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siniylev [52]
3 years ago
13

GIVING BRAINLIEST PLEASE HELP ME!!!

Physics
1 answer:
kumpel [21]3 years ago
4 0

Answer:

I'm Pretty sure the answer your looking for is C

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In your own words, describe the purpose of using a chemical equation.
Scilla [17]
Chemical equations show how chemicals interact when a reaction occurs. They use symbols and formulas to show the chemical reaction so that chemical names do not need to be written out. They also help keep track of all elements and the number of atoms of each element on each side of the equation.
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3 years ago
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The inductance of a solenoid with 450 turns and a length of 24 cm is 7.3 mH. (a) What is the cross-sectional area of the solenoi
Novay_Z [31]

Answer: 0.43 V

Explanation:

L = [μ(0) * N² * A] / l

Where

L = Inductance of the solenoid

N = the number of turns in the solenoid

A = cross sectional area of the solenoid

l = length of the solenoid

7.3*10^-3 = [4π*10^-7 * 450² * A] / 0.24

1.752*10^-3 = 4π*10^-7 * 202500 * A

1.752*10^-3 = 0.255 * A

A = 1.752*10^-3 / 0.255

A = 0.00687 m²

A = 6.87*10^-3 m²

emf = -N(ΔΦ/Δt).........1

L = N(ΔΦ/ΔI) so that,

N*ΔΦ = ΔI*L

Substituting this in eqn 1, we have

emf = - ΔI*L / Δt

emf = - [(0 - 3.2) * 7.3*10^-3] / 55*10^-3

emf = 0.0234 / 0.055

emf = 0.43 V

6 0
3 years ago
The position of a 2.2 kg mass is given by x=(2t3â5t2) m, where t is in seconds.
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These types of electromagnetic waves are listed in order of decreasing
viktelen [127]
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4 0
3 years ago
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Three identical resistors are connected in parallel to a battery. If the current of 12. A flows from the battery, how much curre
Doss [256]

Answer:

4 A

Explanation:

We are given that

R_1=R_2=R_3=4\Omega

I=12 A

We have to find the current flowing through each resistor.

We know that in parallel combination current flowing through different resistors are different and potential difference across each resistor is same.

Formula :

\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}

Using the formula

\frac{1}{R}=\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{3}{4}

R=\frac{4}{3}\Omega

V=IR

Substitute the values

V=12\times \frac{4}{3}=16 V

I_1=\frac{V}{R_1}=\frac{16}{4}=4 A

I_1=I_2=I_3=4 A

Hence, current flows through any one of the resistors is 4 A.

7 0
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