PHYSICS COLLEGE

What frequency corresponds to a period of 4.31s.
T =1/f = 1/4.31s = 0.232hz correct?

Answers

Answer 1

Answer:

Answer:correct

Explanation: Period T is the reciprocal of frequency (i.e T=1/f)

Frequency is the reciprocal of period (i.e F= 1/T)

Therefore if T=4.31s

Frequency F= 1/4.31s=0.232hz

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In a physics laboratory experiment, a coil with 170 turns enclosing an area of 10.9 cm2 is rotated during the time interval 3.50×10−2 s from a position in which its plane is perpendicular to Earth's magnetic field to one in which its plane is parallel to the field. The magnitude of Earth's magnetic field at the lab location is 5.60×10−5 T. What is the total magnitude of the magnetic flux (initial) through the coil before it is rotated?

Answers

$N= 170 turns\nA=10.9cm^2\nt=3.5*10^(-2)s \n\B = 5.6*10^(-5)T$

The Magnetic flow $\Phi_(initial)$ is given by the formula,

$\Phi_(Initial)=BAsin\theta$

Replacing the values

$\Phi_(Initial) =(5.6*10^(-5))(10.9)((10^(-4)m^2)/(1cm^2)) sin90\°$

$\Phi_(Initial) =6.104*10^(-7) Wb$

A series circuit contains a 20-Ω resistor, a 200-mH inductor, a 10-μF capacitor, and an ac power source. At what frequency should the power source drive the circuit in order to have maximum power transferred from the driving source?

Answers

Answer:

f = 113 Hz

Explanation:

In order to have maximum power transferred from the driving source, as the RMS voltage doesn´t depends on frequency, the current I must be maximum.

This condition is met when the circuit behaves a purely resistive, as the impedance is at a minimum.

Such condition is known as resonance, and it satisfies the following equation:

XL = XC ⇒ ω₀ * L = 1 / ω₀*C, where ω₀, is the angular frequency at resonance.

Solving for ω₀, we have:

ω₀ = 1/√LC = 1/√200*10⁻3 H* 10⁻6F = 707 rad/sec

As we need to find the frequency (in cycles/sec), we need to convert from angular frequency to frequency, as follows:

ω₀ = 2*π*f₀ ⇒ f₀ = ω₀ / 2*π = 707 rad/sec / 2*π rad = 113 Hz

The sun does not normally affect the tides to a noticeable degree. however, under certain circumstances the gravitational pull of the sun does affect the tides. under what two (2) conditions.the sun does not normally affect the tides to a noticeable degree. however, under certain circumstances the gravitational pull of the sun does affect the tides. under what two (2) conditions.

Answers

The answers are :
1) when the sun, moon, and earth are in a line only
2) when the gravitational forces of the Moon and the Sun are
perpendicular to one another with respect to the Earth.

Answer:

High and low tides are result of combined effect of gravitational pull of the sun and the moon. When the two align in a straight line, the range of tides is maximum. This happens on new moon and full moon day.

On the other hand, when the sun and the moon align at right angles, the effect of gravity is minimum and the range of the tides is minimum.

A circular coil lies flat on a horizontal surface. A bar magnet is held above the center of the coil with its north pole pointing downward, and is released from rest. What is the direction of the induced current in the coil, as viewed from above, as the magnet approaches the coil in free fall?a. clockwise
b. counterclockwise
c. There is no induced current in the coil.

Answers

Answer:

Option B

Explanation:

As per the Lenz’s law of electromagnetism the current induced in a conductor due to any change has a tendency to oppose the change which is causing this induces current.

Thus, when a constant magnetic field with an electric circuit is varied, it produces and induced current which flow in a direction such that its sets a magnetic field that tries to restore the flux

Hence, option B is correct

A boy flies a kite with the string at a 30∘ angle to the horizontal. The tension in the string is 4.5 N. Part A Part complete How much work does the string do on the boy if the boy stands still?

Answers

Answer:

Work done is zero

Explanation:

given data

Angle of kite with horizontal = 30 degree

tension in the string = 4.5 N

WE KNOW THAT

Work = force * distance

horizontal force = $Tcos\theta = 4.5*cos30 = 3.89 N$

DISTANCE = 0 as boy stands still. therefore

work done = 3.89 *0 = 0

A 500-gram mass is attached to a spring and executes simple harmonic motion with a period of 0.25 second. If the total energy of the system is 4J, find the force constant of the spring?