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koban [17]
2 years ago
6

Two long straight wires enter a room through a window. One carries a current of 3.0 ???? into the room while the other carries a

current of 5.0 ???? out. The magnitude in Tm of the path integral ∮ ????⃗ ∙ ???????? around the window frame is:
Physics
1 answer:
Tju [1.3M]2 years ago
5 0

Answer:

\begin{equation}\\\oint_LB.dl\\\end{equation} = -8πx10^{-7}

Explanation:

If you need calculate

\begin{equation}\\\oint_L B.dl\\ \end{equation}

You can use the Ampere's Law

\begin{equation}\\\oint_L B.dl\\ \end{equation} = I_{in}μ

   

      Where I_{in}: Current passing through the window

                   μ : Free space’s magnetic permeability

                   μ = 4πx10^{-7} T.m.A^{-1}

Then

\begin{equation}\\\oint_L B.dl\\ \end{equation} = (3-5)4πx10^{-7}

\begin{equation}\\\oint_L B.dl\\ \end{equation} = -8πx10^{-7}

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Calculate the force of gravity between two objects of masses 1300 kg and 7800 kg, which are 0.23 m apart.
uranmaximum [27]

Answer:

F = Gm1m2/r^2 where G = 6.67x10^-11, m1 =1300, m2 = 7800, r = 0.23m

F = 6.67x10^-11 *1300*7800/(0.23)^2 = 0.0127852N

Explanation:

6 0
2 years ago
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A boat traveled 280 miles downstream and back. The trip downstream took 7 hours. Trip back took 14 hours. What is the speed of t
avanturin [10]
Assuming 280 miles is the total distance travelled:
Let b = boat speed in still water
Let c = current speed.
For the downstream trip the speed is b + c. In 7 hours at the speed of (b + c) mph the boat travels 140 miles.
7(b + c) = 140 .............(1)
For the upstream trip the speed is b - c. In 14 hours at the speed of (b - c) mph the boat travels 140 miles.
14(b - c) = 140 ............(2)
The left hand sides of equations (1) and (2) are equal. Therefore we can write
7b + 7c = 14b - 14c ...........(3)
Rearranging equation (3) we get
21c = 7b
c = b/3 .......................(4)
The value for c obtained in equation (4) should now be substituted into equation (1) which can then be solved to find the value of b.
7 0
2 years ago
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In the ENGR 10 lab (E391), there are 50 long light bulbs (P=100 W) and 30 regular bulbs (P=60 W). How much energy is consumed li
Alenkinab [10]

Answer:

Total energy saving will be 0.8 KWH

Explanation:

We have given there are 50 long light bulbs of power 100 W so total power of 50 bulb = 100×50 = 5000 W = 5 KW

30 bulbs are of power 60 W

So total power of 30 bulbs = 30×60 = 1800 W = 1.8 KW

Total power of 80 bulbs = 1.8+5 = 6.8 KW

Total time = 3 hour

We know that energy E=power\times time=6.8\times 3=20.4KWH

Now power of each CFL bulb = 25 W

So power of 80 bulbs = 80×25 = 2000 W = 2 KW

Energy of 80 bulbs = 2×3 = 6 KWH

So total energy saving = 6.8-6 = 0.8 KWH

6 0
2 years ago
Two long, parallel wires carry unequal currents in the same direction. The ratio of the currents is 3 to 1. The magnitude of the
astraxan [27]

Answer:

3A is the larger of the two currents.

Explanation:

Let the currents in the two wires be I₁ and I₂

given:

Magnitude of the electric field, B = 4.0μT = 4.0×10⁻⁶T

Distance, R = 10cm = 0.1m

Ratio of the current = I₁ : I₂ = 3 : 1

Now, the magnitude of a magnetic field at a distance 'R' due to the current 'I' is given as

B = \frac{\mu_oI}{2\pi R}

Where \mu_o is the magnitude constant = 4π×10⁻⁷ H/m

Thus, the magnitude of a magnetic field due to I₁ will be

B_1 = \frac{\mu_oI_1}{2\pi R}

B_2 = \frac{\mu_oI_2}{2\pi R}

given,

B = B₁ - B₂ (since both the currents are in the same direction and parallel)

substituting the values of B, B₁ and B₂

we get

4.0×10⁻⁶T =  \frac{\mu_oI_1}{2\pi R} - \frac{\mu_oI_2}{2\pi R}

or

4.0×10⁻⁶T =  \frac{\mu_o}{2\pi R}\times (I_1-I_2 )

also

\frac{I_1}{I_2} = \frac{3}{1}

⇒I_1 = 3\times I_2

substituting the values in the above equation we get

4.0×10⁻⁶T =  \frac{4\pi\times 10^{-7}}{2\pi 0.1}\times (3 I_2-I_2)

⇒I_2 = 1A

also

I_1 = 3\times I_2

⇒I_1 = 3\times 1A

⇒I_1 = 3A

Hence, the larger of the two currents is 3A

3 0
2 years ago
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