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JulsSmile [24]
2 years ago
11

Monochromatic light is incident on a metal surface, and electrons are ejected. If the intensity of the light increases, what wil

l happen to the ejection rate of the electrons?
Physics
1 answer:
drek231 [11]2 years ago
4 0

Answer:The rate of ejection of photoelectrons will increase

Explanation:

If the frequency of incident monochromatic light is held constant and its intensity is increased, the rate of ejection of photoelectrons from the metal surface increases with increase in intensity of the monochromatic light. More current flows due to more ejection of photoelectrons.

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An object is in free fall. After 10 seconds calculate the distance fallen? *
Citrus2011 [14]

Answer:

9.81 × 10 = 98.1 meters

vertical displacement is s=1/2 at^2 + vt

initial vertical velocity is 0 so s=1/2 at^2

a in this instance is gravitational acceleration so 60m= 1/2 (9.81)t^2

solve for t, t = 3.497s. //I corrected this answer as just now I misread horizontal as vertical.

3 0
3 years ago
Read 2 more answers
A rocket sled with an initial mass of 3 metric tons, invluding 1 ton of fuel, rests on a level section of track. At t=0, the sol
ELEN [110]

Answer:

v = 719.2 m / s and     a = 83.33 m / s²

Explanation:

This is a rocket propulsion system where the system is made up of the rocket plus the ejected mass, where the final velocity is

           v - v₀ = v_{e} ln (M₀ / M)

where v₀ is the initial velocity, v_{e} the velocity of the gases with respect to the rocket and M₀ and M the initial and final masses of the rocket

In this case, if fuel burns at 75 kg / s, we can calculate the fuel burned for the 10 s

            m_fuel = 75 10

            m_fuel = 750 kg

As the rocket initially had a mass of 3000 kg including 1000 kg of fuel, there are still 250 kg, so the mass of the rocket minus the fuel burned is

              M = 3000 -750 = 2250 kg

let's calculate

            v - 0 = 2500 ln (3000/2250)

            v = 719.2 m / s

To calculate the acceleration, let's use the concept of the rocket thrust, which is the force of the gases on it. In the case of the rocket, it is

             Push = v_{e} dM / dt

let's calculate

             Push = 2500  75

             Push = 187500 N

            If we use Newton's second law

             F = m a

             a = F / m

let's calculate

              a = 187500/2250

              a = 83.33 m / s²

7 0
2 years ago
You place a 0.17 kg can of soup and a 0.31 kg jar of pickles on the kitchen counter, separated by a distance of 0.42 m. What is
tangare [24]

1.984 \times 10^{-11} \mathrm{N} \text { is the force of gravity exerted on the jar of pickles. }

<u>Explanation</u>:

According to Newton's third law that each force has an equal and opposite reaction force in this case both of the jars will exert the same force an each other

. The force is given by

\mathrm{F}=\frac{G \times M_{1} \times M_{2}}{r^{2}}

Where, F = force, G=\text { gravitational constant }=\left(6.67 \times 10^{-11}\right), mass \left(\mathrm{M}_{1}\right)=0.17 \mathrm{kg}, mass \left M_{2}= 0.31 \mathrm{kg} and Distance(r) = 0.42 m.

Substitute the values in the formula.

\mathrm{F}=\frac{6.67 \times 10^{-11} \times 0.17 \times 0.31}{0.42^{2}}

\mathrm{F}=\frac{3.51 \times 10^{-12}}{0.176}

\mathrm{F}=1.984 \times 10^{-11} \mathrm{N}

\text { The force of gravity exerted on the jar of pickles is } 1.984 \times 10^{-11} \mathrm{N} \text { . }

3 0
3 years ago
,how do charged objects react???quiet urgent
son4ous [18]
Any charged object can<span> exert the force upon other objects ... i think tell me if im right</span>
8 0
2 years ago
Read 2 more answers
D section a only still need help on this
SIZIF [17.4K]

Acceleration means speeding up, slowing down, or changing direction. The graph doesn't show anything about direction, so we just have to examine it for speeding up or slowing down ... any change of speed.

The y-axis of this graph IS speed. So the height of a point on the line is speed. If the line is going up or down, then speed is changing.

Sections a, c, and d are all going up or down. Section b is the only one where speed is not changing. So we can't be sure about b, because we don't know if the track may be curving ... the graph can't tell us that. But a, c, and d are DEFINITELY showing acceleration.

5 0
3 years ago
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