Physics
23.09.2021 13:51
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the radius of the earth is 6380 km and the height of the mount everest is 8848m

the radius of the earth is 6380 km and the height of the mount everest is 8848m . if the value of the acceleration due to gravity on the top of the mount everst is 9.77m/s²,calculate the value of acceleration due to gravity on the surface of the earth .what is the weight of a body of mass 50 kg on the top of mt.everest?
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ghari112345
ghari112345
4,8(63 marks)

Explanation:

g = GM / R²

9.77 m/s² = GM / (6.38×10⁶ m + 8848 m)²

GM = 3.99×10¹⁴ m³/s²

At the surface of the earth:

g = GM / R²

g = (3.99×10¹⁴ m³/s²) / (6.38×10⁶ m)²

g = 9.80 m/s²

At the top of Mount Everest, the weight of a 50 kg object is:

F = mg

F = (50 kg) (9.77 m/s²)

F = 489 N

cyanezc1313
cyanezc1313
4,4(98 marks)

Torque, \tau=1.27\times 10^{-5}\ N-m

Explanation:

Given that,

Number of turns in a circular coil, N = 400

Radius of the circular coil, r = 0.65 cm

Current in the coil, I = 1.2 mA

Uniform magnetic field in the circular coil, B = 0.2 T

Torque in the circular coil is given by :

\tau=NIAB\ \sin\theta

For maximum torque, \sin\theta=1

\tau=400\times 1.2\times 10^{-3}\times \pi (0.65\times 10^{-2})^2\times 0.2\\\\\tau=1.27\times 10^{-5}\ N-m

So, the net torque about the vertical axis is 1.27\times 10^{-5}\ N-m. Hence, this is the required solution.

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