Does anyone know these four?! I really need help and fast pls

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

Peter designed a road with a curve of radius 30 m that is banked so that a 950 kg car traveling at 40.0 km/h can round it even i

spayn [35]

Answer:

$v = 15.56 m/s$

$v = 56 km/h$

Explanation:

When coefficient of friction is approximately zero then we have

$F_ncos\theta = mg$

$F_n sin\theta = \frac{mv^2}{R}$

$tan\theta = \frac{v^2}{Rg}$

here we know that

$v = 40 km/h = 11.11 m/s$

R = 30 m

$tan\theta = \frac{11.11^2}{30\times 9.81}$

$\theta = 22.75 degree$

now when friction coefficient is 0.30 then we have

$F_n cos\theta = mg + F_f sin\theta$

$F_f cos\theta + F_n sin\theta = \frac{mv^2}{R}$

now we have

$v = \sqrt{Rg(\frac{\mu + tan\theta}{1 - \mu tan\theta})}$

$v = \sqrt{30(9.81)(\frac{0.30 + tan22.75}{1 - (0.30) tan22.75})}$

$v = 15.56 m/s$

$v = 56 km/h$

3 0

3 years ago

What is the speed of a sound wave that takes 0.5 s to travel 750 m?

kvasek [131]

Answer:

C. 1500.

Explanation:

750 / .5 = 1500.

Hope this helps & best of luck!

Feel free to message me if you need more help! :)

4 0

3 years ago

What is the pressure at 5000 km below the surface of the earth

attashe74 [19]

Answer:

The centre of the earth is harder to study than the centre of the sun." Temperatures in the lower mantle the reach around 3,000-3,500 degrees Celsius and the barometer reads about 125 gigapascals, about one and a quarter million times atmospheric pressure.

Explanation:

7 0

2 years ago

Describe how the mass, luminosity, surface temperature, and radius of main-sequence stars change in value going from the “bottom

Sladkaya [172]

Answer:

1. Least massive stars are the coolest and least luminous, lower right of main sequence, on HR diagram.

2. Most massive are the hottest and most luminous, upper left of main sequence on Hr Diagram.

3. The radius of stars are related to their sprectral type. having the O being the hottest upper left and M being the coolest bottom right.

4 0

3 years ago