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4 years ago

How does being on the moon effect gravitational field strength

1 answer:

5 0

Your being on the moon has no effect on the moon's
gravitational field strength, or on the Earth's for that
matter.

However, YOU notice a change on YOU when YOU move
from one to the other, because of the effect of the gravitational
field strength on you and your internal organs.

If you could stand on the moon, you would experience an incredible
sense of lightness, since the forces of attraction between the moon
and anything else are only 16% as great as the same forces are on
Earth.

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A coil has N turns enclosing an area of A. In a physics laboratory experiment, the coil is rotated during the time interval delt

zmey [24]

Answer:

a) Magnetic flux before the plane is rotated = (BA) Wb

b) Magnetic flux after the plane is rotated = 0 Wb

c) Magnitude of the average induced emf =

(NBA)/Δt

Explanation:

The magnetic flux is given as

Φ = BA cos θ

where B = magnetic field strength

A = Cross sectional Area of the loop enclosed

θ = the angle in the equation above is between the line normal to plane (NOT the plane itself!) and the magnetic field.

a) Given that the position of the plane of each turn is perpendicular to Earth's magnetic field before being rotated.

Before the plane is rotated, θ = 0°

Φ = BA cos θ = BA cos 0° = BA

b) Given that the position of the plane of each turn is parallel to Earth's magnetic field after being rotated.

After the plane is rotated, θ = 90°

Φ = BA cos θ = BA cos 90° = 0

c) According to the Faraday's law of electromagnetic induction,

E = - N (ΔΦ/Δt) (minus sign to indicate that the direction of the induced emf is opposite the direction of the change of magnetic flux)

ΔΦ = (final Φ) - (initial Φ) = (0 - BA) = - BA

E = - N (- BA)/Δt

E = (NBA)/Δt

Hope this Helps!!!

8 0

3 years ago

Read 2 more answers

An ice cube melts when its mechanical energy increases.

Sergeu [11.5K]

False it only melts when its thermal energy increases. plus an ice cube has no mechanical energy.

7 0

4 years ago

Read 2 more answers

Consider the differential equation

Zielflug [23.3K]

Yp(t) = A1 t^2 + A0 t + B0 t e(4t)

=> y ' = 2A1t + A0 + B0 [e^(4t) +4 te^(4t) ]

   y ' = 2A1t + A0 + B0e^(4t) + 4B0 te^(4t)

=> y '' = 2A1 + 4B0e(4t) + 4B0 [ e^(4t) + 4te^(4t)

   y '' = 2A1 + 4B0e^(4t) + 4B0e^(4t) + 16B0te^(4t)

Now substitute the values of y ' and y '' in the differential equation:

<span>y′′+αy′+βy=t+e^(4t)

</span> 2A1 + 4B0e^(4t) + 4B0e^(4t) + 16B0te^(4t) + α{2A1t + A0 + B0e^(4t) + 4B0 te^(4t) } + β{A1 t^2 + A0 t + B0 t e(4t)} = t + e^(4t)

Next, we equate coefficients

1) Constant terms of the left side = constant terms of the right side:

2A1+ 2αA0 = 0 ..... eq (1)

2) Coefficients of e^(4t) on both sides

8B0 + αB0 = 1 => B0 (8 + α) = 1 .... eq (2)

3) Coefficients on t

2αA1 + βA0 = 1 .... eq (3)

4) Coefficients on t^2

βA1 = 0 ....eq (4)

given that A1 ≠ 0 => β =0

5) terms on te^(4t)

16B0 + 4αB0 + βB0 = 0 => B0 (16 + 4α + β) = 0 ... eq (5)

Given that B0 ≠ 0 => 16 + 4α + β = 0

Use the value of β = 0 found previously

16 + 4α = 0 => α = - 16 / 4 = - 4.

Answer: α = - 4 and β = 0

3 0

4 years ago

in a certain nuclear reactor, neutrons suddenly collide with carbon nuclei, which have 12 times the mass of neutrons. in a head-

castortr0y [4]

carbon atomic mass Equals m2

neutron mass = M,

as mg 12m1 = the speed of the neutron following a collision

m1 - m2 / m1 + m2 * u1 = v1

= [(-12m1+m1) / 13m1] * u1

= [(-12+1) 13] * u1

v1 = -11/13*u1

Neutron speed equals 11/13 starting speed.

Response: 11/13

A nuclear reactor is a device used to start and regulate nuclear chain reactions involving fission or nuclear fusion. Nuclear reactors are used in nuclear power plants to produce electricity and in nuclear propulsion for ships.

The neutron, symbol n or n0, is a subatomic particle with a neutral charge and slightly more mass than a proton. Atomic nuclei consist of protons and neutrons.

The element with atomic number 6 carbon has small, straightforward atoms. Six positively charged protons make up the nucleus, and there are six electrons outside the nucleus that are split between two shells.

To know more about nuclear reactor, click on the link below:

brainly.com/question/23160065

#SPJ4

5 0

1 year ago

An AC source of maximum voltage V0 = 30 V is connected to a resistor R = 50 Ω, an inductor L = 0.6 H, and a capacitor C = 20 µF.

ivolga24 [154]

Hello!

We can begin by solving for the resonance ANGULAR frequency of the circuit.

For an RCL circuit, the resonance angular frequency is given as:
$\omega_0^2 = \frac{1}{LC}\\\\w_0 = \sqrt{\frac{1}{LC}}$

ω₀ = resonance angular frequency (rad/s)

L = Inductance (0.6 H)
C = Capacitance (20 μF)

Plug in the values and solve.

$\omega_0 = \sqrt{\frac{1}{(0.6)(0.00002)} } = 288.675 \frac{rad}{s}$

For an AC power source, the output is usually expressed as:

$V(t) = V_{max}sin(\omega_0 t})$

So, using the appropriate values and setting the source angular frequency equivalent to the circuit's resonance angular frequency:

$V(t) = 30sin(288.675t)$

To find the maximum charge on the capacitor when the frequency of the source is equivalent to the resonance frequency of the circuit (or the angular frequencies are equal), we can begin by finding the maximum voltage across the capacitor.

To find this, however, we must solve for the maximum current across the circuit by finding the total impedance of the circuit. When the circuit is at resonance, the impedance is equivalent to the resistance of the RESISTOR.

So, solve for the maximum current in the circuit using Ohm's Law:

$i = \frac{V}{R}$

In this instance AT RESONANCE:

$I_{Max} = \frac{V_Max}{R}\\\\I_{Max} = \frac{30}{50} = 0.6 A$

Now, we must solve for the capacitive reactance in order to find the maximum voltage across the capacitor. Using the following equation for capacitive reactance:
$X_c = \frac{1}{\omega C}\\\\X_c = \frac{1}{(288.675)(0.00002)} = 173.205 \Omega$

Now that we found the maximum current and capacitive reactance, we can now solve for the maximum voltage across the capacitor:
$V_{C, max} = X_C I_{Max}\\\\V_{C, max} = 173.205 * 0.6 = 103.923 V$

Finally, we can easily solve for the maximum charge on the capacitor using the relationship:
$C = \frac{Q}{V}\\\\Q = CV$

Plug in the values solved for above.

$Q = (0.00002)(103.923) = 0.00208 C = \boxed{2.078 mC}$

6 0

2 years ago