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# Tutor profile: Nitesh B.

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Nitesh B.
Master , Certified Teacher, Educator and Tutoring for 7 Years , Academic Writer
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## Questions

### Subject:Physics (Newtonian Mechanics)

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Question:

A 733-kg satellite is in a circular orbit about Earth at a height above Earth equal to Earth's mean radius.v (a) Find the satellite's orbital speed. ___m/s (b) Find the period of its revolution. ___h (c) Find the gravitational force acting on it. ___N

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Nitesh B.

Ans : R = radius of the orbit in m (2*6371*10^3) T = orbit time in seconds M = mass of earth in kg (6*10^24) m = mass of satellite in kg G = gravitation constant (6,67*10^-11 m^3/(kgs^2)) v = orbital velocity m/s Now setting the equations for the centrifugal force and the gravitational force equal mv^2/R = G*M*m/R^2 yields the equation for the period time: T = √(4pi^2*a^3/(G(M+m))) and plugging in the values for an satellite orbiting the earth: T = 166*10^-6 min.*√((h (km) + 6371)^3) b) T = 238,76 minutes = 14325 seconds. ---- a) The velocity = s/v = 2pir/T v = 2pi*2*6371*10^3/14325 = 5588,85 m/s so speed = 5588,85 m/s ------- c) The gravitational force between the earth and the satellite is equal to the centrifugal force = mv^2/R F = 733*5588,85^2/(2*6371*10^3) F = 1796,8 N (using F = G*M*m)/R^2 yields 1806 N)

### Subject:Physics (Electricity and Magnetism)

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Question:

A 6×10^-3kg bullet is fired horizontally into a 20kg block of wood suspended by a rope from the ceiling. the block swings in an arc rising 3×10^-3m above its lowest position. what was the velocity of the bullet?

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Nitesh B.

Ans : As per the question At the start, only energy in the system is the KE in the bullet right after impact. h= 3×10^-3m = 0.003m So After impact the velocity of the bullet is v =√ (2gh) = √ (2*9.8*0.003) = 0.24 m/s By the conservation of momentum m(bullet)v(bullet,before) = m(bullet)v + m(block)v v(bullet,before) = [m(bullet) + m(block)]v/m(bullet) (0.003 kg + 6 kg)(0.24 m/s)/(0.003 kg) 6.003 * 0.24 = 0.003* v1 v1 = 480.24 m/s = So, the final answer is 4.8024 × 102 m/s

### Subject:Physics

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Question:

What is the strength of the electric field 4.4 cm from a small plastic bead that has been charged to -7.2 nC ? Express your answer to two significant figures and include the appropriate units.

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Nitesh B.

Ans : As per question I use E = kQ/r^2 so k = 9 x 10^ 9 Q = -7.2 nC = -0.0000072 C r = 4.4 cm so E = 9 x 10^ 9 *-0.0000072 C /19.36 E = 3.34711 × 10^3

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