1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
ANEK [815]
2 years ago
8

The outermost electron of an atom has a binding energy of 2.5 eV. The atom is exposed to light of a high enough frequency to cau

se exactly one electron to be ejected. The ejected electron is found to have a KE of 2.0 eV. 23. How much energy did photons of the incoming light contain? A) 0.50 eV B) 0.80 eV C) 4.5 eV D) 5.0 eV
Physics
1 answer:
Varvara68 [4.7K]2 years ago
4 0

Answer: C) 4.5 eV

Explanation: In order to explain this problem we have to consider the photoelectric effect,

The energy balance in this phenomenum is given by:

h*ν= W+ KE where h*ν is the energy of the incident photon, W is the work function and KE is the kinetic energy of the photoelectrons.

We consider W for outer electrons equal to 2.5 eV , also we consider that the KE is found to be 2.0 eV, then we have:

h*ν=2.5 eV+KE =2.5 eV+2.0 eV0 4.5 eV

You might be interested in
If an object absorbs light energy, its thermal energy __________.
german
D), increases. The object absorbs light energy which in turn (energy is energy) usually involves absorbing heat as well.
8 0
2 years ago
A magnetic field is uniform over a flat, horizontal circular region with a radius of 1.50 mm, and the field varies with time. In
atroni [7]

Answer:

The average induced emf around the border of the circular region is 8.48\times 10^{-5}\ V.

Explanation:

Given that,

Radius of circular region, r = 1.5 mm

Initial magnetic field, B = 0

Final magnetic field, B' = 1.5 T

The magnetic field is pointing upward when viewed from above, perpendicular to the circular plane in a time of 125 ms. We need to find the average induced emf around the border of the circular region. It is given by the rate of change of magnetic flux as :

\epsilon=\dfrac{-d\phi}{dt}\\\\\epsilon=A\dfrac{-d(B'-B)}{dt}\\\\\epsilon=\pi (1.5\times 10^{-3})^2\times \dfrac{1.5}{0.125}\\\\\epsilon=8.48\times 10^{-5}\ V

So, the average induced emf around the border of the circular region is 8.48\times 10^{-5}\ V.

6 0
3 years ago
Read 2 more answers
A car moves forward up a hill at 12 m/s with a uniform backward acceleration of 1.6 m/s2. What is its displacement after 6 s?
Romashka [77]

Answer:

The displacement of the car after 6s is 43.2 m

Explanation:

Given;

velocity of the car, v = 12 m/s

acceleration of the car, a = -1.6 m/s² (backward acceleration)

time of motion, t = 6 s

The displacement of the car after 6s is given by the following kinematic equation;

d = ut + ¹/₂at²

d = (12 x 6) + ¹/₂(-1.6)(6)²

d = 72 - 28.8

d = 43.2 m

Therefore, the displacement of the car after 6s is 43.2 m

6 0
2 years ago
The element carbon exists in several different forms. Pure carbon can be found in the form of diamonds, coal, and the graphite i
Artist 52 [7]
B is false.

not all carbon atoms have 6 neutrons. Carbon 13 for example has 7 neutrons
5 0
3 years ago
Read 2 more answers
A 30-g bullet is fired with a horizontal velocity of 460 m/s and becomes embedded in block B which has a mass of 3 kg. After the
ICE Princess25 [194]

Answer:

energy loss due to friction and the impacts = 2.97 J; The impact loss due to AB impacting the carrier is =25.72J; The impact loss at first impact is 6,316.64J

Explanation:

First find the velocity of the bullet after the first impact using

M1V1 + 0 = (M1 + M2)v'

Where M1 is the mass of the bullet

M2 is the mass of the block B

M3 is the mass of the carrier

v' is the velocity

v' = M1V1/(M1 + M2)

v'= (30 × 10^-3 kg)(460 m/s) / (30 × 10^-3 kg + 3 kg)

v' = 13.8/3.03

v'= 4.55m/s

Also calculate final velocity of the carrier v2'

v2' = M1V1/(M1 + M2 + M3)

v2'= (30 × 10 kg)(460 m/s) / (30 × 10 kg + 3 kg + 30kg)

v2' =0.42m/s

Now to calculate energy loss due to friction

Normal force

N= W1 + W2 = (m1 + m2)g

Where W1 and W2 is the weight of the bullet and block respectively

g is gravitational acceleration for taken as 9.81m/s

= (0.030 kg + 3 kg)(9.81 m/s) 29.724 N

Friction force = coefficient of kinetic× normal force

Where coefficient of kinetic = 0.2

Ff = (0.2)(29.724)= 5.945 N

Now

Energy loss due to friction = frictional force × distance

Assume distance is 0.5 m.

Energy loss due to friction = 5.945 N × 0.5 m

= 2.97J

Kinetic energy of block with embedded bullet immediately after first impact:

1/2 × (m1 + m2)(v')^2

1/2 × (30 × 10^-3 kg + 3 kg)(4.55m/s)^2

= 1/2 × (3.03kg) × (4.55m/s)^2

= 31.36 J

Final kinetic energy of bullet, Block, and Carrier together

1/2 × (m1 + m2 + m3)(v2')^2

1/2 × (30 × 10^-3 kg + 3 kg + 30kg) (0.42m/s)^2

1/2 × (33.03kg) × (0.42m/s)^2

= 2.91 J

Therefore

Loss due to friction and stopping impact = Kinetic energy of block with embedded bullet immediately after first impact - Final kinetic energy of bullet, Block, and Carrier together

= 31.36 J - 2.91 J

= 28.69 J

Impact loss due to AB impacting the carrier = loss due to friction- energy due to friction

28.69J - 2.97J

=25.72J

Initial kinetic energy of system ABC = 1/2(m1vo)

=1/2(0.030 kg)(460 m/s)^2 = 6,348J

Therefore

Impact loss at first impact = Initial kinetic energy of system ABC - Kinetic energy of block with embedded bullet immediately after first impact:

= 6,348J - 31.36 J

= 6,316.64J

3 0
3 years ago
Other questions:
  • radar wave is transmitted and later reflected off an aircraft and recieved 1.4×10^3 sec after being sent out. how far is the air
    7·1 answer
  • Four of your friends are in new relationships, but only two of these relationships are healthy. Which of your friends are in unh
    9·2 answers
  • Answers for 13 are
    10·1 answer
  • A thin coil has 17 rectangular turns of wire. When a current of 4 A runs through the coil, there is a total flux of 5 ✕ 10−3 T ·
    13·1 answer
  • The electrons of carbon and hydrogen are _____________ in this molecule of methane.
    7·1 answer
  • Assuming that 1.0g of aspirin dissolves in 450 mL of water at 10o C, how much aspirin would be lost in the 1.4 mL of water added
    14·1 answer
  • Priya is responsible for collecting canned food along three different streets for her school's annual Thanksgiving Food Drive. S
    14·1 answer
  • Simple Pendulum: A 34-kg child on an 18-kg swing set swings back and forth through small angles. If the length of the very light
    8·2 answers
  • Which type of graph would be best for showing the percentage of people in a
    11·2 answers
  • A skier stands at the top of a 50 meter slope. He then skis down the slope. What is his approximate speed at the bottom of the s
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!