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Dmitry_Shevchenko [17]

2 years ago

13

A flat rectangular loop of wire carrying a 4.0-a current is placed in a uniform 0.60-t magnetic field. the magnitude of the torq

ue acting on this loop when the plane of the loop makes a 30° angle with the field is measured to be 1.1 n ∙ m. what is the area of this loop?

1 answer:

ZanzabumX [31]2 years ago

6 0

Torque on a closed current carrying loop is given by formula

$T = NiABsin\theta$

here

N = number of turns = 1

i = current = 4 A

A = area of the loop

B = magnetic field = 0.60 T

$\theta$ = angle between normal of the plane and magnetic field = 90 - 30 = 60 degree

Now we will have

$1.1 = 1* 4 * A * 0.60* sin60$

$1.1 = 2.078*A$

$A = 0.53 m^2$

So area of the loop is 0.53 m^2

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Answer:

Final velocity of the block = 2.40 m/s east.

Explanation:

Here momentum is conserved.

Initial momentum = Final momentum

Mass of bullet = 0.0140 kg

Consider east as positive.

Initial velocity of bullet = 205 m/s

Mass of Block = 1.8 kg

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Initial momentum = 0.014 x 205 + 1.8 x 0 = 2.87 kg m/s

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We need to find final velocity of the block( u )

Final momentum = 0.014 x -103+ 1.8 x u = -1.442 + 1.8 u

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2 years ago

A series circuit has two 10-ohm bulb is added in a series. Technician A says that the three bulbs will be dimmer than when only

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Answer:

Technician A is right. The situation will happens even with only two bulbs in series

Explanation:

We must take into account that

1.- All electric device need its nominal voltage to operate

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3.-Nominal voltage for bulbs are specify for houses voltages you find between fase and neutral wires for instance in Venezuela 120 (v).

4.-In a imaginary circuit of only one bulb, the nominal voltage will be applied and the bulb will operates correctly, but when you add another bulb (in series) the nominal voltage will split between the two bulbs ( we could find a situation such as the first bulb work properly but the second one does not). The voltage split according to Ohms law (in such way that the sum of voltage between the terminal of the first bulb plus the voltage at terminals of the second one are equal to nominal voltage.

For that reason all the bulbs are connected in parallel in wich case all of them will operate with the common voltage

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2 years ago

5.16 An insulated container, filled with 10 kg of liquid water at 20 C, is fitted with a stirrer. The stirrer is made to turn by

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Answer:

a) W=2.425kJ

b) $\Delta E=2.425kJ$

c) $T_f=20.06^{o}C$

d) Q=-2.425kJ

Explanation:

a)

First of all, we need to do a drawing of what the system looks like, this will help us visualize the problem better and take the best possible approach. (see attached picture)

The problem states that this will be an ideal system. This is, there will be no friction loss and all the work done by the object is transferred to the water. Therefore, we need to calculate the work done by the object when falling those 10m. Work done is calculated by using the following formula:

$W=Fd$

Where:

W=work done [J]

F= force applied [N]

d= distance [m]

In this case since it will be a vertical movement, the force is calculated like this:

F=mg

and the distance will be the height

d=h

so the formula gets the following shape:

$W=mgh$

so now e can substitute:

$W=(25kg)(9.7 m/s^{2})(10m)$

which yields:

W=2.425kJ

b) Since all the work is tansferred to the water, then the increase in internal energy will be the same as the work done by the object, so:

$\Delta E=2.425kJ$

c) In order to find the final temperature of the water after all the energy has been transferred we can make use of the following formula:

$\Delta Q=mC_{p}(T_{f}-T_{0})$

Where:

Q= heat transferred

m=mass

$C_{p}$ =specific heat

$T_{f}$ = Final temperature.

$T_{0}$ = initial temperature.

So we can solve the forula for the final temperature so we get:

$T_{f}=\frac{\Delta Q}{mC_{p}}+T_{0}$

So now we can substitute the data we know:

$T_{f}=\frac{2 425J}{(10000g)(4.1813\frac{J}{g-C})}+20^{o}C$

Which yields:

$T_{f}=20.06^{o}C$

d)

For part d, we know that the amount of heat to be removed for the water to reach its original temperature is the same amount of energy you inputed with the difference that since the energy is being removed this means that it will be negative.

$\Delta Q=-2.425kJ$

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2 years ago

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Answer:

B. W is positive and a is negative

Explanation:

As we know that the angular speed of the second clock is in positive direction so as it comes to halt from its initial direction of motion then we have

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now it comes to stop so angular acceleration is taken in opposite to the direction of angular speed

so we will have

$\alpha = negative$

so here correct answer is

B. W is positive and a is negative

8 0

2 years ago

8) A flat circular loop having one turn and radius 5.0 cm is positioned with its plane perpendicular to a uniform 0.60-T magneti

Marrrta [24]

Answer:

EMF induced in the loop is 9.4 V

Explanation:

As we know that initial magnetic flux of the loop is given as

$\phi_1 = B.A$

$\phi_1 = (0.60)(\pi (0.05)^2)$

$\phi_1 = 4.7 \times 10^{-3} Wb$

As soon as the area of the loop becomes zero the final magnetic flux of the loop is ZERO

Now as per faraday's law of electromagnetic induction the EMF is induced due to rate of change in magnetic flux

so we will have

$EMF = \frac{\Delta \phi}{\Delta t}$

so we will have

$EMF = \frac{4.7 \times 10^{-3} - 0}{0.50 \times 10^{-3}}$

$EMF = 9.4 V$

7 0

2 years ago