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vladimir2022 [97]
1 year ago
14

A flat coil of wire has an inductance of 40. 0 mh and a resistance of 6. 00 ω. It is connected to a 21. 2-v battery at the insta

nt t = 0. Consider the moment when the current is 2. 50 a
Physics
1 answer:
MissTica1 year ago
7 0

For a  flat coil of wire has an inductance of 40. 0 mh and a resistance of 6. 00 ω, the rate of energy being delivered is mathematically given as

P= 53 W

<h3>What rate is energy being delivered by the battery?</h3>

Generally, the equation for the Battery power  is mathematically given as

P = I (dt)V

Therefore

P= 2.50 A * 21.2V

P= 53 W

In conclusion, rate of energy being delivered

P= 53 W

Read more about Energy

brainly.com/question/13439286

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Andrej [43]
Energy needs to realease
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2 years ago
What is the lupac<br>name<br>H3C – CHT<br>the H₂ - CH - CH2<br>CH₂<br>CH₂<br>CH3​
dimulka [17.4K]

Answer:??

Explanation:

5 0
2 years ago
At what displacement of a sho is the energy half kinetic and half potential? what fraction of the total energy of a sho is kinet
expeople1 [14]

As we know that KE and PE is same at a given position

so we will have as a function of position given as

KE = \frac{1}{2}m\omega^2(A^2 - x^2)

also the PE is given as function of position as

PE = \frac{1}{2}m\omega^2x^2

now it is given that

KE = PE

now we will have

\frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2}m\omega^2x^2

A^2 - x^2 = x^2

2x^2 = A^2

x = \frac{A}{\sqrt2}

so the position is 0.707 times of amplitude when KE and PE will be same

Part b)

KE of SHO at x = A/3

we can use the formula

KE = \frac{1}{2}m\omega^2(A^2 - x^2)

now to find the fraction of kinetic energy

f = \frac{KE}{TE} = \frac{A^2 - x^2}{A^2}

f = \frac{A^2 - (\frac{A}{3})^2}{A^2}

f_k = \frac{8}{9}

now since total energy is sum of KE and PE

so fraction of PE at the same position will be

f_{PE} = 1 - f_k

f_{PE} = 1 - (8/9) = 1/9

7 0
3 years ago
What is the magnetic field amplitude of an electromagnetic wave whose electric field amplitude is 10v/m?
Nitella [24]

Answer:

3.33\cdot 10^{-8} T

Explanation:

For an electromagnetic wave, the relationship between magnetic field amplitude and electric field amplitude is given by

E=cB

where

E is the amplitude of the electric field

c is the speed of light

B is the amplitude of the magnetic field

For the electromagnetic wave in this problem, we have

E = 10 V/m is the amplitude of the electric field

So if we solve the formula for B, we find the amplitude of the magnetic field:

B=\frac{E}{c}=\frac{10 V/m}{3\cdot 10^8 m/s}=3.33\cdot 10^{-8} T

4 0
3 years ago
While spinning down from 500 rpm to rest, a flywheel does 3.9 kj of work. this flywheel is in the shape of a solid uniform disk
Ksivusya [100]

 

The answer is 4.0 kg since the flywheel comes to rest the kinetic energy of the wheel in motion is spent doing the work. Using the formula KE = (1/2) I w².

Given the following:

I =  the moment of inertia about the axis passing through the center of the wheel; w = angular velocity ; for the solid disk as I = mr² / 2 so KE = (1/4) mr²w². Now initially, the wheel is spinning at 500 rpm so w = 500 * (2*pi / 60) rad / sec = 52.36 rad / sec.

The radius = 1.2 m and KE = 3900 J

3900 J = (1/4) m (1.2)² (52.36)²

m = 3900 J / (0.25) (1.2)² (52.36)²

m = 3.95151 ≈ 4.00 kg

6 0
3 years ago
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