _______ is the coldest layer of the atmosphere

n200080 [17]

The mesosphere is the coldest layer of the atmosphere. It is located above the stratosphere.

4 0

3 years ago

How does the direction of current flow in the coil affect the orientation of the magnetic field produced by the electromagnet

Lynna [10]

Answer:

The magnetic field produced by an electric current is always oriented perpendicular to the direction of the current flow. And.Direction of magnetic field is governed by the 'right hand thumb rule, The right hand rule states that: to determine the direction of the magnetic force on a positive moving charge, ƒ, point the thumb of the right hand in the direction of v, the fingers in the direction of B, and a perpendicular to the palm points in the direction of Force . Similar to the situation with electric field lines, the greater the number of lines (or the closer they are together) in an area the stronger the magnetic field.

8 0

3 years ago

The space shuttle is located exactly half way between the earth and the moon. Which statement is true regarding the gravitationa

sdas [7]

Answer:

The correct answer is option B)

Explanation:

Considering the given question as -

The space shuttle is located exactly half way between the earth and the moon. Which statement is true regarding the gravitational pull on the shuttle? A) The moon pulls more on the shuttle. B) The earth pulls more on the shuttle. C) Both are equal due to equal distances. D) Both are equal due to the mass of the shuttle.

We know that gravitational pull (F) between any two bodies of mass $M_{1}$ and $M_{2}$ is given by -

F = $\dfrac{GM_{1}M_{2} }{r^{2} }$ where 'r' is the distance between the two bodies.

Let ,

$M_{e}$ : Mass of the earth

$M_{m}$ : Mass of the moon

m : Mass of the satellite

$r_{e}$ : Distance of satellite from earth

$r_{m}$ : Distance of satellite from moon

Given that $r_{e}$ = $r_{m}$

Let $r_{e}$ = $r_{m}$ =r

Force on satellite by the earth is -

$F_{e}$ = $\dfrac{GM_{e}m }{r^{2} }$

Force on satellite by the moon is -

$F_{m}$ = $\dfrac{GM_{m}m }{r^{2} }$

∵ Mass of earth ( $M_{e}$ ) > Mass of moon ( $M_{m}$ )

∴ $F_{e}$ > $F_{m}$

∴ The gravitational pull of earth on satellite is more than that of the moon.

4 0

4 years ago

A sound source A and a reflecting surface B move directly toward each other. Relative to the air, the speed of source A is 28.7

aleksandrvk [35]

(a) 1440.5 Hz

The general formula for the Doppler effect is

$f'=(\frac{v+v_r}{v+v_s})f$

where

f is the original frequency

f is the apparent frequency

$v$ is the velocity of the wave

$v_r$ is the velocity of the receiver (positive if the receiver is moving towards the source, negative otherwise)

$v_s$ is the velocity of the source (positive if the source is moving away from the receiver, negative otherwise)

Here we have

f = 1110 Hz

v = 334 m/s

In the reflector frame (= on surface B), we have also

$v_s = v_A = -28.7 m/s$ (surface A is the source, which is moving towards the receiver)

$v_r = +62.2 m/s$ (surface B is the receiver, which is moving towards the source)

So, the frequency observed in the reflector frame is

$f'=(\frac{334 m/s+62.2 m/s}{334 m/s-28.7 m/s})1110 Hz=1440.5 Hz$

(b) 0.232 m

The wavelength of a wave is given by

$\lambda=\frac{v}{f}$

where

v is the speed of the wave

f is the frequency

In the reflector frame,

f = 1440.5 Hz

So the wavelength is

$\lambda=\frac{334 m/s}{1440.5 Hz}=0.232 m$

(c) 1481.2 Hz

Again, we can use the same formula

$f'=(\frac{v+v_r}{v+v_s})f$

In the source frame (= on surface A), we have

$v_s = v_B = -62.2 m/s$ (surface B is now the source, since it reflects the wave, and it is moving towards the receiver)

$v_r = +28.7 m/s$ (surface A is now the receiver, which is moving towards the source)

So, the frequency observed in the source frame is

$f'=(\frac{334 m/s+28.7 m/s}{334 m/s-62.2 m/s})1110 Hz=1481.2 Hz$

(d) 0.225 m

The wavelength of the wave is given by

$\lambda=\frac{v}{f}$

where in this case we have

v = 334 m/s

f = 1481.2 Hz is the apparent in the source frame

So the wavelength is

$\lambda=\frac{334 m/s}{1481.2 Hz}=0.225 m$

8 0

3 years ago

Which of the following situations would produce and average velocity of zero

enyata [817]

There are no situations listed. Try adda picture or more text for your question to be answered.

3 0

4 years ago