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3 years ago

Which event in a solar cell uses the photoelectric effect?

Physics

2 answers:

Helen [10]3 years ago

8 0

Explanation:

Photovoltaic energy is a clean, renewable source of energy that uses solar radiation to produce electricity. It is based on the photoelectric effect—the emission of electrons when electromagnetic radiation (i.e. light) hits a material.

kirza4 [7]3 years ago

4 0

Answer:

electromagnetic waves strike a semiconductor , removing some of the electrons ,which form a current

Explanation:

aPe~X!

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An object with a mass M = 250 g is on a plane inclined at 30º above the horizontal and is attached by a string to a mass m = 150

malfutka [58]

Answer:

0.495 m/s

Explanation:

$T$ = tension force in the string connecting the two objects

$M$ = Mass of the object on inclined plane = 250 g = 0.250 kg

$m$ = Mass of the hanging object = 150 g = 0.150 kg

$a$ = acceleration of each object

From the force diagram, force equation for the motion of the object on the inclined plane is given as

$T - Mg Sin30 = Ma\\T = Mg Sin30 + Ma$

From the force diagram, force equation for the motion of the hanging object on the inclined plane is given as

$mg - T = ma\\T = mg - ma$

Using the above two equations

$Mg Sin30 + Ma = mg - ma$

$(0.250)(9.8) Sin30 + (0.250) a = (0.150) (9.8) - (0.150)a$

$a = 0.6125 ms^{-2}$

$h$ = height dropped by the hanging object = 10 cm = 0.10 m

$v$ = Speed gained by the object

Speed gained by the object can be given as

$v = sqrt(2ah)\\v = sqrt(2(0.6125)(0.20))\\v = 0.495 ms^{-1}$

5 0

3 years ago

A stone is thrown from the top of a building with an initial velocity of 20 m/s downward. The top of the building is 60 m above

erastova [34]

Answer:

t = 2 s

Explanation:

In order to find the time taken by the stone to fall from the top of the building to the ground we can use 2nd equation of motion. 2nd equation of motion is as follows:

s = Vit + (0.5)gt²

where,

t = time = ?

Vi = Initial Velocity = 20 m/s

s = height of building = 60 m

g = 9.8 m/s²

Therefore,

60 m = (20 m/s)t + (0.5)(9.8 m/s²)t²

4.9t² + 20t - 60 = 0

solving this quadratic equation we get:

t = -6.1 s (OR) t = 2 s

Since, the time cannot be negative in magnitude.

Therefore,

6 0

3 years ago

Rudy plans to conduct an experiment using three rosebushes of the same variety and size. His hypothesis is that the plant

Gala2k [10]

Answer: B) He has too many independent variables.

Explanation: Nothing in the experiment is a constant or being measured.

5 0

3 years ago

Which of the fundamental forces explains the structure of atoms and molecules? gravitational force

Gnoma [55]

Answer:

Strong nuclear force

Explanation:

Strong nuclear force (which is the strongest of the forces of the universe) is responsible for the attractive force between quarks to form nucleons (protons and neutrons). It is the reason why the protons (that are positive in charge) do not fly apart due to electromagnetic repulsion in the nucleus of an atom.

5 0

3 years ago

Read 2 more answers

In class we calculated the range of a projectile launched on flat ground. Consider instead, a projectile is launched down-slope

zysi [14]

Answer:

With an initial speed of 10m/s at an angle 30° below the horizontal, and a height of 8m, the projectile travels 7.49m horizontally before it lands.

Explanation:

Since the horizontal motion is independent from the vertical motion, we can consider them separated. The horizonal motion has a constant speed, because there is no external forces in the horizontal axis. On the other hand, the vertical motion actually is affected by the gravitational force, so the projectile will be accelerated down with a magnitude $g$ .

If we have the initial velocity $v_o$ and its angle $\theta$ , we can obtain the vertical component of the velocity $v_{oy}$ using trigonometry:

$v_{oy}=v_osin\theta$

Therefore, if we know the height at which the projectile was launched, we can obtain the final velocity using the formula:

$v_{fy}^{2} =v_{oy}^{2}+2gy\\\\ v_{fy}=\sqrt{v_{oy}^{2}+2gy }$

Next, we compute the time the projectile lasts to reach the ground using the definition of acceleration:

$g=\frac{v_{fy}-v_{oy}}{\Delta t} \\\\\Delta t= \frac{v_{fy}-v_{oy}}{g}=\frac{\sqrt{v_{oy}^{2}+2gy} -v_{oy}}{g}$

Finally, from the equation of horizontal motion with constant speed, we have that:

$x=v_{ox}\Delta t= v_{ox}\frac{\sqrt{v_{oy}^{2}+2gy} -v_{oy}}{g}$

For example, if the projectile is launched at an angle 30° below the horizontal with an initial speed of 10m/s and a height 8m, we compute:

$v_{ox}=10\frac{m}{s} cos30=8.66\frac{m}{s}\\v_{oy}=10\frac{m}{s} sin30=5\frac{m}{s}\\\\x=8.66\frac{m}{s} \frac{\sqrt{(5\frac{m}{s}) ^{2}+2(9.8\frac{m}{s^{2}})8m}-5\frac{m}{s} }{9.8\frac{m}{s^{2} } } =7.49m$

In words, the projectile travels 7.49m horizontally before it lands.

8 0

4 years ago