in a cricket match there are 5000 spectators counted 10 by 10 the number of significant figure in the measurement will be

A coil with an inductance of 2.3 H and a resistance of 14 Ω is suddenly connected to an ideal battery with ε = 100 V. At 0.13 s

klemol [59]

Given Information:

Resistance = R = 14 Ω

Inductance = L = 2.3 H

voltage = V = 100 V

time = t = 0.13 s

Required Information:

(a) energy is being stored in the magnetic field

(b) thermal energy is appearing in the resistance

(c) energy is being delivered by the battery?

Answer:

(a) energy is being stored in the magnetic field ≈ 219 watts

(b) thermal energy is appearing in the resistance ≈ 267 watts

(c) energy is being delivered by the battery ≈ 481 watts

Explanation:

The energy stored in the inductor is given by

$U = \frac{1}{2} Li^{2}$

The rate at which the energy is being stored in the inductor is given by

$\frac{dU}{dt} = Li\frac{di}{dt} \: \: \: \: eq. 1$

The current through the RL circuit is given by

$i = \frac{V}{R} (1-e^{-\frac{t}{ \tau} })$

Where τ is the the time constant and is given by

$\tau = \frac{L}{R}\\ \tau = \frac{2.3}{14}\\ \tau = 0.16$

$i = \frac{110}{14} (1-e^{-\frac{t}{ 0.16} })\\i = 7.86(1-e^{-6.25t})\\\frac{di}{dt} = 49.125e^{-6.25t}$

Therefore, eq. 1 becomes

$\frac{dU}{dt} = (2.3)(7.86(1-e^{-6.25t}))(49.125e^{-6.25t})$

At t = 0.13 seconds

$\frac{dU}{dt} = (2.3) (4.37) (21.8)\\\frac{dU}{dt} = 219.11 \: watts$

(b) thermal energy is appearing in the resistance

The thermal energy is given by

$P = i^{2}R\\P = (7.86(1-e^{-6.25t}))^{2} \cdot 14\\P = (4.37)^{2}\cdot 14\\P = 267.35 \: watts$

(c) energy is being delivered by the battery?

The energy delivered by battery is

$P = Vi\\P = 110\cdot 4.37\\P = 481 \: watts$

4 0

3 years ago

Which term describes atoms with different atomatic masses due to varying numbers of neutrons

frez [133]

Isotopes is the answer.

C

8 0

3 years ago

Read 2 more answers

In the lab downstairs physics majors use a rotating mirror to measure the speed of light within a few percent of the actual valu

iris [78.8K]

The number of complete cycles the rotating mirror goes through before the angular velocity gets to zero is approximately 1166.8 revs

<h3>What is angular velocity?</h3>

Angular velocity is the ratio of the angle turned to the time taken.

The kinematic equation for angular velocity are presented as follows;

ω = ω₀ + α·t

θ = θ₀ + ω₀·t + 0.5·α·t²

Where;

θ₀ = The initial angle turned = 0

ω₀ = The initial angular velocity of the mirrors = 115 rad/s clockwise

α = The angular acceleration = (115 - (-115))rad/s/(85 s) = -46/17 m/s²

t = The duration of the motion;

When the angular velocity, ω is zero, we get;

0 = 115 - 46/17·t

t = 85/2

Which indicates;

θ = 0 + 115× (85/2) + 0.5×(46/17) ×(85/2)² = 7331.25

θ = 7331.25 radians

θ = 7331.25/(2×π) ≈ 1166.8 rev

The mirrors would have turned through approximately 1166.8 revolutions when the angular gets to zero

Learn more about angular velocity and acceleration here:

brainly.com/question/13014974

#SPJ1

7 0

11 months ago

Why is the Earth generally warmer at the equator than at the poles?

Elenna [48]

Because of the earths curve and temperature. The temperature in an area depends on the amount of the Sun's energy reaching the surface in that area. The equator tilts closer to the sun than the poles do. I hope this helps!

5 0

2 years ago

Read 2 more answers

An oil droplet is sprayed into a uniform electric field of adjustable magnitude. The 0.11 g droplet hovers

ohaa [14]

Answer:

The direction of the field is downward, and negatively charged particles will experience an upwards force due to the field.

F = N e E where E is the value of the field and N e the charge Q

M g = N e E and M g is the weight of the drop

N = M g / (e E)

N = 1.1E-4 * 9.8 / (1.6E-19 * 370) = 1.1 * 9.8 / (1.6 * 370) * E15 = 1.82E13

.00011 kg is a very large drop

Q = N e = M g / E = .00011 * 9.8 / 370 = 2.91E-6 Coulombs

Check: N = Q / e = 2.91E-6 / 1.6E-19 = 1.82E13 electrons

7 0

2 years ago