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2 years ago

What is the purpose of the antireflective coating on the glass layer of a solar

Physics

2 answers:

bonufazy [111]2 years ago

7 0

Answer:

The answer is D if you have something that is reflective how are you going to absorb any energy.

Explanation:

A P E X

gizmo_the_mogwai [7]2 years ago

3 0

Answer:

I think d. is the answer I'm not conform

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Hideki is having trouble completing long workouts. What does he need to improve?

KonstantinChe [14]

He needs to improve his endurance time.

3 0

3 years ago

How far will an object move in 6 seconds if its average speed during that time is 45 m/s

babunello [35]

Answer:

270 Meters

Explanation:

45×6=270

3 0

3 years ago

Read 2 more answers

A ball is thrown with an initial speed vi at an angle i with the horizontal. The horizontal range of the ball is R, and the ball

adell [148]

Answer:

Part a)

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

$v_x = \frac{\sqrt{3Rg}}{2}$

Part c)

$v_y = \sqrt{Rg/3}$

Part d)

$v = \frac{1}{2}\sqrt{13Rg}$

Part e)

$\theta_i = 33.7 degree$

Part f)

$H = \frac{13R}{8}$

Part g)

$X = \frac{13R}{4}$

Explanation:

Initial speed of the launch is given as

initial speed = $v_i$

angle = $\theta_i$ degree

Now the two components of the velocity

$v_x = v_i cos\theta_i$

similarly we have

$v_y = v_i sin\theta_i$

Part a)

Now we know that horizontal range is given as

$R = \frac{v_i^2 (2sin\theta_icos\theta_i)}{g}$

maximum height is given as

$H = \frac{R}{6} = \frac{v_i^2 sin^2\theta_i}{2g}$

so we have

$v_i sin\theta = \sqrt{Rg/3}$

time of flight is given as

$T = \frac{2v_isin\theta_i}{g}$

$T = \frac{2\sqrt{Rg/3}}{g}$

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

Now the speed of the ball in x direction is always constant

so at the peak of its path the speed of the ball is given as

$R = v_x T$

$R = v_x 2\sqrt{\frac{R}{3g}}$

$v_x = \frac{\sqrt{3Rg}}{2}$

Part c)

Initial vertical velocity is given as

$v_y = v_i sin\theta_i$

$v_i sin\theta = \sqrt{Rg/3}$

Part d)

Initial speed is given as

$v = \sqrt{v_x^2 + v_y^2}$

so we will have

$v = \sqrt{Rg/3 + 3Rg/4}$

$v = \frac{1}{2}\sqrt{13Rg}$

Part e)

Angle of projection is given as

$tan\theta_i = \frac{v_y}{v_x}$

$tan\theta_i = \frac{\sqrt{Rg/3}}{\sqrt{3Rg}/2}$

$\theta_i = 33.7 degree$

Part f)

If we throw at same speed so that it reach maximum height

then the height will be given as

$H = \frac{v^2}{2g}$

$H = \frac{13R}{8}$

Part g)

For maximum range the angle should be 45 degree

so maximum range is

$X = \frac{v^2}{g}$

$X = \frac{13R}{4}$

3 0

3 years ago

Scientific way of thinking

noname [10]

Answer:

huh,? can you explain the question more please

7 0

3 years ago

Read 2 more answers

State characteristics of the images formed by pin - hole camera

Leokris [45]

Answer:

Real image

Explanation:

The picture is real, but it is reversed and tiny. An picture generated by a pinhole camera has certain features. As compared item, the image created by a pinhole camera is normally pretty small and looks reversed both on the vertically and horizontally axis.

8 0

3 years ago