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2 years ago

A solenoid of length 10 cm has 247 turns and a magnetic field at the center of 1.2 T. What is the current through the solenoid?

Physics

1 answer:

elena55 [62]2 years ago

6 0

Answer:

386.6 A

Explanation:

We are given;

Solenoid length; L = 10 cm = 0.1 m

Number of turns; N = 247

Magnetic field; B = 1.2 T

Now, at the centre of a long solenoid of N turns/metre carrying a current(I), the formula for the magnetic field is given as;

B = μ_oNI/L

Making I the subject we have;

I = BL/μ_oN

Where μ_o is the vacuum of permeability and has a constant value of 4π × 10^(−7) H/m

Thus;

I = (1.2 × 0.1)/(4π × 10^(−7) × 247)

I = 386.6 A

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Light travels in a straight line at a constant speed of 300000 m/s. What is the acceleration of light?

mafiozo [28]

Answer:

As light travels in a straight line at a constant speed, it's acceleration is <u>0 m/s²</u>.

There is no rate of change of speed, so there is no acceleration.

<u>0 m/s²</u> is the right answer.

4 0

3 years ago

61. A physics student has a single-occupancy dorm room. The student has a small refrigerator that runs with a current of 3.00 A

Mademuasel [1]

Answer:

Part a)

$percentage = 21.3$ %

Part b)

$percentage = 2.13 \times 10^{-5}$ %

Explanation:

As we know that total power used in the room is given as

$P = P_1 + P_2 + P_3 + P_4$

here we have

$P_1 = (110)(3) = 330 W$

$P_2 = 100 W$

$P_3 = 60 W$

$P_4 = 3 W$

$P = 330 + 100 + 60 + 3$

$P = 493 W$

Part a)

Since power supply is at 110 Volt so the current obtained from this supply is given as

$110\times i = 493$

$i = 4.48 A$

now resistance of transmission line

$R = \frac{\rho L}{A}$

$R = \frac{(2.8 \times 10^{-8})(10\times 10^3)}{\pi(4.126\times 10^{-3})^2}$

$R = 5.23 \ohm$

now power loss in line is given as

$P = i^2 R$

$P = (4.48)^2(5.23)$

$P = 105 W$

Now percentage loss is given as

$percentage = \frac{loss}{supply} \times 100$

$percentage = \frac{105}{493} \times 100$

$percentage = 21.3$ %

Part b)

now same power must have been supplied from the supply station at 110 kV, so we have

$110 \times 10^3 (i ) = 493$

$i = 4.48\times 10^{-3} A$

now power loss in line is given as

$P = i^2 R$

$P = (4.48 \times 10^{-3})^2(5.23)$

$P = 1.05 \times 10^{-4} W$

Now percentage loss is given as

$percentage = \frac{loss}{supply} \times 100$

$percentage = \frac{1.05 \times 10^{-4}}{493} \times 100$

$percentage = 2.13 \times 10^{-5}$ %

6 0

3 years ago

PLEASE HELP ASAP!!! 49 POINTS AND BRAINLIEST FOR BEST AND FASTEST ANSWER!!! (DO NOT ANSWER IF YOU DON'T KNOW, OR I WILL REPORT Y

Colt1911 [192]

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5 0

2 years ago

Read 2 more answers

Situation 6.1 A 13.5-kg box slides over a rough patch 1.75 m long on a horizontal floor. Just before entering the rough patch, t

GenaCL600 [577]

B) 14.0 N

The way to solve this problem is to determine the kinetic energy the box had before and after the rough patch of floor. The equation for kinetic energy is:

E = 0.5 M V^2

where

E = Energy

M = Mass

V = velocity

Substituting the known values, let's calculate the before and after energy.

Before:

E = 0.5 M V^2

E = 0.5 13.5kg (2.25 m/s)^2

E = 6.75 kg 5.0625 m^2/s^2

E = 34.17188 kg*m^2/s^2 = 34.17188 joules

After:

E = 0.5 M V^2

E = 0.5 13.5kg (1.2 m/s)^2

E = 6.75 kg 1.44 m^2/s^2

E = 9.72 kg*m^2/s^2 = 9.72 Joules

So the box lost 34.17188 J - 9.72 J = 24.451875 J of energy over a distance of 1.75 meters. Let's calculate the loss per meter by dividing the loss by the distance.

24.451875 J / 1.75 m = 13.9725 J/m = 13.9725 N

Rounding to 1 decimal place gives 14.0 N which matches option "B".

8 0

2 years ago

A rigid tank whose volume is unknown is divided into two parts by a partition. One side of the tank contains an ideal gas at 935

a_sh-v [17]

Answer:

2805 °C

Explanation:

If the gas in the tank behaves as ideal gas at the start and end of the process. We can use the following equation:

$P=RTn/V$

The key issue is identify the quantities (P,T, V, n) in the initial and final state, particularly the quantities that change.

In the initial situation the gas have an initial volume $V_{i}$ , temperature $T_{i}$ , and pressure $P$ ,.

And in the final situation the gas have different volume $V_{f}$ and temeperature $T_{f}$ , the same pressure $P$ ,, and the same number of moles $n$ ,.

We can write the gas ideal equation for each state:

$P=RT_{i}n/V_{i}$ and $P=RT_{f}n/V_{f}$ , as the pressure are equals in both states we can write

$RT_{i}n/V_{i} = RT_{f}n/V_{f}$

solving for $T_{f}$

$T_{f} = T_{i}/V_{i} * V_{f}$ (*)

We know $T_{i}$ = 935 °C, and that the $V_{f}$ (the complete volume of the tank) is the initial volume $V_{i}$ plus the part initially without gas which has a volume twice the size of the initial volume (read in the statement: the other side has a volume twice the size of the part containing the gas). So the final volume $V_{f}= V_{i} + 2V_{i}=3V_{i}$

Replacing in (*)

$T_{f} = 935/V_{i} * 3V_{i} = 935*3= 2805$

7 0

2 years ago