Q6. When a girl steps on to a weighing machine, it shows a reading of 42kg.

An experiment is conducted on a long straight wire of diameter d. A constant current is sent through the wire and the magnetic f

soldi70 [24.7K]

Answer:

D.

Explanation:

To solve the exercise it is necessary to apply the concepts related to the Magnetic Field described by Faraday.

The magnetic field is given by the equation:

$B = \frac{\mu_0 I}{2\pi d}$

Where,

$\mu =$ Permeability constant

d = diameter

I = Current

For the given problem we have a change in the diameter, twice that of the initial experiment, therefore we define that:

$B_1 = \frac{\mu_0 I}{2\pi d}$

$B_2 = \frac{\mu_0 I}{2\pi 2d}$

The ratio of change between the two is given by:

$\frac{B_2}{B_1} = \frac{\frac{\mu_0 I}{2\pi d}}{\frac{\mu_0 I}{2\pi 2d}}$

$\frac{B_2}{B_1} = \frac{d}{2d}$

$\frac{B_2}{B_1} = \frac{1}{2}$

$B_2 = B_1 \frac{1}{2}$

Therefore the correct answer is D.

4 0

2 years ago

Which of the following causes a current in an electrical circuit?

Len [333]

C. because you need voltage in a circuit to cause current.

8 0

3 years ago

A seismographic station receives S and P waves from an earthquake, separated in time by 17.8 s. Assume the waves have traveled o

Georgia [21]

Answer:

D = 230.2 Km

Explanation:

let distance between seismograph and focus of quake is D

From time distance formula we can calculate the time taken by the S wave

$T_1 =\frac{D}{4.5}$

From time distance formula we can calculate the time taken by the P wave

$T_2 =\frac{D}{6.90}$

It is given in equation both waves are seperated from each other by 17.8 sec

so we have

$T_1 - T_2 = 17.8$ sec

Putting both time value to get distance value

$\frac{D}{4.5} - \frac{D}{6.90} = 17.8$

D = 230.2 Km

8 0

3 years ago

Find a numerical value for ρearth, the average density of the earth in kilograms per cubic meter. use 6378km for the radius of t

Andre45 [30]

Answer:

5501 kg/m^3

Explanation:

The value of g at the Earth's surface is

$g=\frac{GM}{R^2}=9.70 m/s^2$

where G is the gravitational constant

M is the Earth's mass

$R=6378km = 6.378 \cdot 10^6 m$ is the Earth's radius

Solving the formula for M, we find the value of the Earth's mass:

$M=\frac{gR^2}{G}=\frac{(9.81 m/s^2)(6.378\cdot 10^6 m)^2}{6.67\cdot 10^{-11}}=5.98\cdot 10^{24}kg$

The Earth's volume is (approximating the Earth to a perfect sphere)

$V=\frac{4}{3}\pi r^3 = \frac{4}{3}\pi (6.378\cdot 10^6 m)^3=1.087\cdot 10^{21} m^3$

So, the average density of the Earth is

$\rho = \frac{M}{V}=\frac{5.98\cdot 10^{24} kg}{1.087\cdot 10^{21} m^3}=5501 kg/m^3$

4 0

3 years ago

Read 2 more answers

A dragster going at 5 m/s north increases its velocity to 23 m/s north in 4 seconds. What is its acceleration during this time i

m_a_m_a [10]

Acceleration = (change in speed) / (time for the change)

Acceleration = (23 - 5) m/s / 4s = 18/4 m/s² = <em>4.5 m/s²</em>

6 0

3 years ago

Read 2 more answers