All categories

3 years ago

HELP ASAP!!

Physics

1 answer:

Nimfa-mama [501]3 years ago

7 0

The strength of the magnetic field is equal to 20 Tesla.

<h3>Explanation:</h3>

Given:

Induced potential difference = V = 12 V

Length of wire = L = 0.20 m

Speed of the moving wire = V = 3.0 m/s

Magnetic field strength = B = ?

A conductor, placed in a uniform magnetic field; when moved at a constant speed with respect to the field, leads to a changing magnetic flux, generating an electromotive force (EMF). Using, Faraday's law of magnetic induction, a moving conductor's induced EMF in terms of the magnetic field strength is given by :

$E = L\times V\times B$ ......................(1)

where

E = Induced potential difference (EMF)

L = Length of the conductor

V = Speed of the conductor moved with respect to the magnetic field

B = Strength of uniform magnetic field

Rewriting equation (1) for B, we get

$B = \frac{E}{L\times V} = \frac{12}{0.2\times 3} = \frac{12}{0.6} = 20\ T$

The strength of magnetic field is equal to 20 Tesla.

You might be interested in

State how a ray of light can travel from air into glass without bending?

Katyanochek1 [597]

Answer:

If you shine a beam of light (a bundle of parallel rays) through the air, it will travel in a straight line. Rays of light usually travel in straight lines until they hit something. If a ray of light hits the surface of a sheet of glass, some light will be reflected by the surface of the glass.

Explanation:

STREAM LALISA AND MONEY

6 0

2 years ago

A group of students prepare for a robotic competition and build a robot that can launch large spheres of mass M in the horizonta

jeyben [28]

Answer:

$V_0$

Explanation:

Given that, the range covered by the sphere, $M$ , when released by the robot from the height, $H$ , with the horizontal speed $V_0$ is $D$ as shown in the figure.

The initial velocity in the vertical direction is $0$ .

Let $g$ be the acceleration due to gravity, which always acts vertically downwards, so, it will not change the horizontal direction of the speed, i.e. $V_0$ will remain constant throughout the projectile motion.

So, if the time of flight is $t$ , then

$D=V_0t\; \cdots (i)$

Now, from the equation of motion

$s=ut+\frac 1 2 at^2\;\cdots (ii)$

Where $s$ is the displacement is the direction of force, $u$ is the initial velocity, $a$ is the constant acceleration and $t$ is time.

Here, $s= -H, u=0,$ and $a=-g$ (negative sign is for taking the sigh convention positive in $+y$ direction as shown in the figure.)

So, from equation (ii),

$-H=0\times t +\frac 1 2 (-g)t^2$

$\Rightarrow H=\frac 1 2 gt^2$

$\Rightarrow t=\sqrt {\frac {2H}{g}}\;\cdots (iii)$

Similarly, for the launched height $2H$ , the new time of flight, $t'$ , is

$t'=\sqrt {\frac {4H}{g}}$

From equation (iii), we have

$\Rightarrow t'=\sqrt 2 t\;\cdots (iv)$

Now, the spheres may be launched at speed $V_0$ or $2V_0$ .

Let, the distance covered in the $x-$ direction be $D_1$ for $V_0$ and $D_2$ for $2V_0$ , we have

$D_1=V_0t'$

$D_1=V_0\times \sqrt 2 t$ [from equation (iv)]

$\Rightarrow D_1=\sqrt 2 D$ [from equation (i)]

$\Rightarrow D_1=1.41 D$ (approximately)

This is in the $3$ points range as given in the figure.

Similarly, $D_2=2V_0t'$

$D_2=2V_0\times \sqrt 2 t$ [from equation (iv)]

$\Rightarrow D_2=2\sqrt 2 D$ [from equation (i)]

$\Rightarrow D_2=2.82 D$ (approximately)

This is out of range, so there is no point for $2V_0$ .

Hence, students must choose the speed $V_0$ to launch the sphere to get the maximum number of points.

7 0

3 years ago

Where do the electrons that form the auroras enter the magnetosphere? a. Through holes c. At the equator b. Between the magnetic

marta [7]

I think the answer is d. In the magnetotail. I hope this helps! :)

7 0

3 years ago

Read 2 more answers

A spacecraft orbits the Earth at height of 1600km. Calculate the escape velocity for the spacecraft.

guapka [62]

Answer:

what do u need help with

Explanation:

Calculate the escape velocity for the spacecraft. [G= 6.67×10^-11Nm^2kg^-2, mass of the Earth= 5.97×10^24kg, radius of the Earth= ...

7 0

2 years ago

Find the GCF 12, 18, and 84

ahrayia [7]

To find the answer, plot down the factors for every number.

12: 1, 2 ,3 ,4, 6, 12

18: 1, 2, 3, 6, 9, 18

84: 1, 2, 3, 4, 6, 7, 12

If you noticed, the number that was common to the 3 numbers, were 1, 2, 3, and 6
And 6 is the bigger number
So 6 is your GCF

5 0

3 years ago

Read 2 more answers