Refer to the diagram. Which answer correctly labels the water between the equator and the North Pole?

"Younger teens often brag about a lack of sleep, but that usually stops when they get older. Why do you think that is?"

svet-max [94.6K]

as we grow up we kinda experience some age staging so a teen got a lotta stuff going on they wanna be social active they be working on their popularity. so i guess thats it

3 0

3 years ago

(a) Neil A. Armstrong was the first person to walk on the moon. The distance between the earth and the moon is . Find the time i

a_sh-v [17]

Answer:

a) It took 1.28 seconds to Neil Armstrong's voice to reach the Earth via radio waves.

b) The minimum time that will be required for a message from Mars to reach the Earth via radio waves is 192 seconds.

Explanation:

The electromagnetic spectrum is the distribution of radiation due to the different frequencies at which it radiates and its different intensitie. That radiation is formed by electromagnetic waves, which are transverse waves formed by an electric field and a magnetic field perpendicular to it.

The distribution of the radiation in the electromagnetic spectrum can also be given in wavelengths, but it is more frequent to work with it at frequencies:

Gamma rays

X-rays

Ultraviolet rays

Visible region

Infrared

Microwave

Radio waves.

Any radiation that belongs to electromagnetic spectrum has a speed in vacuum of $3x10^{8}m/s$ .

a) Find the time it took for his voice to reach the Earth via radio waves.

To know the time that took for Neil Armstrong's voice to reach the Earth via radio waves, the following equation can be used:

$c = \frac{d}{t}$ (1)

Where v is the speed of light, d is the distance and t is the time.

Notice that t can be isolated from equation 1.

$t = \frac{d}{c}$ (2)

The distance from the Earth to the Moon is $3.85x10^{8} m$ , therefore.

$t = \frac{3.85x10^{8} m}{3x10^{8}m/s}$

$t = 1.28s$

Hence, it took 1.28 seconds to Neil Armstrong's voice to reach the Earth via radio waves.

b) Determine the minimum time that will be required for a message from Mars to reach the Earth via radio waves.

The distance from the Earth to the Mars at its closest approach is $5.76x10^{10}m$ , therefore.

$t = \frac{5.76x10^{10}m}{3x10^{8}m/s}$

$t = 192s$

Hence, the minimum time that will be required for a message from Mars to reach the Earth via radio waves is 192 seconds.

3 0

3 years ago

On a 10 kg cart (shown below), the cart is brought up to speed with 50N of force for 7m, horizontally. At this point (A), the ca

pav-90 [236]

Answer: Letter B! Is your answer

3 0

3 years ago

A thin coil has 17 rectangular turns of wire. When a current of 4 A runs through the coil, there is a total flux of 5 ✕ 10−3 T ·

Alex73 [517]

Answer:

Inductance, L = 0.0212 Henries

Explanation:

It is given that,

Number of turns, N = 17

Current through the coil, I = 4 A

The total flux enclosed by the one turn of the coil, $\phi=5\times 10^{-3}\ Tm^2$

The relation between the self inductance and the magnetic flux is given by :

$L=\dfrac{N\phi}{I}$

$L=\dfrac{17\times 5\times 10^{-3}}{4}$

L = 0.0212 Henries

So, the inductance of the coil is 0.0212 Henries. Hence, this is the required solution.

7 0

3 years ago

A star is moving away from an observer at 1% of the speed of light. At what wavelength would the observer find an emission line

Ivan

Answer:

λ = 5940 Angstroms

Explanation:

This is an exercise of the relativistic Doppler effect

f’= f √((1- v / c) / (1 + v / c))

Where the speed in between the strr and the observer is positive if they move away

Let's use the relationship

c = λ f

f = c /λ

We replace

c /λ’ = c /λ √ ((1- v / c) / (1 + v / c))

λ = λ’ √ ((1- v / c) / (1 + v / c))

Let's calculate

v = 0.01 c

v = 0.01 3 10⁸

v= 3 10⁶ m / s

λ = 6000 √ [(1- 3 10⁶/3 10⁸) / (1+ 3 10⁶/3 10⁸)]

λ = 6000 √ [0.99 / 1.01]

λ = 5940 Angstroms

6 0

3 years ago