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anyanavicka [17]
3 years ago
8

At what distance will the weight of a body be halved , Earth's radius=6.4×10^6

Physics
1 answer:
Sholpan [36]3 years ago
5 0

A body of mass m has weight

F = GMm/r²

on the surface of the Earth, where G is the universal gravitational constant, M is the mass of the Earth, and r is it's radius.

If the weight is to be halved, then we have

1/2 F = 1/2 GMm/r² = (1/√2)² GMm/r² = GMm/(√2 r²)

so the distance between the body and the planet's center needs to be

√2 × 6.4 × 10⁶ m ≈ 9.1 × 10⁶ m

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A point charge Q is held at a distance r from the center of a dipole that consists of two charges ±q separated by a distance s.
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Answer:

a) the magnitude of the force is

F= Q(\frac{kqs}{r^3}) and where k = 1/4πε₀

F = Qqs/4πε₀r³

b)  the magnitude of the torque on the dipole

τ = Qqs/4πε₀r²

Explanation:

from coulomb's law

E = \frac{kq}{r^{2} }

where k = 1/4πε₀

the expression of the electric field due to dipole at a distance r is

E(r) = \frac{kp}{r^{3} } , where p = q × s

E(r) = \frac{kqs}{r^{3} } where r>>s

a) find the magnitude of force due to the dipole

F=QE

F= Q(\frac{kqs}{r^3})

where k = 1/4πε₀

F = Qqs/4πε₀r³

b) b) magnitude of the torque(τ) on the dipole is dependent on the perpendicular forces

τ = F sinθ × s

θ = 90°

note: sin90° = 1

τ = F × r

recall  F = Qqs/4πε₀r³

∴ τ = (Qqs/4πε₀r³) × r

τ = Qqs/4πε₀r²

8 0
3 years ago
In a closed system, a cart with a mass of 1.5 kg is rolling to the right at 1.4 m/s, while another cart of mass 1.0 kg is rollin
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Answer: The correct option is (c.).

Explanation:

Mass of the cart A= 1.5 kg

Velocity of Cart A = 1.4 m/s towards right

Mass of the cart B = 1.0 kg

Velocity of Cart B = 1.4 m/s towards left

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P_A=1.5 kg\times 1.4 m/s=2.1 kg m/s

P_B=1.0 kg\times (-1.4m/s)=-1.4 kg m/s

(Negative sign means velocity of the cart is in opposite direction of that of the cart A)

Total Momentum =P_A+P_B=2.10 kg m/s-1.40 kg m/s=0.70 kg m/s

Hence, the correct option is (c.).

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