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3 years ago

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The sun delivers an average power of 1.499 w/m2 to the top of neptune's atmosphere. find the magnitudes of vector e max and vect

or b max for the electromagnetic waves at the top of the atmosphere.

1 answer:

sesenic [268]3 years ago

4 0

See attachment for answer:

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Humans have three types of cone cells in their eyes, which are responsible for color vision. Each type absorbs a certain part of

xxMikexx [17]

Answer:

Frequency of the light will be equal to $6.97\times 10^{1}Hz$

Explanation:

We have given wavelength of the light $\lambda =430nm=430\times 10^{-9}m$

Velocity of light is equal to $v=3\times 10^8m/sec$

We have to find the frequency of light

We know that velocity is equal to $v=\lambda f$ , here $\lambda$ is wavelength and f is frequency of light

So frequency of light will be equal to $f=\frac{v}{\lambda }=\frac{3\times 10^8}{430\times 10^{-9}}=6.97\times 10^{1}Hz$

So frequency of the light will be equal to $6.97\times 10^{1}Hz$

5 0

3 years ago

In this lab you will use a cart and track to explore various aspects of motion. You will measure and record the time it takes th

Keith_Richards [23]

It is given that by using track and cart we can record the time and the distance travelled and also the speed of the cart can be recorded. With all this data we can solve questions on the laws of motion.

Like using the first law of motion we can determine the force of gravity acting on the cart that has moved a certain distance and the velocity or the speed of card has already been registered and since time is known putting the values in formula would help us calculate the gravitational pull acting on cart.

3 0

3 years ago

Read 2 more answers

A sprinter accelerates from rest to 10.0 m/s in 1.28 s . Part A Part complete What is her acceleration in m/s2? a a = 7.81 m/s2

Mashutka [201]

Explanation:

It is given that,

Initial speed of sprinter, u = 0

Final speed of sprinter, v = 10 m/s

Time taken, t = 1.28 s

a. We need to find the acceleration of sprinter. It can be calculated using first equation of motion as :

$a=\dfrac{v-u}{t}$

$a=\dfrac{10\ m/s}{1.28\ s}$

$a=7.81\ m/s^2$

b. Final speed of the sprinter, v = 36 km/h

Time, t = 0.000355 h

Acceleration, $a=\dfrac{36}{0.000355}$

$a=101408.45\ km/h^2$

Hence, this is the required solution.

3 0

3 years ago

In which medium does light travel faster: one with a critical angle of 27.0° or one with a critical angle of 32.0°? Explain. (Fo

Eddi Din [679]

Answer:

Among those two medium, light would travel faster in the one with a reflection angle of $32^{\circ}$ (when light enters from the air.)

Explanation:

Let $v_{1}$ denote the speed of light in the first medium. Let $v_{\text{air}}$ denote the speed of light in the air. Assume that the light entered the boundary at an angle of $\theta_{1}$ to the normal and exited with an angle of $\theta_{\text{air}}$ . By Snell's Law, the sine of $\theta_{1}\!$ and $\theta_{\text{air}}\!$ would be proportional to the speed of light in the corresponding medium. In other words:

$\displaystyle \frac{v_{1}}{v_{\text{air}}} = \frac{\sin(\theta_{1})}{\sin(\theta_{\text{air}})}$ .

When light enters a boundary at the critical angle $\theta_{c}$ , total internal reflection would happen. It would appear as if the angle of refraction is now $90^{\circ}$ . (in this case, $\theta_{\text{air}} = 90^{\circ}$ .)

Substitute this value into the Snell's Law equation:

$\begin{aligned}\frac{v_{1}}{v_{\text{air}}} &= \frac{\sin(\theta_{1})}{\sin(\theta_{\text{air}})} \\ &= \frac{\sin(\theta_{c})}{\sin(90^{\circ})} \\ &= \sin(\theta_{c})\end{aligned}$ .

Rearrange to obtain an expression for the speed of light in the first medium:

$v_{1} = v_{\text{air}} \cdot \sin(\theta_{1})$ .

The speed of light in a medium (with the speed of light slower than that in the air) would be proportional to the critical angle at the boundary between this medium and the air.

For $0 < \theta < 90^{\circ}$ , $\sin(\theta)$ is monotonically increasing with respect to $\theta$ . In other words, for $\!\theta$ in that range, the value of $\sin(\theta)\!$ increases as the value of $\theta\!$ increases.

Therefore, compared to the medium in this question with $\theta_{c} = 27^{\circ}$ , the medium with the larger critical angle $\theta_{c} = 32^{\circ}$ would have a larger $\sin(\theta_{c})$ . such that light would travel faster in that medium.

4 0

3 years ago

A copper wire 20 m long and 4mm in diameter is attached to the Ceiling and a 400 N

Fiesta28 [93]

Answer:

A

Explanation:

5 0

2 years ago