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tensa zangetsu [6.8K]
3 years ago
5

A very long solenoid with a circular cross section and radius r1= 1.30 cm with ns= 260 turns/cm lies inside a short coil of radi

us r2= 4.60 cm and Nc= 23 turns. If the current in the solenoid is ramped at a constant rate from zero to Is= 1.90 A over a time interval of 88.0 ms, what is the magnitude of the emf in the outer coil while the current in the solenoid is changing?
Physics
1 answer:
allochka39001 [22]3 years ago
5 0

Answer:

0.00851 volts

Explanation:

radius r1= 1.30 cm with ns= 260 turns/cm

radius r2= 4.60 cm and Nc= 23 turns

constant rate from zero to Is= 1.90 A

time interval of 88.0 ms

Area of the solenoid

A_1=\pi*r_1^2

=\pi*0.013^2

A_1=5.31*10^-^4 m^2

Mututal Inductance

M=uo*n*N*A1

M=(4\pi*10^-^7)*23*(260*10^2)*(5.31*10^-^4)

M=3.99*10^-^4 H

a)

EMF induced in the outer coil

E=M(dIs/dt)

E=(3.99*10^-^4)*(1.9/88*10^-^3)

E=0.00861 Volts

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3 years ago
14. a ball is thrown horizontally from the roof of a building 75 m tall with a speed of 4.6 m/s. a. how much later does the ball
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The time taken to hit the ground is 3.9 s, the range is 18m and the final velocity is 42.82 m/s

<h3>Motion Under Gravity</h3>

The motion of an object under gravity is the vertical motion of the object under the influence of acceleration due to gravity.

Given that a ball is thrown horizontally from the roof of a building 75 m tall with a speed of 4.6 m/s.

a. how much later does the ball hit the ground?

The time can be calculated by considering the vertical component of the motion with the use of formula below.

h = ut + 1/2gt²

Where

  • Height h = 75 m
  • Initial velocity u = 0 ( vertical velocity )
  • Acceleration due to gravity g = 9.8 m/s²
  • Time t = ?

Substitute all the parameters into the formula

75 = 0 + 1/2 × 9.8 × t²

75 = 4.9t²

t² = 75/4.9

t² = 15.30

t = √15.3

t = 3.9 s

b. how far from the building will it land?

The range can be found by using the formula

R = ut

Where u = 4.6 m/s ( horizontal velocity )

R = 4.6 × 3.9

R = 18 m

c. what is the velocity of the ball just before it hits the ground?

The final velocity will be

v = u + gt

v = 4.6 + 9.8 × 3.9

v = 4.6 + 38.22

v = 42.82 m/s

Therefore, the answers are 3.9 s, 18 m and 42.82 m/s

Learn more about Vertical motion here: brainly.com/question/24230984

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A boat is traveling at an initial velocity of 2.7 meters per second in the positive direction. It accelerates at a rate of 0.15
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Answer:

\boxed {\boxed {\sf 4.5 \ m/s \ in \ the  \ positive \ direction}}

Explanation:

We are asked to find the final velocity of the boat.

We are given the initial velocity, acceleration, and time. Therefore, we will use the following kinematic equation.

v_f= v_i + at

The initial velocity is 2.7 meters per second. The acceleration is 0.15 meters per second squared. The time is 12 seconds.

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  • a= 0.15 m/s²
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Substitute the values into the formula.

v_f = 2.7 \ m/s + (0.15 \ m/s^2)(12 \ s)

Multiply the numbers in parentheses.

v_f= 2.7 \ m/s + (0.15 \ m/s/s * 12 \ s)

v_f = 2.7 \ m/s + (0.15 \ m/s *12)

\v_f=2.7 \ m/s + (1.8 \ m/s)v_f=2.7 \ m/s + (1.8 \ m/s)

Add.

v_f=4.5 \ m/s

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Answer:

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3 years ago
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