Obami | Digital Stakeholder Experiences

SECTION A

ONE Word Items - Give ONE word/term for each of the following descriptions. Write only the word/term

Marks: 5

Question 1:

The rate of change of velocity

Submitted Answer

Model Answer:

Acceleration

Question 2:

The distance between two consecutive points in phase on a wave

Submitted Answer

Model Answer:

Wavelength

Question 3:

A region of space in which an electric charge experiences an electrostatic force

Submitted Answer

Model Answer:

Electric field

Question 4:

The type of electromagnetic wave with the shortest wavelength

Submitted Answer

Model Answer:

Gamma / 𝛾 (rays)

Question 5:

The minimum frequency of light needed to remove an electron from the surface of a metal

Submitted Answer

Model Answer:

Threshold (frequency)

MULTIPLE-CHOICE QUESTIONS - Four options are provided as possible answers to the following questions. Each question has only ONE correct answer.

Marks: 20

Question 1:

Which ONE of the following physical quantities is equal to the product of force and constant velocity?

Submitted Answer

Model Answer:

Power

Question 2:

A 30 kg iron sphere and a 10 kg aluminium sphere with the same diameter fall freely from the roof of a tall building. Ignore the effects of friction.

When the spheres are 5 m above the ground, they have the same ...

Submitted Answer

Model Answer:

acceleration

Question 3:

The free-body diagram below shows the relative magnitudes and directions of all the forces acting on an object moving horizontally in an easterly direction.

The kinetic energy of the object ...

Submitted Answer

Model Answer:

decreases.

Question 4:

The hooter of a vehicle travelling at constant speed towards a stationary observer, produces sound waves of frequency 400 Hz. Ignore the effects of wind.
Which ONE of the following frequencies, in hertz, is most likely to be heard by the observer?

Submitted Answer

Model Answer:

480

Question 5:

When two waves meet at a point, the amplitude of the resultant wave is the algebraic sum of the amplitudes of the individual waves.
This principle is known as …

Submitted Answer

Model Answer:

superposition.

Question 6:

A parallel plate capacitor, X, with a vacuum between its plates is connected in a circuit as shown below. When fully charged, the charge stored on its plates is equal to Q.

Capacitor X is now replaced with a similar capacitor, Y, with the same dimensions but with paper between its plates. When fully charged, the charge stored on the plates of capacitor Y is …

Submitted Answer

Model Answer:

larger than Q.

Question 7:

Which ONE of the following graphs best represents the relationship between the electrical power and the current in a given ohmic conductor?

Submitted Answer

Model Answer:

D

Question 8:

In a vacuum, all electromagnetic waves have the same …

Submitted Answer

Model Answer:

speed.

Question 9:

In the sketch below, a conductor carrying conventional current, I, is placed in a magnetic field.

Which ONE of the following best describes the direction of the magnetic force experienced by the conductor?

Submitted Answer

Model Answer:

Out of the page perpendicular to the direction of the magnetic field

Question 10:

An atom in its ground state absorbs energy E and is excited to a higher energy state. When the atom returns to the ground state, a photon with energy ...

Submitted Answer

Model Answer:

E is released.

SECTION B - Show the formulae and substitutions in ALL calculations. Round off your final numerical answers to a minimum of TWO decimal places.

Vertical Projectile Motion - A ball of mass 0,15 kg is thrown vertically downwards from the top of a building to a concrete floor below. The ball bounces off the floor. The velocity versus time graph below shows the motion of the ball. Ignore the effects of air friction. TAKE DOWNWARD MOTION AS POSITIVE.

Marks: 19

Attached Section Resource:

98890184-7ec3-4a84-a6ac-191b265da2c0.png

Question 1:

From the graph, write down the magnitude of the velocity at which the ball bounces off the floor.

Submitted Answer

Model Answer:

15 m.s^-1

Question 2:

Is the collision of the ball with the floor ELASTIC or INELASTIC? Refer to the data on the graph to explain the answer.

Submitted Answer

Model Answer:

Option 1:

Inelastic.

The speed/velocity at which the ball leaves the floor is less / different than that at which it strikes the floor. OR The speed/velocity of the ball changes during the collision.

Therefore the kinetic energy changes/is not conserved.

Option 2:

Collision is inelastic.

ΔK = ½mv_f² - ½mv_i²
= ½(0,15)(15)² - ½(0,15)(20)²

= - 13,13 J

K_i ≠ K_f / ΔK ≠ 0

Option 3:

Collision is inelastic.

K_f = ½mv_f²

= ½(0,15)(15)²
= 16,88 J
K_i = ½mv_i²
= ½(0,15)(20)²
= 30 J
K_f ≠ K₁ / ΔK ≠ 0

Question 3:

Calculate the height from which the ball is thrown.

Submitted Answer

Model Answer:

OPTION 1:
v_f² = v_i² + 2aΔy
(20)² = (10)² + 2(9,8)Δy
∴ Δy = 15,31 m

OPTION 2:
W_net = ΔK
F_netΔycos θ = ½ m(v_f² – v_i²)
m(9,8)Δycos0° = ½ m(20² – 10²)
Δy = 15,31 m

OPTION 3:
(E_p + E_k)_top = (E_p + E_k)_bottom
(mgh + ½ mv²)_top = (mgh + ½ mv²)_^bottom
m(9,8)h + ½m(10)² = m(9,8)(0) + ½m(20)²
h = 15,31 m

Option 4:

v_f = v_i + aΔt
20 = 10 + (9,8)(Δt)
∴ Δt = 1,02 s

Δy = v_iΔt + ½aΔt²
= (10)(1,02) + ½(9,8)(1,02)²
∴ Δy = 15,3 m

Option 5:

Option 6:

v_f = v_i + aΔt
20 = 10 + (9,8)(Δt)
∴ Δt = 1,02 s

Height = area between graph & t axis

= ½(sum ║ sides)h_⊥
= ½(10 + 20)1,02
= 15,3 m

Option 7:

v_f = v_i + aΔt
20 = 10 + (9,8)(Δt)
∴ Δt = 1,02 s

Height = area between graph & t axis

= lb + ½bh = ½(10 + 20)1,02
= (1,02)(10) + ½(1,02)(10)
= 15,3 m

Option 8:

Question 4:

Calculate the magnitude of the impulse imparted by the floor on the ball.

Submitted Answer

Model Answer:

Question 5:

Calculate the magnitude of the displacement of the ball from the moment it is thrown until time t.

Submitted Answer

Model Answer:

Option 1:

Displacement from floor to max. height:

Total displacement:

= - 11,48 + 15,3

= 3,82 m / 3.83 m

Option 2:

v_f = v_i + aΔt
0 = -15 + (9,8)Δt
Δt = 1,53 s

Δy = v_iΔt + ½ aΔt²

= (-15)(1,53) + ½ (9,8)(1,53)²
= -11,48 m

Total displacement:

= - 11,48 + 15,3

= 3,82 m

Option 3:

v_f = v_i + aΔt
0 = -15 + (9,8)Δt
Δt = 1,53 s

Total displacement:

= - 11,48 + 15,3

= 3,82 m

Option 4:

v_f = v_i + aΔt
0 = -15 + (9,8)Δt
Δt = 1,53 s

Area = ½ bh
= ½ (1,53)(-15)
= -11,48 m

Total displacement:

= - 11,48 + 15,3

= 3,82 m

Option 5:

E_M(initial) = E_M(final)
(E_p + E_k)_initial = (E_p + E_k)_final
(mgh + ½ mv²)_initial = (mgh + ½ mv²)_final
(0,15)(9,8)(0) + ½ (0,15)(15)² = (0,15)(9,8)h + ½ (0,15)(0)²
h = 11,48 m

Total displacement:

= - 11,48 + 15,3

= 3,82 m

Option 6:

W_net = ΔK
F_netΔycos θ = ½ m(v_f² – v_i²)
m(9,8)Δycos180° = ½ m(0² – 15²)
Δy = 11,48 m

Total displacement:

= - 11,48 + 15,3

= 3,82 m

Option 7:

Total displacement:

= - 11,48 + 15,3

= 3,82 m

Question 6:

Sketch a position versus time graph for the motion of the ball from the moment it is thrown until it reaches its maximum height after the bounce. USE THE FLOOR AS THE ZERO POSITION.
Indicate the following on the graph:
• The height from which the ball is thrown
• Time t

Submitted Answer

Model Answer:

009fac6a-4b4e-4a4c-92f7-84da8701658c.png

Motion in Two Dimensions - A boy on ice skates is stationary on a frozen lake (no friction). He throws a package of mass 5 kg at 4 m·s-1 horizontally east as shown below. The mass of the boy is 60 kg. At the instant the package leaves the boy's hand, the boy starts moving.

Marks: 13

Attached Section Resource:

5a94f47f-eeae-4d0a-8276-abb45750f7ff.png

Question 1:

In which direction does the boy move? Write down only EAST or WEST.

Submitted Answer

Model Answer:

West

Question 2:

Which ONE of Newton's laws of motion explains the direction in which the boy experiences a force when he throws the package? Name and state this law in words.

Submitted Answer

Model Answer:

(Newton's) Third Law (of Motion).
When object A exerts a force on object B,
object B exerts a force equal in magnitude on object A, but opposite in direction.

Question 3:

Calculate the magnitude of the velocity of the boy immediately after the package leaves his hand. Ignore the effects of friction.

Submitted Answer

Model Answer:

Option 1:

East as positive:
Σp_i = Σp_f
0 = (60)v_f + (5)(4)
∴ v_f = - 0,33
∴ v_f = 0,33 m·s^-1

Option 2:

West as positive:
Σp_i = Σp_f
0 = (60)v_f + (5)(-4)
∴ v_f = - 0,33
∴ v_f = 0,33 m·s^-1

Option 3:

East as positive:

Δp_A = -Δp_B
(60)v_f – 0 = -[(5)(4) – 0]
∴ v_f = - 0,33
∴ v_f = 0,33 m·s^-1

Option 4:

Δp_A = -Δp_B
(60)v_f – 0 = -[(5)(-4) – 0]
∴ v_f = - 0,33
∴ v_f = 0,33 m·s^-1

Option 5:

East as positive:

Option 6:

West as positive:

Question 4:

How will the answer to QUESTION 3 be affected if the boy throws the same package at a higher velocity in the same direction.

Write down INCREASES, DECREASES or REMAINS THE SAME.

Submitted Answer

Model Answer:

Increases

Question 5:

How will the answer to QUESTION 3 be affected if the boy throws a package of double the mass at the same velocity as in QUESTION 3.

Write down INCREASES, DECREASES or REMAINS THE SAME an explain the answer.

Submitted Answer

Model Answer:

Option 1:

Increases.
• Δp package increases, thus Δp boy increases.
• For the same mass of boy, v will be greater.

Option 2:

Increases.

From the equation in QUESTION 3: -m_Av_Af = m_Bv_Bf
• If mass of package/B doubles/increases, the momentum of the boy / A doubles / increases.
• For same mass of boy / A, the velocity of boy / A doubles/increases.

Option 3:

Increases
-m_Bv_Bf = m_pv_pf
v_B = (-m_pv_pf)/(m_B) for same m_B, if m_P doubles, then v_B doubles

Conservation of Mechanical Energy - A 5 kg rigid crate moves from rest down path XYZ as shown below (diagram not drawn to scale). Section XY of the path is frictionless. Assume that the crate moves in a straight line down the path.

Marks: 15

Attached Section Resource:

0ce183fa-56e5-41b6-bef2-e13bf4b732f5.png

Question 1:

State, in words, the principle of the conservation of mechanical energy

Submitted Answer

Model Answer:

Option 1:

The total mechanical energy remains constant / is conserved in a closed / isolated system / in absence of external forces /non-conservative forces.

Option 2:

The sum of the potential and kinetic energy of a system remains constant in a closed/isolated system.

Option 3:

When the work done on an object by the non-conservative forces is zero, the total mechanical energy is conserved.

Question 2:

Use the principle of the conservation of mechanical energy to calculate the speed of the crate when it reaches point Y.

Submitted Answer

Model Answer:

Option 1:

E_{mechanical at X} = E_{mechanical at Y}
(E_p + E_k)_X = (E_p + E_k)_Y
(mgh + ½ mv²)_X= (mgh + ½ mv²)_Y
5(9,8)(5) + ½(5)(0²) = (5)(9,8)(1) + ½(5)v_f²
v = 8,85 m∙s^-1

Option 2:

E_{mechanical at X} = E_{mechanical at Y}
(E_p + E_k)_X = (E_p + E_k)_Y
(mgh + ½ mv²)_X= (mgh + ½ mv²)_Y
5(9,8)(4) + ½(5)(0²) = (5)(9,8)(0) + ½(5)v_f²
v = 8,85 m∙s^-1

Question 3:

On reaching point Y, the crate continues to move down section YZ of the path. It experiences an average frictional force of 10 N and reaches point Z at a speed of 4 m∙s^-1.

APART FROM FRICTION, write down the names of TWO other forces that act on the crate while it moves down section YZ.

Submitted Answer

Model Answer:

Weight / gravitational (force) / (force of) gravity

Normal force

Question 4:

In which direction does the net force act on the crate as it moves down section YZ? Write down only from 'Y to Z' or from 'Z to Y'.

Submitted Answer

Model Answer:

Z to Y

Question 5:

Use the WORK-ENERGY THEOREM to calculate the length of section YZ.

Submitted Answer

Model Answer:

OPTION 1:
W_net = ΔK
Ww + W_f = ½m(v_f² – v_i²)
mgΔycos0° + fΔxcos180° = ½ m(v_f² – v_i²)
(5)(9,8)(1)(1) + (10)Δx(-1) = ½(5)(4² – 8,85²)
Δx = 20,48 m

OPTION 2:
W_net = ΔK
W_w + W_f = ½m(v_f² – v^_i2)
-ΔE_p + W_f = ½m(v_f² – v_i²)
-(0 – mgh) + fΔxcos180° = ½ m(v_f² – v_i²)
(5)(9,8)(1) + (10)Δx(-1) = ½(5)(4² – 8,85²)
Δx = 20,48 m

OPTION 3:
W_net = ΔK
W_w + W_f = ½m(v_f² – v_i²)
-ΔE_p + W_f = ½m(v_f² – v_i²)
-(0 – mgh) + fΔxcos180° = ½ m(v_f² – v_i²)
(5)(9,8)(5) + (10)Δx(-1) = ½(5)(4² – 0²)

Δx = 20,48 m

OPTION 4:

W_net = ΔK
W_w + W_f = ½m(v_f² – v_i²)

mgΔxcos(90^o – θ) + fΔxcos180° = ½ m(v_f² – v_i²)
mgΔxsinθ + fΔxcos180° = ½ m(v_f² – v_i²)

mgΔx (1/Δx) + fΔxcos180° = ½ m(v_f² – v_i²)

(5)(9,8) + (10)Δx(-1) = ½(5)(4² – 8,85²)
Δx = 20,48 m

OPTION 5:
W_net = ΔK
W_w|| + W_f = ½m(v_f² – v_i²)
mgsinθΔxcosθ + fΔxcosθ = ½ m(v_f² – v_i²)

mg (1/Δx)Δxcos0°+ fΔxcos180° = ½ m(v_f² – v_i²)

(5)(9,8) + (10)Δx(-1) = ½(5)(4² – 8,85²)
Δx = 20,48 m

OPTION 6:

W_net = ΔK
FnetΔxcosθ = ½m(v_f² – v_i²)
(10 - 49sinθ)Δxcos180° = ½ m(v_f2 – v_i²)
(10- 49(1/Δx)) Δxcos180° = ½ m(v_f² – v_i²)

(10Δx – 49)(-1) = ½(5)(4² – 8,85²)
Δx = 20,48 m

OPTION 7:
W_nc = ΔE_p + ΔE_k
fΔxcosθ = (mgh_f – mgh_i) + ( ½ mv_f² – ½ mv_i²)
(10)Δxcos180° = [0 – (5)(9,8)(1)] + [ ½ (5)(4)⁴ – ½ (5)(8,85)²]
Δx = 20,48 m

Question 6:

Another crate of mass 10 kg now moves from point X down path XYZ.

How will the velocity of this 10 kg crate at point Y compare to that of the 5 kg crate at Y? Write down only GREATER THAN, SMALLER THAN or EQUAL TO.

Submitted Answer

Model Answer:

Equal to

Doppler Effect - An ambulance approaches a stationary observer at a constant speed of 10,6 m∙s-1, while its siren produces sound at a constant frequency of 954,3 Hz. The stationary observer measures the frequency of the sound as 985 Hz.

Marks: 9

Question 1:

Name the medical instrument that makes use of the Doppler effect.

Submitted Answer

Model Answer:

Doppler flow meter

Question 2:

Calculate the velocity of sound.

Submitted Answer

Model Answer:

Question 3:

How would the wavelength of the sound wave produced by the siren of the ambulance change if the frequency of the wave were higher than 954,3 Hz? Write down only INCREASES, DECREASES or STAYS THE SAME.

Submitted Answer

Model Answer:

Decreases

Question 4:

Give a reason for the answer to QUESTION 3.

Submitted Answer

Model Answer:

Option 1:

For a constant velocity of sound / speed if the frequency increases, λ decreases.

Option 2:

λ α 1/f or f α 1/λ at constant velocity/speed

Dispersion of Light - Learners investigate how the broadness of the central bright band in a diffraction pattern changes as the wavelength of light changes. During the investigation, they perform two experiments. The slit width and the distance between the slit and the screen are kept constant.

Marks: 13

Attached Section Resource:

00a60312-9a69-41d5-80e8-a29ad8233d9d.png

Question 1:

In the first experiment, they pass light from a monochromatic source through a single slit and obtain pattern P on a screen. In the second experiment, they pass light from a different monochromatic source through the single slit and obtain pattern Q on the screen.

Define the term diffraction.

Submitted Answer

Model Answer:

Option 1:

The bending of waves around obstacles / corners / through an opening / aperture.

Option 2:

The spreading of waves around the edge of a barrier/through an opening/aperture.

Question 2:

Which ONE of the two patterns (P or Q) was obtained using the monochromatic light of a longer wavelength?

Submitted Answer

Model Answer:

P

Question 3:

For this investigation, write down the dependent variable.

Submitted Answer

Model Answer:

Broadness of the central bright band / diffraction pattern / angle of diffraction / degree of diffraction / sin θ / position of the first minimum

Question 4:

For this investigation, write down the investigative question.

Submitted Answer

Model Answer:

Example of answer:
What is the relationship between the broadness of the central band and the wavelength (of light used)?

Criteria that need to met to answer this question:

Dependent and independent variables correctly identified.

Question about the relationship between the independent and dependent variables correctly formulated.

Question 5:

In ONE of their experiments, they use light of wavelength 410 nm and a slit width of 5 x 10^-6 m.

Calculate the angle at which the SECOND MINIMUM will be observed on the screen.

Submitted Answer

Model Answer:

Option 1:

Option 2:

Question 6:

The single slit is now replaced with a double slit. Describe the pattern that will be observed on the screen.

Submitted Answer

Model Answer:

Light (bright) and dark bands.
Light /dark bands of equal width.

Electrodynamics - In the diagram shown below, point charge A has a charge of +16 μC. X is a point 12 cm from point charge A.

Marks: 14

Attached Section Resource:

99c1ab99-2c50-43c1-838f-76f26348cb2a.png

Question 1:

Draw the electric field pattern produced by point charge A.

Submitted Answer

Model Answer:

cefe98f4-124c-40e9-bc53-6919a7677d54.png

Question 2:

Is the electric field in QUESTION 1 UNIFORM or NON-UNIFORM?

Submitted Answer

Model Answer:

Non-uniform

Question 3:

Calculate the magnitude and direction of the electric field at point X due to point charge A.

Submitted Answer

Model Answer:

E=kQ/r²

= [(9x10⁹)(16x10^-6)]/(0.12)²

= 1x10⁷ N.C^-1

east

Question 4:

Another point charge B is now placed at a distance of 35 cm from point charge A as shown below. The NET electric field at point X due to point charges A and B is 1 x 107 N·C^-1 west.

Is point charge B POSITIVE or NEGATIVE?

Submitted Answer

Model Answer:

Positive

Question 5:

Calculate the magnitude of point charge B.

Submitted Answer

Model Answer:

Option 1:

Option 2:

Electronics - A learner wants to use a 12 V battery with an internal resistance of 1 Ω to operate an electrical device. He uses the circuit below to obtain the desired potential difference for the device to function. The resistance of the device is 5 Ω. When switch S is closed the device functions at its maximum power of 5 W.

Marks: 16

Attached Section Resource:

fb6917b1-d24b-48fe-a2eb-047a3582d374.png

Question 1:

Explain, in words, the meaning of an emf of 12 V.

Submitted Answer

Model Answer:

12 J of energy are transferred to / work done on
each coulomb (of charge) / per C charge passing through the battery.

Question 2:

Calculate the current that passes through the electrical device.

Submitted Answer

Model Answer:

Option 1:

P = I²R
5 = I²(5)
∴I = 1 A

Option 2:

P = V²/R

5 = V²/5

V = 5V

P = VI
5 = (5)I
I = 1 A

Option 3:

P = V²/R

5 = V²/5

V = 5V

V = IR
5 = I(5)
I = 1 A

Question 3:

Calculate the resistance of resistor R_x.

Submitted Answer

Model Answer:

OPTION 1:
Emf = I(R + r)
12 = (1)(R + 1)
R = 11 Ω
R_p = 11 – 5 = 6 Ω

Then:

Or:

OPTION 2:
Emf = I(R + r)

12 = (1)(Rp + 5 + 1)
∴ Rp = 6 Ω

Then:

Or:

OPTION 3:
V = I R_T
12 = (1)R
R_T = 12 Ω

R_p = R_T – (5 + 1)
= 12 – 6
= 6 Ω

Then:

Or:

Option 4:

V_5Ω = IR
= (1)(5)
= 5 V

V_internal = Ir
= (1)(1)
= 1 V

V_parallel = 12 – (1 + 5)
= 6 V

V_parallel = IR
6 = I(12)
∴ I = 0,5 A

IR_x = 1 – 0,5
= 0,5 A

V = IR
6 = (0,5)(4 + R_x)
∴R_x = 8 Ω

Question 4:

Switch S is now opened. Will the device still function at maximum power? Write down YES or NO. Explain the answer without doing any calculations.

Submitted Answer

Model Answer:

No.
Total resistance (R) increases.
Current (I) decreases.
(For a constant R) power (P = I²R) decreases.

Electromagnetic Radiation - The simplified sketch, shown below, represents an AC generator. The main components are labelled A, B, C and D.

Marks: 14

Attached Section Resource:

e41d3c3e-0ba3-43e4-97af-03912dc880c1.png

Question 1:

Write down the name of component:

A

B

Submitted Answer

Model Answer:

A - slip rings
B - brush(es)

Question 2:

Write down the function of component B.

Submitted Answer

Model Answer:

Maintains electrical contact with the slip rings.

OR

To take current out/in of the coil.

Question 3:

State the energy conversion which takes place in an AC generator.

Submitted Answer

Model Answer:

Mechanical /kinetic energy to electrical energy.

Question 4:

A similar coil is rotated in a magnetic field. The graph below shows how the alternating current produced by the AC generator varies with time.

How many rotations are made by the coil in 0,03 seconds?

Submitted Answer

Model Answer:

1.5

Question 5:

Calculate the frequency of the alternating current.

Submitted Answer

Model Answer:

Option 1:

Option 2:

Question 6:

Will the plane of the coil be PERPENDICULAR TO or PARALLEL TO the magnetic field at t = 0,015 s?

Submitted Answer

Model Answer:

Parallel to

Question 7:

If the generator produces a maximum potential difference of 311 V, calculate its average power output.

Submitted Answer

Model Answer:

Option 1:

Answers in the range: 3298,13 – 3299,18 W are acceptable

Option 2:

Option 3:

Option 4:

Option 5:

Optical Phenomena and Properties of Matter - In the simplified diagram shown below, light is incident on the emitter of a photocell. The emitted photoelectrons move towards the collector and the ammeter registers a reading.

Marks: 12

Attached Section Resource:

64d00d82-3572-4c71-b10a-4c60b67caf42.png

Question 1:

Name the phenomenon illustrated above.

Submitted Answer

Model Answer:

Photo-electric effect

Question 2:

The work function of the metal used as emitter is 8,0 x 10-19 J. The incident light has a wavelength of 200 nm.

Calculate the maximum speed at which an electron can be emitted.

Submitted Answer

Model Answer:

Option 1:

Option 2:

Question 3:

Incident light of a higher frequency is now used.
How will this change affect the maximum kinetic energy of the electron emitted in QUESTION 2? Write down only INCREASES, DECREASES or REMAINS THE SAME.

Submitted Answer

Model Answer:

Increases

Question 4:

The intensity of the incident light is now increased.
How will this change affect the speed of the electron calculated in QUESTION 2? Write down INCREASES, DECREASES or REMAINS THE SAME. Give a reason for the answer.

Submitted Answer

Model Answer:

Option 1:

Remains the same.
Intensity only affects number of photoelectrons emitted per second.

Option 2:

Remains the same.
The kinetic energy of the emitted photoelectrons remains the same.

Option 3:

Remains the same.
Only the frequency/wavelength of the incident light affects the maximum kinetic energy.

Question 5:

A metal worker places two iron rods, A and B, in a furnace. After a while he observes that A glows deep red while B glows orange.
Which ONE of the rods (A or B) radiates more energy? Give a reason for the answer.

Submitted Answer

Model Answer:

Opton 1:

B.
Orange light has a higher frequency than red light.

Option 2:

B.

Orange light has smaller wavelength than red light.

Question 6:

Neon signs illuminate many buildings. What type of spectrum is produced by neon signs?

Submitted Answer

Model Answer:

Line emission (spectra)