Answer:
t = 4.17 hours
Explanation:
given,
The distance between Sun and Neptune, d = 4.5 billion Km
= 4.5 x 10⁹ Km
= 4.5 x 10¹¹ m
The velocity of light, c = 3 x 10⁸ m/s
The velocity is always equal to displacement by the time.
<em>V = d / t m/s</em>
∴ t = d / V
= 4.5 x 10¹¹ m / 3 x 10⁸ m/s
= 15,000 s
= 4.17 h
Hence, the time taken by the light rays to reach the Neptune is, t = 4.17 h
By Newton's second law, the net vertical force acting on the object is 0, so that
<em>n</em> - <em>w</em> = 0
where <em>n</em> = magnitude of the normal force of the surface pushing up on the object, and <em>w</em> = weight of the object. Hence <em>n</em> = <em>w</em> = <em>mg</em> = 196 N, where <em>m</em> = 20 kg and <em>g</em> = 9.80 m/s².
The force of static friction exerts up to 80 N on the object, since that's the minimum required force needed to get it moving, which means the coefficient of <u>static</u> friction <em>µ</em> is such that
80 N = <em>µ</em> (196 N) → <em>µ</em> = (80 N)/(196 N) ≈ 0.408
Moving at constant speed, there is a kinetic friction force of 40 N opposing the object's motion, so that the coefficient of <u>kinetic</u> friction <em>ν</em> is
40 N = <em>ν</em> (196 N) → <em>ν</em> = (40 N)/(196 N) ≈ 0.204
And so the closest answer is C.
(Note: <em>µ</em> and <em>ν</em> are the Greek letters mu and nu)
Answer:
66.2 sec
Explanation:
C₁ = 1.0 F
C₂ = 1.0 F
ΔV = Potential difference across the capacitor = 6.0 V
C = parallel combination of capacitors
Parallel combination of capacitors is given as
C = C₁ + C₂
C = 1.0 + 1.0
C = 2.0 F
R = resistance = 33 Ω
Time constant is given as
T = RC
T = 33 x 2
T = 66 sec
V₀ = initial potential difference across the combination = 6.0 Volts
V = final potential difference = 2.2 volts
Using the equation


t = 66.2 sec
Answer:
The magnitude of the electric field and direction of electric field are
and 75.36°.
Explanation:
Given that,
First charge 
Second charge
Distance between two corners r= 50 cm
We need to calculate the electric field due to other charges at one corner
For E₁
Using formula of electric field

Put the value into the formula


For E₂,
Using formula of electric field

Put the value into the formula


We need to calculate the horizontal electric field



We need to calculate the vertical electric field



We need to calculate the net electric field

Put the value into the formula



We need to calculate the direction of electric field
Using formula of direction



Hence, The magnitude of the electric field and direction of electric field are
and 75.36°.