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postnew [5]
3 years ago
9

If the frequency of this beam is increased while the intensity is held constant, does the number of electrons ejected per second

from the metal surface increase, decrease, or stay the same?
Physics
1 answer:
Ivan3 years ago
8 0

Answer:

if the intensity of photons is constant then number of ejected electrons will remain same

Explanation:

As per photoelectric effect we know that when light of sufficient frequency fall on the surface of metal then electrons get ejected out of the surface with certain kinetic energy

Here the energy of photons is used to eject out the electrons from metal surface and to give the kinetic energy to the ejected electrons

so we have

h\nu = W + KE

here W = work function of metal which shows the energy required to eject out electrons from metal surface

KE = kinetic energy of ejected electrons

now if we increase the frequency of the photons that incident on the metal surface then in that case the incident energy will increase

So the electrons will eject out with more kinetic energy while if the number of photon is constant or the intensity of photons is constant then number of ejected electrons will remain same

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Why do astronomers find it difficult to detect individual exoplanets?
Maurinko [17]
Light from the stars, because the orbits make it difficult to see them. 
6 0
3 years ago
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A ball with 100 J of PE is released from a height of 10 m. What will be the KE of the ball at 5
harkovskaia [24]

Answer:

The kinetic energy is: 50[J]

Explanation:

The ball is having a potential energy of 100 [J], therefore

PE = [J]

The elevation is 10 [m], and at this point the ball is having only potential energy, the kinetic energy is zero.

E_{p} =m*g*h\\where:\\g= gravity[m/s^{2} ]\\m = mass [kg]\\m= \frac{E_{p} }{g*h}\\ m= \frac{100}{9.81*10}\\\\m= 1.01[kg]\\\\

In the moment when the ball starts to fall, it will lose potential energy and the potential energy will be transforme in kinetic energy.

When the elevation is 5 [m], we have a potential energy of

P_{e} =m*g*h\\P_{e} =1.01*9.81*5\\\\P_{e} = 50 [J]\\

This energy is equal to the kinetic energy, therefore

Ke= 50 [J]

8 0
3 years ago
You are planning to make an open rectangular box from an 8-inch by 15-inch piece of cardboard by cutting congruent squares from
Phoenix [80]

Let us say that x is the cut that we will make on the sides to make a box, therefore the new dimensions are:

l = 15 – 2x

w = 8 – 2x

It is 2x since we cut on two sides.

 

We know that volume is:

V = l w x

V = (15 – 2x) (8 – 2x) x

V = 120x – 30x^2 – 16x^2 + 4x^3

V = 120x – 46x^2 + 4x^3

 

Taking the 1st derivative:

dV/dx = 120 – 92x + 12x^2

 

Set dV/dx = 0 to get maxima:

120 – 92x + 12x^2 = 0

 

Divide by 12:

x^2 – (92/12)x + 10 = 0

(x – (92/24))^2 = -10 + (92/24)^2

x - 92/24 = ±2.17

x = 1.66, 6

We cannot have x = 6 because that will make our w negative, so:

x = 1.66 inches

 

So the largest volume is:

V = 120x – 46x^2 + 4x^3

V = 120(1.66) – 46(1.66)^2 + 4(1.66)^3

V = 90.74 cubic inches

4 0
4 years ago
What is the specific heat of an object if it takes 1200 J to raise the temperature of a 20 kg object by 6 degrees C?
ch4aika [34]

Answer:

c=10\ J/kg^{\circ} C

Explanation:

Given that,

Heat required, Q = 1200 J

Mass of the object, m = 20 kg

The increase in temperature, \Delta T=6^{\circ} C

We need to find the specific heat of the object. The heat required to raise the temperature is given by :

Q=mc\Delta T\\\\c=\dfrac{Q}{m\Delta T}\\\\c=\dfrac{1200}{20\times 6}\\\\c=10\ J/kg^{\circ} C

So, the specific heat of the object is 10\ J/kg^{\circ} C.

5 0
3 years ago
A meteorologist plans to release a weather balloon from ground level, to be used for high-altitude atmospheric measurements. The
Slav-nsk [51]

Answer:

563.86 N

Explanation:

We know the buoyant force F = weight of air displaced by the balloon.

F = ρgV where ρ = density of air = 1.29 kg/m³, g = acceleration due to gravity = 9.8 m/s² and V = volume of balloon = 4πr/3 (since it is a sphere) where r = radius of balloon = 2.20 m

So, F = ρgV = ρg4πr³/3

substituting the values of the variables into the equation, we have

F =  1.29 kg/m³ × 9.8 m/s² × 4π × (2.20 m)³/3

= 1691.58 N/3

= 563.86 N

8 0
3 years ago
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