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d1i1m1o1n [39]
3 years ago
6

What holds the moon in place, orbiting around the Earth

Physics
1 answer:
AnnZ [28]3 years ago
4 0

The Earth's gravity keeps the Moon orbiting us. It keeps changing the direction of the Moon's velocity. This means gravity makes the Moon accelerate all the time, even though its speed remains constant.

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Mechanical energy is the sum of<br> energy and potential energy.
MAVERICK [17]

Answer:

True, the total amount of mechanical energy is merely the sum of the potential energy and the kinetic energy. This sum is simply referred to as the total mechanical energy.

Explanation: Hope it helps you:))))

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7 0
2 years ago
If i = 1.70 A of current flows through the loop and the loop experiences a torque of magnitude 0.0760 N ⋅ m , what are the lengt
Alik [6]

Answer:

Length of the sides of the square loop is given by

s = √[(τ)/(NIB sin θ)]

Explanation:

The torque, τ, experienced by a square loop of area, A, with N number of turns around the loop and current of I flowing in the wire, with a magnetic field presence, B, and the plane of the loop tilted at angle θ to the x-axis, is given by

τ = (N)(I)(A)(B) sin θ

If everything else is given, the length of a side of the square loop, s, can be obtained from its Area, A.

A = s²

τ = (N)(I)(A)(B) sin θ

A = (τ)/(NIB sin θ)

s² = (τ)/(NIB sin θ)

s = √[(τ)/(NIB sin θ)]

In this question, τ = 0.076 N.m, I = 1.70 A

But we still need the following to obtain a numerical value for the length of a side of the square loop.

N = number of turnsof wire around the loop

B = magnetic field strength

θ = angle to which the plane of the loop is tilted, measured with respect to the x-axis.

6 0
3 years ago
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Fittoniya [83]

Answer:

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5 0
2 years ago
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What is the kinetic energy of a 5-kg object moving at 7 m/s?
Sunny_sXe [5.5K]
To answer this question, we would need the formula for the kinetic energy which is Kinetic Energy = ½ x m x v^2

Where the following means: m is the mass of the object and v is the velocity

So At 7 m/s: Kinetic Energy = ½ x 5 x 7^2 = 122.5 J is the answer
6 0
3 years ago
Two long current-carrying wires run parallel to each other and are separated by a distance of 5.00 cm. If the current in one wir
Darya [45]

Answer:

The magnitude of the force per unit length is 2.145 x 10⁻⁵ N/m and the direction of the force is outward or repulsive since the current in the two parallel wires are flowing in opposite direction.

Explanation:

Given;

distance between the parallel wires, r = 5.0 cm = 0.05 m

current in the first wire, I₁ = 1.65 A

current in the second wire, I₂ = 3.25 A

The magnitude of the force per unit length between the two wires is calculated as follows;

\frac{F}{l} =\frac{\mu_0 I_1 I_2}{2\pi r} \\\\\frac{F}{l} =\frac{4\pi \times 10^{-7} \times 1.65 \times 3.25}{2\pi \times 0.05} \\\\\frac{F}{l} = 2.145 \times 10^{-5} \ N/m

Therefore, the magnitude of the force per unit length is 2.145 x 10⁻⁵ N/m and the direction of the force is outward or repulsive since the current in the two parallel wires are flowing in opposite direction.

5 0
3 years ago
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