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postnew [5]
2 years ago
9

What equal amounts of positive charge would have to be placed on the Earth and on the Moon to neutralize their gravitational att

raction? Do you need to know the Moon’s distance to solve this problem? Why or why not? (b) How many metric tons of hydrogen would be needed to provide the positive charge calculated in part (a)? The molar mass of hydrogen is 1.008 g/mol.
Physics
1 answer:
jonny [76]2 years ago
6 0

Answer:

Fg = G M m / R^2       gravitational attraction

Fe = k Q^2 / R^2       electric repulsion

G M m = k Q^2

Q = (G M m / k)^1/2

Since m = .0123 M

Q = (.0123 G / k)^1/2 * M

Q = (.0123 * 6.67 * 10E-11 / 9 * 10E9)^1/2 * 5.98 * 10E24

Q = 5.71 * 10E13 C       charge required on each body

n = 5.71 * 10E13 / 1.6 * 10E-19 = 3.57 * 10E32 atoms

N = 3.57 * 10E32 / 6.02 * 10E23  = 5.93 * 10E8 g-mol = 5.93 * 10E*5 kg-mol

Since 1 metric ton = 1000 kg

One would need 593 metric tons of hydrogen

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2 years ago
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A student pushes a 0.2 kg box against a spring causing the spring to compress 0.15 m. When the spring is released, it will launc
german

Answer:

The maximum height the box will reach is 1.72 m

Explanation:

F = k·x

Where

F = Force of the spring

k = The spring constant = 300 N/m

x  = Spring compression or stretch = 0.15 m

Therefore the force, F of the spring = 300 N/m×0.15 m = 45 N

Mass of box = 0.2 kg

Work, W, done by the spring = \frac{1}{2} kx^2 and the kinetic energy gained by the box is given by KE = \frac{1}{2} mv^2

Since work done by the spring = kinetic energy gained by the box we have

\frac{1}{2} mv^2 =  \frac{1}{2} kx^2  therefore we have v = \sqrt{\frac{kx^2}{m} } = x\sqrt{\frac{k}{m} } = 0.15\sqrt{\frac{300}{0.2} } = 5.81 m/s

Therefore the maximum height is given by

v² = 2·g·h or h = \frac{v^2}{2g} = \frac{5.81^{2} }{2*9.81} = 1.72 m

6 0
3 years ago
A flat coil having 160 turns, each with an area of 0.20 m 2, is placed with the plane of its area perpendicular to a magnetic fi
S_A_V [24]

Answer:

10.2 Watt

Explanation:

N  = number of turns in flat coil = 160

A  = area = 0.20 m²

B₀= initial magnetic field = 0.40 T

B  = final magnetic field = - 0.40 T

Change in magnetic field is given as

ΔB = B - B₀ = - 0.40 - 0.40 = - 0.80 T

t  = time taken for the magnetic field to change = 2.0 s

Induced emf is given as

E = \frac{- N A \Delta B}{t}

E = \frac{- (160) (0.20) (- 0.80)}{2}

E = 12.8 volts

R = Resistance of the coil = 16 Ω

Power is given as

P = \frac{E^{2}}{R}

P = \frac{(12.8)^{2}}{16}

P = 10.2 Watt

6 0
3 years ago
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