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Combined decks Flashcards

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1
Q

What changes could be made to a convex lense to produce an enlarged and upright image?

A
  1. Move the lens closer to the object so that it lies within the focal length of the lens to produce a virtual, enlarged and upright image.
  2. Use a thinner/less curved lens with a larger radius of curvature, so that the object lies within the focal length.
  3. Decrease the refractive index of the glass so its focal length increases, so that the object lies within the focal length.
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2
Q

Distance of an object away from a concave mirror

A
  • Distance will be negative, as image formed is virtual
  • The image has been magnified 3x, so the image distance is 3x the object distance from the mirror
  • Therefore, di = -3do
  • Substitute values into 1/f = 1/do + 1/di
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3
Q

Difference between images produced in concave/convex mirrors (between f and c)

A
  • The image formed by the convex mirror is upright, diminished and virtual.
  • Convex mirrors are diverging mirrors, which means that no real light rays intersect on the same side of the mirror as the light source. This means all images in a convex mirror are virtual. All virtual images are upright and diminished.
  • The image formed by the convex mirror is diminished because virtual light rays travel less before they appear to meet.
  • The image formed by the concave mirror is inverted, enlarged and real.
  • Concave mirrors are converging mirrors, which means that real light rays will intersect (if the object is beyond the focal point) on the same side of the mirror as the light source.
  • The image formed by the concave mirror is enlarged because real reflected light rays travel longer before they meet.
  • The largest image will be produced when the object is as close to f as possible. At this position, the reflected rays are nearly parallel, causing a large di and hi.
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4
Q

Distinguishing between real and virtual images

A
  • A real image occurs where light rays converge, whereas a virtual image occurs where light rays only appear to converge.
  • A virtual image is upright, and cannot be projected onto a screen.
  • A real image is inverted, and can be projected onto a screen.
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5
Q

Light shining through two slits and producing a pattern of multiple slits.

A
  • Bright bands are due to constructive interference, which are regions where the crest of one wave combines with the crest of another. The path
    difference is either zero (central band) or a whole
    number of wavelengths.
  • Dark bands are due to destructive interference,
    which are regions where the crest of one wave combines with the
    trough of another wave. The path difference is
    either half a wavelength or an odd mulitiple of half
    wavelengths.
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6
Q

Sound heard walking past two speakers

A

The person walking in a straight line closer to the speakers will hear the loud and soft sounds at shorter distances as compared to the person walking in a straight line further away. This is because the antinodal lines fan out from the speakers, and therefore the further away one is from the speakers, the further apart the loud and soft sounds heard will be.

  • Louder sound is heard where waves from two sources constructively interfere. This is because the path difference between the two sound waves is nλ, so the total amplitude of the resultant wave increases, creating a louder sound.
  • Quieter sound is heard where waves from two sources deconstructively interfere. This is because the path difference between the two sound waves is 1/2λ, so the total amplitude of the resultant wave decreases, creating a quieter sound.
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7
Q

Find if point x is a node or antinode

A
  1. Calculate the frequency (f = 1/T)
  2. Calculate the wavelength (v = fλ)
  3. Calculate the path difference (distance to point x from 1 opening - distance to point x from other opening)
  4. Path difference / wavelength
  5. If the result is half a wavelength, point x must be a node. If the result is a whole wavelength, point x must be an antinode.
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8
Q

Images formed by concave lenses

A

Regardless of the position of the object, the image will always be virtual, upright and diminished. The image is located on the object’s side of the lens.

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9
Q

Images formed by convex lenses

A

(Same as images formed in concave mirrors)

  1. The object is beyond C
    An inverted, real and diminished image will be formed.
  2. The object is at C
    An inverted, real image will be formed, that is the same size as the object.
  3. The object is between C & F
    An inverted, real and enlarged image will be formed.
  4. The object is at F
    No image will be formed.
  5. The object is in front of F
    An upright, virtual and enlarged image will be formed, that is located on the object’s side of the lens.
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10
Q

Images formed in concave mirrors

A
  1. The object is beyond C
    An inverted, real and diminished image will be formed.
  2. The object is at C
    An inverted, real image will be formed, that is the same size as the object.
  3. The object is between C & F
    An inverted, real and enlarged image will be formed.
  4. The object is at F
    No image will be formed.
  5. The object is in front of F
    An upright, virtual and enlarged image will be formed, that is behind the mirror (eg. dentist mirror).
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11
Q

Images formed in convex mirrors

A

Regardless of the position of the object, the image will always be virtual, upright and diminished. The image is located behind the mirror.

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12
Q

Equation for critical angle

A

sinθc = n2 sin90°/n1

Where,

  • θc is the critical angle
  • n2 is the absolute refractive indices for medium 2
  • n1 is the absolute refractive indices for medium 1
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13
Q

Total internal reflection

A

For total internal reflection to occur, the angle of incidence has to be greater than the critical angle, and the light ray must be travelling from a medium of a higher refractive index (optical density) to one of a lower refractive index.

Therefore, when rays of light at a very large angle of incidence hit the cold/hot air interface, the angle of incidence in greater than the critical angle, which causes all the light rays to undergo total internal reflection.

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14
Q

Critical angle

A

The critical angle is the angle of incidence (θ1), when the angle of refraction is 90°.

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15
Q

Refractive index

A

If two different mediums (water/air/glass) have different refractive indices, light will refract differently at the boundary between them.

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16
Q

When refraction occurs…

A

The frequency of the wave does not change when crossing the boundary. From v = fλ, velocity is directly proportional to wavelength. Therefore, as the frequency remains constant, and the wave velocity increases, the wavelength will increase also.

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17
Q

Refraction

A
  • Refraction is the change in the speed of a light wave as it travels from one medium into another, causing it to change direction.
  • When light travels into a medium of greater density, it refracts towards the normal as the medium of higher density has a higher refractive index.
  • When light travels into a medium of less density, it defracts away from the normal as the medium of less density has a lower refractive index.
  • Rays of light from (the bottom of the pool) are refracted (bent) before reaching a persons eye. Our brain makes us think that the rays are coming straight backwards from where they are, making the object appear closer to the surface of the water than it actually is.
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18
Q

Equation for path difference

with nodal lines

A

PS1 - PS2 = (n - 1/2) λ

Where,

  • PS1 is the number of lines between source 1 and the point
  • PS2 is the number of lines between source 2 and the point
  • n is the nodal line number
  • λ is the wavelength

Note - Nodal line number is taken from the central dotted line

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19
Q

Equation for path difference

with anti-nodal lines

A

PS1 - PS2 = nλ

Where,

  • PS1 is the number of lines between source 1 and the point
  • PS2 is the number of lines between source 2 and the point
  • n is the antinodal line number
  • λ is the wavelength

Note - Antinodal line number is taken after the central dotted line (middle line = 0)

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20
Q

Diffraction

A
  • Diffraction is the process of waves spreading as they pass through a gap
  • Maximum diffraction occurs when the wavelength of the wave is approximately equal to the size of the gap.
  • The wavelength of a wave does not change when it is diffracted.

eg. Sound waves have a longer wavelength compared to light waves. Because of this longer wavelength, the soundwaves are more able to diffract around barriers (as the openings are around the same order of size as the wavelength of the sound waves). Comparatively, lightwaves cannot diffract as much around barriers. Therefore, often people can hear objects, but not see them.

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21
Q

Reflection and transmission of a pulse

Heavy string to light string

A

Heavy string to light string

  • Pulse moves slower along heavy string and faster along the light string.
  • A small pulse reflected and the same way up as the original pulse moves back along the heavy string.
  • The amplitude will be smaller when reflected due to a decrease in energy (as not all the energy is reflected).
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22
Q

Reflection and transmission of a pulse

Light string to heavy string

A

Light string to heavy string

  • Pulse moves slower along heavy string and faster along the light string.
  • A small pulse reflected and upside down to the original pulse moves back along the light string.
  • The amplitude will be smaller when reflected due to a decrease in energy (as not all the energy is reflected).
  • As the pulses in the light string are travelling faster, they are further from the boundary when reflected.
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23
Q

Effect of adding an ammeter in parallel

A
  • Adding an ammeter in parallel will short circuit the connected bulb, as all the current will go through the ammeter instead due to the low resistance of the ammeter (0V across connected bulb). This will cause the connected bulb to go out, and result in the total resistance of the circuit decreasing.
  • This will cause the overall current in the circuit to increase (I = V/R)
  • Therefore, the voltage of any bulbs connected in series will increase (V = IR) and the voltage of any bulbs connected in parallel will decrease (Voltage is directly proportional to resistance (V = IR), therefore when the current is the same, as resistance decreases, voltage decreases also).
  • Therefore, the power used by any bulbs connected in series will have more power, and any bulbs connected in parallel will have less power (P = IV).
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24
Q

Why bulb A in parallel will be brighter than bulb B if it has a higher resistance.

A
  • Brightness is determined by power.
  • As they are in parallel, the voltage across each bulb is the same. However, the current through bulb B is higher, due to having a lower resistance, as I = V/R. Therefore, bulb B will use more power, and therefore be brighter, as P = IV.
25
Q

How to change the velocity of a particle in an electric field

A
  1. Increase the charge (q)
    - This causes the charge to experience a greater force as F = Eq. This causes it to have greater acceleration (assuming the increase in charge adds negligible mass) and a higher velocity.
  2. Increase the electric field strength (E) by decreasing the distance between the plates or by increasing the voltage across the plates
    - This causes the charge to experience a greater force as F = Eq. This causes it to have greater acceleration and a higher velocity.
26
Q

Force on electromagnetic loop (flying fox)

A
  • The force on the wire connected to the positive
    terminal is equal and opposite the force on the
    wire connected to the negative terminal. Therefore the magnetic force has a net force of 0, and will not change the time to complete the flying
    fox.
  • This force is so small, even if it was not
    canceled out it would be negligible.
27
Q

Affect of adding another bulb in parallel to a circuit on the same branch as another component

A
  • Adding another bulb on the same branch as another component (in series) will cause the resistance of the branch to increase, and the overall resistance of the circuit to increase ((1/RT = 1/R1 + 1/R2).
  • This will cause the overall current in the circuit to decrease (I = V/R)
  • Therefore, the voltage of any bulbs connected in series will decrease (V = IR) and the voltage of any bulbs connected in parallel will increase (Voltage is directly proportional to resistance (V = IR), therefore when the current is the same, as resistance increases, voltage increases also).
  • As the bulb and the other component on the same branch are in series, they only get half of what is received by the component on the other branch.
  • Increased voltage causes more power dissipation, which will cause the component on the other branch to become brighter (P = V^2/R). As a result, it is more likely to blow, as it will receive more than its initial voltage.
28
Q

Size of force on rod sliding left to right

A
  • The electric field between the plates is constant, as the distance between each plate and voltage across the plates is constant (E = V/d). Therefore, the force is constant.
  • The force acts uniformly across the rod, as the charge is uniformly distributed. This means that each part of the rod will experience the same force and no bending will occur (F = Eq).
29
Q

Charge of an electron

A

-1.6 x 10^-19

30
Q

Unit for electric field strength

A

Vm^-1

31
Q

Resistance of components in parallel, where A uses more power than B

A
  • Because both components are in parallel, they have the same voltage. Therefore, as the voltage is constant and A uses more power than B, there is more current passing through A (P = IV). As a result, A has less resistance (R = V/I).
32
Q

Force on a charged particle in a magnetic field

A
  • Magnetic fields need to be perpendicular (90°) to the direction of motion of a charged particle to maximise the force acting upon it. Otherwise, the magnetic force will have little to no effect.
33
Q

Current in electromagnetic loop (swing)

A
  • When the entire loop is inside the magnetic field, the current is 0, as both sides of the loop are cutting across the magnetic field (not parallel) and have the same induced voltage. One voltage pushes the charges clockwise, while the other voltage pushes the charges anti clockwise. This results in a net voltage of 0.
  • If the speed doubles, the induced voltage doubles (V = Bvl). Assuming the resistance is constant, this will cause the current to double also (I = V/R).
34
Q

Conversion from mV or kV to V

A
  • Divide mV by 1000

- Multiply kV by 1000

35
Q

Electric field in a wire

A
  • The electric force on an electron is F = Eq, and therefore if the charge and electric field strength are constant, the force will be constant also
  • If another battery is added in series, this will double the voltage, and if the wire length is halved, this will double the electric field strength (E = V/d). Therefore, the electric force on the wire will be 4x as large.
36
Q

Calculations for a rod connected to wire moving across a magnetic field

A
  1. Calculate the induced voltage (V = BvL)
  2. Calculate the induced current through the rod (I = V/R)
  3. Calculate the force applied to the rod (F = BIL)
37
Q

Rod connected to wire moving across a magnetic field

A
  • The wires used to connect a light are composed of free to move charges, which are moving perpendicular to a magnetic field. From F = Bvq, these charges will experience a magnetic force. This will cause positive charges to experience a force to the top of the wire, and negative charges to the bottom. As work is done to cause this separation of charges, which forms an electric field. This will result in an induced voltage.
  1. If one conductor is outside the magnetic field
    - As one conductor is outside the magnetic field no voltage is induced in the conductor, however, a voltage is induced in the moving conductor as it moves across the magnetic field (V = BvL), so a current would flow (in a clockwise direction).
  2. If both conductors are in the same magnetic field
    - As both conductors are inside the magnetic field, of the same length and with the same speed, the same voltage is induced in both conductors, so no current would flow. (Polarity in both conductors is the same (positive at the top), meaning that no current will flow).
  3. If one conductor is not moving, but the other is
    - As one conductor is not moving, no voltage is induced in the conductor, (V = BvL), however, a voltage is induced in the moving conductor, so a current would flow (in a clockwise direction).
38
Q

Affect of adding another bulb in parallel to a circuit on a separate branch

A
  • Adding another bulb in parallel will cause the overall resistance of the circuit to decrease ((1/RT = 1/R1 + 1/R2).
  • This will cause the overall current in the circuit to increase (I = V/R)
  • Therefore, the voltage of any bulbs connected in series will increase (V = IR) and the voltage of any bulbs connected in parallel will decrease (Voltage is directly proportional to resistance (V = IR), therefore when the current is the same, as resistance decreases, voltage decreases also).
  • Therefore, the power used by any bulbs connected in series will have more power, and any bulbs connected in parallel will have less power (P = IV).
39
Q

Power of components

A
  • The energy per second of each component is the power output
  • Power relates to voltage and current (P = IV)

Parallel
If both components are on the same branch…
- They both have the same current through them
- Voltage is directly proportional to resistance (V = IR), therefore when the current is the same, as resistance increases, voltage increases also. Therefore, the voltage with the highest resistance would have a greater voltage across it and therefore use more energy per second.

40
Q

Significant figures

A
  • The final answer cannot be any more accurate than the least number of significant figures in the question.
41
Q

Calculating voltage across components

A
  • In series, the voltage splits between components and the current remains the same
  • In parallel, the voltage remains the same between components and the current splits equally between the number of electrical pathways (unless the resistances on each path are different, in which case it can be a different current across each pathway)

(With component in parallel circuit)

  1. Calculate the total resistance of the circuit
  2. Calculate the total current in the circuit
  3. Calculate the total voltage in the circuit
  4. Calculate the voltage across the other component(s) you are not trying to calculate
  5. Subtract the answer from the total voltage in the circuit

(With component in series circuit)
1. Use V = IR with the resistance of component and the total current

42
Q

A charge moving through a uniform electric field

A
  • The forces acting on the charge are electrical force (upwards) and weight force caused by gravity (downwards)
  • When the charge is suspended between the plates, the electrical force is equal and opposite to the weight force (Eq = mg)
  • To allow for suspension, the electrical force must be acting upwards, and therefore the type of charge must be opposite to the charge on the top plate.
  • If the distance between the plates is decreased while the charge and voltage across the plates remains constant, this will increase the electric field strength (E = V/d) and therefore increase the electric force acting on the charge. This will cause the negative charge to accelerate towards the positive plate, as the forces acting on it are unbalanced (electric force > gravitational force).
43
Q

Electron in an electric field.

A
  • When the electron is near the negatively charged plate, it has electric potential energy.
  • When the electron is free to move, it is accelerated by the electric field towards the positive plate and the electric potential energy stored inside the particle is transformed into kinetic energy.

NOTE - Only movement parallel to the field lines affects the electric potential energy. Movement perpendicular to the field lines does not affect a particle’s potential energy.

To move a particle in the opposite direction, work must be done to overcome the electrostatic forces on the particle. The work done on the particle is stored as electric potential energy.

The change in a particle’s electric potential energy can be calculated using ΔEp = Eqd

44
Q

Effect of an electric field on a particle entering the field perpendicular to the field lines

A

If a particle experiences an electrostatic force, the force will make the particle accelerate. If a particle enters the electric field perpendicular to the field lines, it will travel along a parabolic path. This is because whilst the particle’s velocity parallel to the field increases, the particle’s velocity perpendicular to the field remains unchanged.

45
Q

Calculating the induced current

A
  1. Calculate the induced voltage (V = BvL)

2. Calculate the induced current (V = IR)

46
Q

Force that causes acceleration in circular motion

A

The tension in the cord provides centripetal force, which is the unbalanced force that causes acceleration.

47
Q

Equation to calculate centripetal force and acceleration

A 
Fc = mv^2 / r
Ac = v^2 / r

Where,
v is the constant speed
r is the radius of the circle

  • To calculate v use v = d/t,

Where,
d is the circumference of the circle
t is the time for one revolution

48
Q

How an object on a slope can remain stationary

A

As the object is stationary, this means that the forces are balanced (in equilibrium). Therefore, the component of gravity down the slope is equal to the friction force, and the component of gravity into the slope is equal to the reaction force.

49
Q

Calculate power to decelerate from Vf, Vi and time

A

Using P = △E/t

  1. △E is change in kinetic energy (1/2mv^2)
  2. Solve for kinetic energy using Vf and Vi separately
  3. Subtract the initial velocity solution from the final velocity solution to find △E
  4. Substitute back into main equation with time to solve

Using P = W/t

  1. Solve for acceleration (a = △v/△t)
  2. Solve for force (F = ma)
  3. Solve for distance (Vf^2 = Vi^2 + 2ad)
  4. Solve for work (W = F x d)
  5. Substitute back into main equation with time to solve
50
Q

Circular motion

A
  • There is a centripetal force which acts towards the centre of the curve which causes the object to accelerate towards the centre at a constant speed along a circular path, by changing the direction of the objects velocity.
  • As the net force is 0, it will continue to travel at a constant speed at a tangent to the curve in a straight line (acting at 90 degrees to the direction of travel of object, so it does not change the size of the velocity).
51
Q

Change in momentum (with protection) question

A
  • When the event occurs, the same change in momentum occurs. However, with the protection the stopping time is greater, than without the protection (the protection gets hit first before body).
  • △p = F△t, therefore, as the impulse is the same, as the time increases, the force will decrease, and therefore the event with the protection will produce less force. This means that it is less likely that an injury will be sustained.
  • This assumes that the speed is the same so the mass of the protection does not significantly affect the momentum of the body, so the impulse will be constant.

(eg. or bending knees when jumping)

52
Q

Work done on a spring

A

Ep = 1/2 kx^2

Where,
Ep is the elastic potential energy
k is the spring constant
x is the distance extended

53
Q

Conservation of momentum

A

Momentum is only conserved if there is no external net force (eg. gravity or air resistance). It can be assumed that there is no friction with ice.

54
Q

Unit for impulse (△p)

A

The unit for impulse is N s

55
Q

Unit for momentum (p)

A

The unit for momentum is kgms^-1

56
Q

Projectile motion calculations

A
  1. Travel time
    Calculate the time take to reach the peak of its motion, then double it.
    - Use initial vertical velocity, final vertical velocity and acceleration
    (Vf = Vi + at)
  2. Distance travelled
    - Use initial horizontal velocity and travel time
    (v = d/t)
  3. Height reached
    - Use initial vertical velocity, travel time and acceleration
    (d = Vit + 1/2at^2)
57
Q

Projectile motion

A
  • Follows a parabolic curve
  • Gravity is the only force acting on the projectile, applying -9.8ms^-2 to the projectile in a downwards direction. Because of this…
  • As the projectile moves upwards, the force of gravity acts in the opposite direction to the vertical motion of the projectile and decreases its vertical velocity. This causes the projectile to experience a negative acceleration (deceleration), until the vertical velocity of the projectile is equal to 0.
  • At this point, the motion of the projectile has covered half of its total distance, and taken half of its total time.
  • As the projectile moves downwards, the force of gravity acts in the same direction to the vertical motion of the projectile and increases its vertical velocity. This causes the projectile to experience a positive acceleration.
  • As there is no horizontal force acting on the projectile, the horizontal velocity of the projectile remains constant and therefore, there will be no horizontal acceleration.
58
Q

Conversion from kWh to J

A

Multiply kWh by 3.6 x 10^6

59
Q

Conversion from mT or μT to T

A
  • Divide mT by 1000

- Divide μT by 1,000,000

Science (12 decks)

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