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Lemur [1.5K]
4 years ago
6

Suppose that a rectangular toroid has 1,500 windings and a self-inductance of 0.060 H. If the height of the rectangular toroid i

s h = 0.80 m, what is the current (in A) flowing through the rectangular toroid when the energy in its magnetic field is 9.0 ✕ 10−6 J?
Physics
1 answer:
V125BC [204]4 years ago
5 0

Answer:

Current in the toroid will be I=17.32\times 10^{-3}A

Explanation:

We have given number of winding in rectangular toroid N = 1500

Self inductance of toroid L = 0.06 H

Magnetic energy stored in toroid E=9\times 10^{-16}J

We have to find the current in the toroid

Magnetic energy stored is equal to E=\frac{1}{2}Li^2

9\times 10^{-6}=\frac{1}{2}\times 0.06\times I^2

I=17.32\times 10^{-3}A

So current in the toroid will be I=17.32\times 10^{-3}A

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a driver travels 135 km, east in 1.5 h, stops for 45 minutes for lunch, and then resumes driving for the next 2.0 h through a di
daser333 [38]

Answer:

v = 98.75 km/h

Explanation:

Given,

The distance driver travels towards the east, d₁ = 135 km

The time period of the travel, t₁ = 1.5 h

The halting time, tₓ = 46 minutes

The distance driver travels towards the east, d₂ = 215 km

The time period of the travel, t₁ = 2 h

The average speed of the vehicle before stopping

                                    v₁ = d₁/t₁

                                        = 135/1.5

                                       = 90 km/h

The average speed of vehicle after stopping

                                    v₂ = d₂/t₂

                                         = 215/2

                                        = 107.5 km/h

The total average velocity of the driver

                                      v = (v₁ +v₂) /2

                                         = (90 + 107.5)/2

                                         = 98.75 km/h

Hence, the average velocity of the driver, v = 98.75 km/h

7 0
4 years ago
When a 5.0kg cart undergoes a 2.2m/s increase in speed, what is the impulse of the cart
Karolina [17]

Answer:

11.0 kg m/s

Explanation:

The impulse exerted on the cart is equal to its change in momentum:

I=\Delta p=m\Delta v

where

m = 5.0 kg is the mass of the cart

\Delta v=2.2 m/s is its change in speed

Substituting numbers into the equation, we find

I=(5.0kg)(2.2 m/s)=11 kg m/s

7 0
3 years ago
Read 2 more answers
In a scale model of the Solar System, if the Sun is represented by a ball with a diameter of about 8 inches, about how large wou
Evgesh-ka [11]

Answer:

size of popcorn seed may meet the scaled earth size of x = 0.07312 in.  

Explanation:

Given:-

- The diameter of actual sun, Ds = 1.3927 * 10^6 km

- The diameter of model sun ball, ds = 8 in

- The diameter of the actual earth, De = 12,742 km

Find:-

How large would be the earth model

Solution:-

- We can use the actual diameters of the earth and sun to develop the ratio of earth diameter to that of sun. The direct ratio can be used to model the diameter of earth for the scale model

                                          Sun                   Earth

                     Actual :     1392700 km     12,742 km

                      Scale :           8 in                    x

- Use direct relations and solve for x:

                               x = (8 in)*(12,742 km ) / ( 1392700 km )

                              x = 0.07312 in.

- From the given options apple, grapefruit and marble diameters are larger than 1 inch. So only the size of popcorn seed may meet the scaled earth size of x = 0.07312 in.  

4 0
3 years ago
Read 2 more answers
Answer these please.
PolarNik [594]

Explanation:

hope this helps you

..........

5 0
4 years ago
he drag characteristics of a torpedo are to be studied in a water tunnel using a 1 : 7 scale model. The tunnel operates with fre
dusya [7]

Answer:20.03 m/s

Explanation:

Given

L_r=1:7

velocity of Prototype v_p=53 m/s

Taking Froude number same for both flow as it is a dimensionless number for different flow regimes in open Flow

(\frac{v_m}{\sqrt{L_mg}})=(\frac{v_p}{\sqrt{L_pg}})

v_m=v_p\times \sqrt{\frac{L_m}{L_p}}

v_m=53\times \frac{1}{\sqrt{7}}

v_m=20.03 m/s

           

4 0
3 years ago
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