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MatroZZZ [7]
2 years ago
15

Math Exploration

Physics
1 answer:
Airida [17]2 years ago
5 0

Question: What is the frequency of a wave that has a wave speed of 120 m/s and a wavelength of 0.40 m?

Answer: The equation that relates frequency of a wave to a waves speed and wavelength is Speed of Wave= Frequency X Wavelength. Since you are given speed and wavelength, you plug those two known numbers into the equation, 120= Frequency X 0.40. You then divide 120 by .4 to get your frequency of 300.

Explanation: this might help for

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Explanation:

Consider a dipole having magnetic moment 'm' is placed in magnetic field \vec{B} then the torque exerted by the field on the dipole is

\tau = m\times B

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Now to rotate the dipole in the field to its final position the work required to be done is

U=\int \tau d\alpha

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Minimum energy mB is for the case when m is anti parallel to B.

Minimum energy -mB is for the case when m is parallel to B.

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A 56 kg diver runs and dives from the edge of a cliff into the water which is located 4.0 m below. If she is moving at 8.0 m/s t
Reil [10]

Answer:

1) 2197.44 J

2) 0 J

3) 2197.44 J = Constant

4) 2197.44 J

5) Approximately 8.86 m/s

Explanation:

The given parameters are;

The mass of the diver, m = 56 kg

The height of the cliff, h = 4.0 m

The speed with which the diver is moving, vₓ = 8.0 m/s

The gravitational potential energy = Mass, m × Height of the cliff, h × Acceleration due to gravity, g

1) Her gravitational potential energy = 56 × 4.0 × 9.81 = 2197.44 J

2) The kinetic energy = 1/2·m·u²

Where;

u = Her initial velocity = 0 when she just leaves the cliff

Therefore;

Her kinetic energy when she just leaves the cliff = 1/2 × 56 × 0² = 0 J

3) The total mechanical energy = Kinetic energy + Potential energy

The total mechanical energy is constant

Her total mechanical energy relative to the water surface when she leaves the cliff = Her gravitational potential energy = 2197.44 J = Constant

4) Her total mechanical energy relative to the water surface just before she enters the water = 2197.44 J

5) The speed with which she enters the water, v, is given from, v² = u² + 2·g·h

Where;

u = The initial velocity at the top of the cliff before she jumps= 0 m/s

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v = √78.48 ≈ 8.86 m/s

The speed with which she enters the water, v ≈ 8.86 m/s

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