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sergij07 [2.7K]
3 years ago
10

at what position an object be placed in front of a concave mirror of radius of curvature 40cm so that an erect image of magnific

ation of 3 be produced ?
Physics
1 answer:
Thepotemich [5.8K]3 years ago
8 0
  • C=40cm
  • f=C/2=40/2=20

  • m=3

\\ \sf\longmapsto m=\dfrac{f}{f-u}

\\ \sf\longmapsto 3=\dfrac{20}{20-u}

\\ \sf\longmapsto 3(20-u)=20

\\ \sf\longmapsto 60-3u=20

\\ \sf\longmapsto 3u=40

\\ \sf\longmapsto u=13.3cm

  • m=magnification
  • f is focal length
  • C is radius of curvature
  • u is object distance
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Answer:

d)

Explanation:

Newton's 3rd Law tells us that for every action we will have a reaction equal in magnitude and opposite in direction. In a collision between obects A and B this means that the force on A applied by B will be equal in magnitude than the force on B applied by A, which leaves d as the true answer and the rest as false.

8 0
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An LC circuit is built with a 20 mH inductor and an 8.0 PF capacitor. The capacitor voltage has its maximum value of 25 V at t =
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Answer:

a) the required time is 0.6283 μs

b) the inductor current is 0.5 mA

Explanation:

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we determine the angular velocity;

ω = 1 / √( LC )

ω = 1 / √( ( 20 × 10⁻³ H ) × ( 8.0 × 10⁻¹² F) )

ω = 1 / √( 1.6 × 10⁻¹³  )

ω = 1 / 0.0000004

ω = 2.5 × 10⁶ s⁻¹

a) How much time does it take until the capacitor is fully discharged for the first time?

V_m =  V₀sin( ωt )

we substitute

25V =  25V × sin( 2.5 × 10⁶ s⁻¹ × t )

25V =  25V × sin( 2.5 × 10⁶ s⁻¹ × t )

divide both sides by 25 V

sin( 2.5 × 10⁶ × t ) = 1

( 2.5 × 10⁶ × t ) = π/2

t = 1.570796 / (2.5 × 10⁶)

t = 0.6283 × 10⁻⁶ s

t = 0.6283 μs

Therefore, the required time is 0.6283 μs

b) What is the inductor current at that time?

I(t) = V₀√(C/L) sin(ωt)

{ sin(ωt) = 1 )

I(t) = V₀√(C/L)

we substitute

I(t) = 25V × √( ( 8.0 × 10⁻¹² F ) / ( 20 × 10⁻³ H ) )

I(t) = 25 × 0.00002

I(t) = 0.0005 A

I(t) = 0.5 mA

Therefore, the inductor current is 0.5 mA

8 0
3 years ago
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R=W=mg=(65 kg)(9.8 m/s^2)=637 N

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