The image distance is 33.3 cm while the image height is 14.2 cm.
<h3>What is a converging lens?</h3>
A converging lens will always have a positive focal length hence, we have to find the object distance as follows;
1/f = 1/v + 1/u
1/12 = 1/v + 1/20
1/v = 1/12 - 1/20
1/v = 0.08 - 0.05
v =33.3 cm
Now;
Magnification = 33.3 cm/20.0 cm =1.67
M = Image height/Object height
1.67 = Image height/8.50 cm
Image height = 1.67 * 8.50 cm
Image height = 14.2 cm
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Answer:
200N appox
Explanation:
Step one:
Given data
mass of ball= 0.143kg
initial velocity of ball u= 42m/s
final velocity of the ball= 49m/s
time of impact= 0.005s
Step two
From the relation Ft=mΔv
we can find the average force as
F=mΔv/t
substitute
F=0.143*(49-42)/0.005
F=0.143*7/0.005
F=1.001/0.005
F=200.2
F= 200N appox
The average force is 200N
Its accelarating from 1-10 i know that because the line is going up. i hope i helped
Answer:
<u>Amplitude - remains the same</u>
<u>Frequency - increases</u>
<u>Period - decreases</u>
<u>Velocity - remains the same.</u>
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Explanation:
The amplitude of the wave remains the same since you are not changing the distance your hand moves and the amplitude of the wave depends on how much distance your hand covers while moving.
The frequency of your wave increases since now you are moving your hand more number of times in the same period i.e. your hand is moving faster in one second. So, the frequency of your wave increases.
The period is the time taken by the wave to travel a certain distance. Since your hand is now moving faster, the wave will travel faster and will take less time to cover the same distance hence, we can say that its period will decrease.
The velocity of a wave depends on the medium in which it is travelling. Your wave was previously travelling in air and the new wave is also travelling in the same medium so the velocity of the wave remains unchanged.
The potential difference across the parallel plate capacitor is 2.26 millivolts
<h3>Capacitance of a parallel plate capacitor</h3>
The capacitance of the parallel plate capacitor is given by C = ε₀A/d where
- ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m,
- A = area of plates and
- d = distance between plates = 4.0 mm = 4.0 × 10⁻³ m.
<h3>Charge on plates</h3>
Also, the surface charge on the capacitor Q = σA where
- σ = charge density = 5.0 pC/m² = 5.0 × 10⁻¹² C/m² and
- a = area of plates.
<h3>
The potential difference across the parallel plate capacitor</h3>
The potential difference across the parallel plate capacitor is V = Q/C
= σA ÷ ε₀A/d
= σd/ε₀
Substituting the values of the variables into the equation, we have
V = σd/ε₀
V = 5.0 × 10⁻¹² C/m² × 4.0 × 10⁻³ m/8.854 × 10⁻¹² F/m
V = 20.0 C/m × 10⁻³/8.854 F/m
V = 2.26 × 10⁻³ Volts
V = 2.26 millivolts
So, the potential difference across the parallel plate capacitor is 2.26 millivolts
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