Which direction does the earth on its axis ?

A 0.45 m radius, 500 turn coil is rotated one-fourth of a revolution in 4.01 ms, originally having its plane perpendicular to a

AysviL [449]

Answer:

B = 0.126 T

Explanation:

As per Faraday's law we know that rate of change in magnetic flux will induce EMF in the coil

So here we can say that EMF induced in the coil is given as

$EMF = \frac{\phi_2 - \phi_1}{\Delta t}$

initially the coil area is perpendicular to the magnetic field

and after one fourth rotation of coil the area vector of coil will be turned by 90 degree

so we can say

$\phi_1 = NBAcos 0 = BA$

$\phi_2 = NBAcos 90 = 0$

now we will have

$EMF = \frac{NBA}{t}$

$10,000 V = \frac{(500)(B)(\pi \times 0.45^2)}{4.01\times 10^{-3}}$

$B = 0.126 T$

3 0

3 years ago

Consider 3 polarizers. Polarizer 1 has a vertical transmission axis and polarizer 3 has a horizontal transmission axis. Taken to

kirill [66]

Answer:

Option D.) 0.09 I₀

Explanation:

The intensity, I, of a polarized light after passing through a polarizing filter is given by $I = I_{0} cos^{2} \theta$

$I_{0} =$ Original Intensity

$\theta =$ Angle between the direction of polarization and the axis of filter

After passing through the first polarizer, the initial intensity, $I_{o}$ is halved, i.e $I_{1} = \frac{I_{0} }{2}$

After passing through the second polarizer, the angle of polarization, θ = 30⁰

$I_{2} = I_{1} cos^{2} \theta$

$I_{2} = \frac{I_{0} }{2} cos^{2} 30\\I_{2} =\frac{3}{8} I_{0} \\I_{2} = 0.375 I_{0}$

After passing through the third polarizer, the angle of polarization, θ = 90⁰-30⁰ = 60⁰

$I_{3} = I_{2} cos^{2} \theta$

$I_{3} = 0.375I_{0} cos^{2} 60$

$I_{2} =0.09 I_{0}$

7 0

4 years ago

The diagram shows the parabolic path of a projectile that leaves the foot of a kicker with a horizontal velocity of 15 m/s and a

rusak2 [61]

Answer:

I would increase the horizontal velocity or the vertical velocity or both to make the ball go the extra distance to cross the goal line.

Explanation:

In order to increase the horizontal distance covered by the ball, we need to examine the variables involved in the formula of range of projectile. The formula for the range of projectile is given as follows:

R = V₀² Sin 2θ/g

where, g is a constant on earth (acceleration due to gravity) and θ is the angle of ball with ground at the time of launching. The value of θ should be 45° for maximum range. In this case we do not know the angle so, we can not tell if we should change it or not.

The only parameter here which we can increase to increase the range is launch velocity (V₀). The formula for V₀ in terms of horizontal and vertical components is as follows:

V₀ = √(V₀ₓ² + V₀y²)

where,

V₀ₓ = Horizontal Velocity

V₀y = Vertical Velocity

Hence, it is clear from the formula that we can increase both the horizontal and vertical velocity to increase the initial speed which in turn increases the horizontal distance covered by the ball.

<u>Therefore, I would increase the horizontal velocity or the vertical velocity or both to make the ball go the extra distance to cross the goal line.</u>

4 0

3 years ago

Squids are the fastest marine invertebrates, using a powerful set of muscles to take in and then eject water in a form of jet pr

tatuchka [14]

Answer:

0.25 m/s

Explanation:

This problem can be solved by using the law of conservation of momentum - the total momentum of the squid-water system must be conserved.

Initially, the squid and the water are at rest, so the total momentum is zero:

$p_i = 0$