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2 years ago

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

Physics

1 answer:

Sholpan [36]2 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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Read 2 more answers

A 0.50 kg object is attached to a spring with force constant 157 N/m so that the object is allowed to move on a horizontal frict

Genrish500 [490]

Explanation:

We have,

Mass of an object is 0.5 kg

Force constant of the spring is 157 N/m

The object is released from rest when the spring is compressed 0.19 m.

(A) The force acting on the object is given by :

F = kx

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3 years ago

At 4.00 l, an expandable vessel contains 0.864 mol of oxygen gas. how many liters of oxygen gas must be added at constant temper

svet-max [94.6K]

Givens
=====
V = 4.00 L
T = 273oK We're assuming the temperature does not change, just the pressure.
n = 0.864 moles
R = 8.314 joules / mole * oK
P = ?????

Formula
======
PV = n*R*T
P = n*R*T/V
P = 0.864 * 8.314 * 273 / 4
P = 490 kpa

You have to add 1.6 – 0.864 = 0.736 moles of gas.

We have to assume that the temperature and pressure remain the same when we add the 0.736 moles of gas. We are now looking for the volume.

PV = n*R*T

V = 0.736 * 8.314 * 273 / 490

V = 3.41 L Remember this is at about 4 atmospheres so we have to convert to Standard Pressure.

Total Volume = 3.41 + 4.00 = 4.41

V1 * P1 = V2 * P2

P1 = 490 kPa

P2 = 101 kPa

V1 = 7.41 L

V2 = ????

7.41* 490 = V2 * 101 V2 = 7.41 * 490 / 101 V2 = 35.94 L