All categories

1 year ago

Which event in a solar cell uses the photoelectric effect?

Physics

2 answers:

monitta1 year ago

5 0

Answer: Light shone on metal expulses electrons from its surface.

musickatia [10]1 year ago

4 0

Answer:

Light shone on metal expulses electrons from its surface. This phenomenon is the photoelectric effect, and the electrons are called photoelectrons. Experiments indicate that by increasing light frequency, the kinetic energy of the photoelectrons increases, and by intensifying the light, the current increases

You might be interested in

If i apply 280 n of force to a 40kg object, what will it's acceleration be?

nika2105 [10]

Answer:

f=ma

f=280N

m=40kg

a=?

280=40a

a=280/40

a=70N/kg

6 0

2 years ago

A force of 16 N to the west is applied to each object below. Which object will

Montano1993 [528]

41kg object that is moving east at 5 m s

7 0

2 years ago

Read 2 more answers

Which of the following remains constant during the motion of a projectile fired from a planet?

trapecia [35]

Answer:

C. Horizontal component of velocity

Explanation:

Object in motion stays in motion,

nothing works against its motion in the horizontal direction, unlike in the vertical direction, gravity pulls object down.

8 0

2 years ago

Read 2 more answers

Sophie says that geologic maps do not matter because she gets no benefits from them. Why is Sophie wrong? a. She can see the ran

astra-53 [7]

The Answer is Option C
I think...
Sorry If i am wrong...

4 0

2 years ago

Read 2 more answers

The design speed of a multilane highway is 60 mi/hr. What is the minimum stopping sight distance that should be provided on the

kicyunya [14]

Answer:

Part a: When the road is level, the minimum stopping sight distance is 563.36 ft.

Part b: When the road has a maximum grade of 4%, the minimum stopping sight distance is 528.19 ft.

Explanation:

Part a

When Road is Level

The stopping sight distance is given as

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}$

Here

SSD is the stopping sight distance which is to be calculated.
u is the speed which is given as 60 mi/hr
t is the perception-reaction time given as 2.5 sec.
a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
G is the grade of the road, which is this case is 0 as the road is level

Substituting values

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0)}\\SSD=220.5 +342.86 ft\\SSD=563.36 ft$

So the minimum stopping sight distance is 563.36 ft.

Part b

When Road has a maximum grade of 4%

The stopping sight distance is given as

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}$

Here

SSD is the stopping sight distance which is to be calculated.
u is the speed which is given as 60 mi/hr
t is the perception-reaction time given as 2.5 sec.
a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
G is the grade of the road, which is given as 4% now this can be either downgrade or upgrade

For upgrade of 4%, Substituting values

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35+0.04)}\\SSD=220.5 +307.69 ft\\SSD=528.19 ft$

<em>So the minimum stopping sight distance for a road with 4% upgrade is 528.19 ft.</em>

For downgrade of 4%, Substituting values

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0.04)}\\SSD=220.5 +387.09 ft\\SSD=607.59ft$

<em>So the minimum stopping sight distance for a road with 4% downgrade is 607.59 ft.</em>

As the minimum distance is required for the 4% grade road, so the solution is 528.19 ft.

3 0

2 years ago