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2 years ago

How can you induce vemf in a loop with a constant cross-sectional area a placed in a uniform magnetic field b?

Physics

1 answer:

AleksandrR [38]2 years ago

3 0

Answer:

Steps are given in explanation section.

Explanation:

These are the steps to induce vemf in a loop with a constant cross-sectional area a placed in a uniform magnetic field b.

(1.) Move the loop in the opposite direction of the magnetic field vector.

(2.) Rotate the loop around its perpendicular axis to the direction of the magnetic field vector.

(3.) Move the loop in the direction of the magnetic field vector.

(4.) Rotate the loop around its axis parallel to the direction of the magnetic field vector.

(5.) Move the loop in a direction perpendicular to the magnetic field vector.

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1 year ago

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A car of mass 1100kg moves at 24 m/s. What is the braking force needed to bring the car to a halt in 2.0 seconds? N

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Explanation:

Given parameters:

Mass = 1100kg

Velocity = 24m/s

time = 2s

unknown:

Braking force = ?

Solution:

The braking force is the force needed to stop the car from moving.

Force = ma = $\frac{mv}{t}$

m is the mass of the car

v is the velocity

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Force brainly.com/question/4033012

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2 years ago

A cylinder rotating about its axis with a constant angular acceleration of 1.6 rad/s2 starts from rest at t = 0. At the instant

OverLord2011 [107]

Answer:

The magnitude of the total linear acceleration is 0.27 m/s²

b. 0.27 m/s²

Explanation:

The total linear acceleration is the vector sum of the tangential acceleration and radial acceleration.

The radial acceleration is given by;

$a_t = ar$

where;

a is the angular acceleration and

r is the radius of the circular path

$a_t = ar\\\\a_t = 1.6 *0.13\\\\a_t = 0.208 \ m/s^2$

Determine time of the rotation;

$\theta = \frac{1}{2} at^2\\\\0.4 = \frac{1}{2} (1.6)t^2\\\\t^2 = 0.5\\\\t = \sqrt{0.5} \\\\t = 0.707 \ s\\\\$

Determine angular velocity

ω = at

ω = 1.6 x 0.707

ω = 1.131 rad/s

Now, determine the radial acceleration

$a_r = \omega ^2r\\\\a_r = 1.131^2 (0.13)\\\\a_r = 0.166 \ m/s^2$

The magnitude of total linear acceleration is given by;

$a = \sqrt{a_t^2 + a_r^2} \\\\a = \sqrt{0.208^2 + 0.166^2} \\\\a = 0.266 \ m/s^2\\\\a = 0.27 \ m/s^2$

Therefore, the magnitude of the total linear acceleration is 0.27 m/s²

b. 0.27 m/s²

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