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barxatty [35]
3 years ago
12

In a photoelectric experiment, you shine light onto an electrode and record a current of 25 μA. When you apply +500 mV to the el

ectrode, the current drops to 19 μA. What is the stopping potential magnitude in V?
Physics
1 answer:
kkurt [141]3 years ago
7 0

Answer:

2.083 V.

Explanation:

Stopping potential is the potential that is required to stop the current to zero . This potential is applied externally to oppose the potential created by the photoelectric effect . It gives the measure the photoelectric potential being generated .

Here current drops to 25 μA to 19 μA by a potential of 500mV

Change in current

= 25 - 19 = 6 μA

Voltage requirement for unit reduction in current

= 500 / 6 μA

To reduce current 0f 25 μA

requirement of V = (500 / 6 )  x 25 =   2083.33 mV = 2.083 V.

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Identify impacts of air pollution
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Answer:

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3 years ago
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A car is cruising at a steady speed of 35 mph. Suddenly, a cuddly puppy runs out into the road. The driver takes 1.7 seconds to
Schach [20]

Answer:

The distance traveled is 0.037 mi

Explanation:

The equation for the position and velocity of an accelerated object is:

x = x0 + v0 * t + 1/2 * a * t²

v = v0 + a * t

where

x = position at time t

x0 = initial position

t = time

a = acceleration

v0 = initial velocity

If the velocity is constant, then a = 0 and the position will be:

x = x0 + v * t where "v" is the velocity

First, let´s find the distance traveled until the driver push the brake:

The speed is constant. Then:

x = x0 + v * t (considering the origin of the reference system to be located at the point at which the driver sees the puppy, x0 = 0)

x = 35 mi/h (1 h / 3600 s) * 1.7 s = 0.017 mi

Then, the drivers moves with constant acceleration until the car stops (v = 0)

From the equation for velocity:

v = v0 + a * t

Since v = 0, we can obtain the acceleration of the car until it stops. With that acceleration, we can calculate how much distance the car moves before it stops.

0 = v0 + a * t

-v0 / t = a

-35 mi/h (1 h / 3600s) / 4.0 s = a

a = -2.4 x 10⁻³ mi/s²

The distance traveled will be:

x = x0 + v0 * t + 1/2 * a * t²

Now x0 will be the distance traveled before the driver slows down.

x = 0.017 mi + 35 mi/h (1 h / 3600s) * 4 s + 1/2 * ( -2.4 x 10⁻³ mi/s²) * (4s)²

x = 0.037 mi

6 0
4 years ago
An airplane flies in a horizontal circle of radius 500 m at a speed of 150 m/s. If the radius were changed to 1000 m, but the sp
laila [671]

Answer:

The centripetal acceleration changed by a factor of 0.5

Explanation:

Given;

first radius of the horizontal circle, r₁ = 500 m

speed of the airplane, v = 150 m/s

second radius of the airplane, r₂ = 1000 m

Centripetal acceleration is given as;

a = \frac{v^2}{r}

At constant speed, we will have;

v^2 =ar\\\\v = \sqrt{ar}\\\\at \ constant\ v;\\\sqrt{a_1r_1} = \sqrt{a_2r_2}\\\\a_1r_1 = a_2r_2\\\\a_2 = \frac{a_1r_1}{r_2} \\\\a_2 = \frac{a_1*500}{1000}\\\\a_2 = \frac{a_1}{2} \\\\a_2 = \frac{1}{2} a_1

a₂ = 0.5a₁

Therefore, the centripetal acceleration changed by a factor of 0.5

7 0
4 years ago
(a) Calculate the magnitude of the gravitational force exerted by Mars on a 80 kg human standing on the surface of Mars. (The ma
andrew11 [14]

Answer:

a) F=1.044\times 10^9\ N

b)F'=1.044\times 10^9\ N

c) F_p=1.0672\times10^{-7}\ N

d) Treat the humans as though they were points or uniform-density spheres.

Explanation:

Given:

  • mass of Mars, M=6.4\times 10^{23}\ kg
  • radius of the Mars, r=3.4\times 10^{6}\ m
  • mass of human, m=80\ kg

a)

Gravitation force exerted by the Mars on the human body:

F=G.\frac{M.m}{r^2}

where:

G=6.67 \times 10^{-11}\ m^3.kg^{-1}.s^{-2} = gravitational constant

F=6.67\times10^{-11}\times \frac{6.4\times 10^{23}\times 80}{(3.4\times 10^{6})^2}

F=1.044\times 10^9\ N

b)

The magnitude of the gravitational force exerted by the human on Mars is equal to the force by the Mars on human.

F'=F

F'=1.044\times 10^9\ N

c)

When a similar person of the same mass is standing at a distance of 4 meters:

F_p=6.67\times10^{-11}\times \frac{80\times 80}{4}

F_p=1.0672\times10^{-7}\ N

d)

The gravitational constant is a universal value and it remains constant in the Universe and does not depends on the size of the mass.

  • Yes, we have to treat Mars as spherically symmetric so that its center of mass is at its geometric center.
  • Yes, we also have to ignore the effect of sun, but as asked in the question we have to calculate the gravitational force only due to one body on another specific body which does not brings sun into picture of the consideration.
4 0
3 years ago
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