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dolphi86 [110]
3 years ago
5

A bar magnet moves through a loop of wire with constant velocity, and the north pole enters the loop first the induced current w

ill be greatest when the magnet is located so that the loop is
Physics
2 answers:
Ksivusya [100]3 years ago
8 0
<span>A bar magnet moves through a loop of wire with constant velocity, and the north pole enters the loop first the induced current will be greatest when the magnet is located so that </span>the loop is near either the north or the south pole.
noname [10]3 years ago
3 0

Answer:

Loop must be near the end of the pole of bar magnet

Explanation:

As we know by Faraday's law of electromagnetic induction the induced EMF or induced current in the loop is given by

EMF = \frac{d\phi}{dt}

here we know that

\phi = B.A

now we know that as the magnet comes closer to the loop the magnetic field due to any of the pole of magnet will increase.

So here the maximum flux will pass through the magnet when magnet is closer to the loop.

So here when A bar magnet moves through a loop of wire with constant velocity, and the north pole enters the loop first the induced current will be greatest when the magnet is located so that the loop is when Loop must be near the end of the pole of bar magnet

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The gravitational force,F, on a rocket at a distance,r, from the center of the earth isgiven byF=kr2wherek= 1013N·km2. (Newton·k
Brrunno [24]

Answer:

The gravitational force changing velocity is

\frac{dF}{dt}=-8\frac{N}{s}

Explanation:

The expression for the gravitational force is

F=\frac{k}{r^{2}}\\\\k=10x10^{13} N*km^{2}\\\\r=10x10^{4} km\\\\V=0.4 \frac{km}{s}

Differentiate the above equation

\frac{dF}{dt}=\frac{k}{r^{2}}\\\frac{dF}{dt}=k*r^{-2}\\\frac{dF}{dt}=-2*k*r^{-3} \frac{dr}{dt}\\\frac{dF}{dt}=\frac{-2k}{r^{3}}\frac{dr}{dt}

The velocity is the distance in at time so

V=\frac{dr}{dt}=0.4 \frac{km}{s}

\frac{dF}{dt}=\frac{-2*k}{r^{3}}*0.4\\\frac{dF}{dt}=\frac{-8*10x^{13}N*km^{2} }{(10x10^{4}) ^{3}} \\\frac{dF}{dt}=\frac{-8x10^{12} }{1x10^{12}}

\frac{dF}{dt}=-8\frac{N}{s}

8 0
3 years ago
An object weighing 150 N and is suspended from the ceiling by a wire. What is the tension in the cord?​
soldi70 [24.7K]

Answer:

<h2>150N</h2>

Explanation:

According to newton's third law of motion, Action and reaction are equal and opposite, hence for an object that weighs 150N suspended by a wire, the tension on the wire is 150N

Ultimately the tension on a string or an object is equal to the mass times   gravity(which is same as the weight of the object)

7 0
3 years ago
What is the weight of a feather (mass = 0.0001 kg) that floats through earth's and the moon's atmospheres?
Dvinal [7]

Weight = (mass) x (acceleration of gravity)

Acceleration of gravity = 9.81 m/s² on Earth, 1.62 m/s² on the Moon.

The feather's weight is . . .

On Earth:  (0.0001 kg) x (9.81 m/s²) = <em>0.000981 Newton </em>

On the Moon:  (0.0001 kg) x (1.62 m/s²) = <em>0.000162 N</em>

The presence or absence of atmosphere makes no difference.  In fact, the numbers would be the same if the feather were sealed in a jar, or spinning wildly in a tornado, or hanging by a thread, or floating in a bowl of water or chicken soup.  Weight is just the force of gravity between the feather and the Earth.  It's not affected by what's around the feather, or what's happening to it.

6 0
3 years ago
The voltage across a membrane forming a cell wall is 74.0 mV and the membrane is 9.20 nm thick. What is the electric field stren
Sindrei [870]

Answer:

7.60× 10^6 V/m

Explanation:

electric field strength can be determined as ratio of potential drop and distance, I.e

E=V/d

Where E= electric field

V= potential drop= 74.0 mV= 0.07 V

d= distance= 9.20 nm = 9.2×10^-9 m

Substitute the values

E= 0.07/ 9.2×10^-9

= 7.60× 10^6 V/m

5 0
3 years ago
There are two identical small metal spheres with charges 38.9 µC and −27.6399 µC. Thedistance between them is 6 cm. The spheres
Greeley [361]

Answer:

2683.3N

Explanation:

According to coulombs law which states that "the force of attraction existing between two charge q1 and q2 is directly proportional to the product of the charges and inversely proportional to the square of the distance (d) between them. Mathematically |F|= k|q1| |q2| /d² where;

F is the force of attraction between the charges

q1 and q2 are the charges

d is the distance between them

k is the coulombs constant

Given |q1|= 38.9 × 10^-6C and |q2| = 27.6399×10^-6C d = 6cm = 0.06m

k = 8.98755 × 109 Nm² /C²

Substituting the given data's in the equation we have;

|F| = 8.98755 × 10^9×38.9×10^-6×27.6399×10^-6/0.06²

|F| = 9.66/0.06²

|F| = 9.66/0.0036

|F| = 2683.3N

The magnitude of the force will be 2683.3N

Note that the modulus of the charges changes negative value of q2 to positive value. The opposite signs of the charges doesn't affect the final calculation, it only tells the force of attraction or repulsion between the charges. Since they are unlike charges, they will attract each other in the field.

4 0
3 years ago
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