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postnew [5]
3 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]3 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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Grace [21]

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7 0
3 years ago
A student attaches a rope to a 20.0 kg box of books. He pulls with a force of 90.0 N at an angle of 30 degrees with the horizont
kolbaska11 [484]

Answer:

The magnitude of the acceleration of the box is 2.01 m/s².

Explanation:

Hi there!

Please, see the attached figure for a graphical description of the problem.

We have the following horizontal forces:

Fr = friction force.

Fx = Horizontal component of the applied force, F.

And we have the following vertical forces:

Fy = vertical component of the applied force.

N = normal force exerted on the box.

W = weight of the box.

According to Newton´s second law:

∑F = m · a

Then, in the horizontal direction:

Fx - Fr = m · a

Where "m" is the mass of the box and "a" its acceleration.

Fx can be obtained by trigonometry (see figure):

Fx = F · cos 30°

Fx = 90.0 N · cos 30°

Fr is calculated as follows:

Fr = μ · N

Where μ is the coefficient of friction and N the normal force.

So, we have to find the magnitude of the normal force.

Using Newton´s second law in the vertical direction:

∑F = N + Fy - W = m · a

Notice that the box has no vertical acceleration, then:

N + Fy - W = 0

Solving for N:

N = W - Fy

The weight is calculated as follows:

W = m · g

Where g is the acceleration due to gravity:

W = 20.0 kg · 9.8 m/s² = 196 N

And the vertical component of the applied force can be obtained by trigonometry:

Fy = F · sin 30°

Fy = 90.0 N · sin 30°

The normal force will be:

N = W - Fy = 196 N - 90.0 N · sin 30°

N = 151 N

Now, we can calculate the friction force:

Fr = μ · N

Fr = 0.250 · 151 N

Fr = 37.8 N

And now, we can obtain the acceleration of the box:

Fx - Fr = m · a

(Fx - Fr) / m = a

(90.0 N · cos 30° - 37.8 N ) / 20.0 kg = a

a = 2.01 m/s²

The magnitude of the acceleration of the box is 2.01 m/s².

8 0
3 years ago
Suppose you wish to construct a motor that produces a maximum torque whose magnitude is 1.7 × 10-2 N·m. The coil of the motor ha
Len [333]

Answer:

86 turns

Explanation:

Parameters given:

Magnetic torque, τ = 1.7 * 10^(-2) Nm

Area of coil, A = 9 * 10^(-4) m²

Current in coil, I = 1.1 A

Magnetic field, B = 0.2 T

The magnetic toque is given mathematically as:

τ = N * I * A * B

Where N = number of turns

To find the number of turns, we make N subject of formula:

N = τ/(I * A * B)

Therefore:

N = (1.7 * 10^(-2)) / (1.1 * 9 * 10^(-4) * 0.2)

N = 85.85 = 86 turns (whole number)

The number of turns must be 86.

3 0
3 years ago
See the person on the right side of the front car. Six reference points could be used to show that the person is in / is NOT in
Rasek [7]

▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪

1. According to person standing on ground ~

  • The person is in motion

2. According to The car ~

  • The person is not in motion

3. According to the Seat ~

  • The person is not in motion

4. According to another person on ride ~

  • The person is not in motion

5. According to the track ~

  • The person is in motion

6. According to the Sun ~

  • The person is in motion

I hope that's what you were looking for, goodluck for your assignment ~

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2 years ago
Using the
xxMikexx [17]

Answer:

The principle of momentum conservation states that if there no external force the total momentum of the system before and after the collision is conserved.

Since momentum is a vector, we should investigate the directions and magnitudes of initial and final momentum.

\vec{P}_{initial} = \vec{P}_{final}\\\vec{P}_{initial} = m_1\vec{v}_1 + 0\\\vec{P}_{final} = m_1\vec{v}_1' + m_2 \vec{v}_2'

If the first ball hits the second ball with an angle, we should separate the x- and y-components of the momentum (or velocity), and apply conservation of momentum separately on x- and y-directions.

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3 years ago
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