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IgorC [24]
2 years ago
14

When a source of dim orange light shines on a photosensitive metal, no photoelectrons are ejected from its surface. what could b

e done to increase the likelihood of producing photoelectrons
Physics
1 answer:
padilas [110]2 years ago
5 0

Answer: Replace the orange light source with a higher frequency light source

Explanation:

To expel electrons from a piece of metal, the incoming light must have a minimum frequency to cause a photoelectric effect, i.e., the ejection of photoelectrons from a metal surface, which is also known as the metal's threshold frequency.

If v = frequency of incident photon and vth= threshold frequency, then,

  • For v < vth, there will be no ejection of photoelectron.
  • For v = vth, photoelectrons are just ejected from the metal surface, in this case, the kinetic energy of the electron ejected is zero
  • For v > vth, then photoelectrons will come out of the surface along with kinetic energy

Therefore, we would have to increase the frequency of incident light so that it becomes greater than the threshold frequency of that surface, and consequently a photoelectric process takes place.

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the cyclist has a mass of 50kg and is acceleratiing at 0.9m/s^2. What is the size of the unbalanced force F acting on the cyclis
stiv31 [10]

Answer:

45 N

Explanation:

The question can be solved by using Newton's second law:

F = ma

where

F is the net (unbalanced) force on the cyclist

m is the mass of the cyclist

a is his acceleration

In this problem:

m = 50 kg

a=0.9 m/s^2

Therefore, the unbalanced force is

F=(50)(0.9)=45 N

4 0
4 years ago
The wavelength is 1. 87 x 10-7 m. What is the frequency?.
jeka57 [31]

Answer:

11.7 im pretty sure. i did the work so it should be correct

3 0
2 years ago
What will be the new volume when 128 mL of gas at 20.0°C is heated to 40.0°C while pressure remains unchanged? What will be the
soldier1979 [14.2K]

The answer will be 137 mL

7 0
3 years ago
The system below uses massless pulleys and ropes. The coefficient of friction is μ. Assume that M1 and M2 are sliding. Gravity i
Archy [21]

Explanation:

Using Newtons second law on each block

F = m*a

Block 1

T_{1} - u*g*M_{1} = M_{1} *a \\\\T_{1} = M_{1}*(a + u*g) ... Eq1

Block 2

T_{2} - u*g*M_{2} = M_{2} *a \\\\T_{2} = M_{2}*(a + u*g) ... Eq2

Block 3

- (T_{1} + T_{2} ) + g*M_{3} = M_{3} *a \\\\T_{1} + T_{2} = M_{3}*( -a + g) ... Eq3

Solving Eq1,2,3 simultaneously

Divide 1 and 2

\frac{T_{1} }{T_{2}} = \frac{M_{1}*(a+u*g)}{M_{2}*(a+u*g)}  \\\\\frac{T_{1} }{T_{2}} = \frac{M_{1} }{M_{2} }\\\\ T_{1} =  \frac{M_{1} *T_{2} }{M_{2} } .... Eq4

Put Eq 4 into Eq3

T_{2} = \frac{M_{3}*(g-a) }{1+\frac{M_{1} }{M_{2} } }  ...Eq5

Put Eq 5 into Eq2 and solve for a

a = \frac{M_{3}*g -u*g*(M_{1} + M_{2}) }{M_{1} + M_{2} + M_{3} }  .... Eq6

Substitute back in Eq2 and use Eq4 and solve for T2 & T1

T_{2} = M_{2}*M_{3}*g*(\frac{1-u}{M_{1} + M_{2}+M_{3}})\\\\T_{1} = M_{1}*M_{3}*g*(\frac{1-u}{M_{1} + M_{2}+M_{3}})\\\\

5 0
4 years ago
Two cars, one of mass 1100 kg, and the second of mass 2500 kg, are moving at right angles to each other when they collide and st
Pie

Answer:

v_{f}=17.47 m/s

Explanation:

Let's use the conservation of momentum to solve it.

p_{initial}= p_{final} (1)

  • The total initial momentum will be: m_{1}v_{1i}+m_{2}v_{2i}
  • The total final momentum will be: m_{1}v_{1f}+m_{2}v_{2f}, but as they stick together after the collision, v1f = v2f = vf.

So we can rewrite (1), using the above information:

m_{1}v_{1i}+m_{2}v_{2i}=m_{1}v_{f}+m_{2}v_{f}

m_{1}v_{1i}+m_{2}v_{2i}=v_{f}(m_{1}+m_{2})

v_{f}=\frac{m_{1}v_{1i}+m_{2}v_{2i}}{m_{1}+m_{2}}

v_{f}=\frac{1100\cdot 14+2500\cdot 19}{1100+2500}

Finally, the magnitude of the velocity of the wreckage of the two cars immediately after the collision is:

v_{f}=17.47 m/s

I hope it helps you!

5 0
4 years ago
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