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Andre45 [30]
2 years ago
5

WILL GIVE BRAINLIEST!!!

Physics
1 answer:
Gemiola [76]2 years ago
3 0

Answer:

1/Re=1/R1  +1/R2= 2x1/R1=2/R1=2/R2=2/50=1/25

1/Re=1/25, so the answer is Re= 25

the answer is A. 25.0

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A solid uniform disk of diameter 3.20 m and mass 42 kg rolls without slipping to the bottom of a hill, starting from rest. If th
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Answer:

A(3.56m)

Explanation:

We have a conservation of energy problem here as well. Potential energy is being converted into linear kinetic energy and rotational kinetic energy.

We are given ω= 4.27rad/s, so v = ωr, which is 6.832 m/s. Place your coordinate system at top of the hill so E initial is 0.

Ef= Ug+Klin+Krot= -mgh+1/2mv^2+1/2Iω^2

Since it is a solid uniform disk I= 1/2MR^2, so Krot will be 1/4Mv^2(r^2ω^2=  v^2).

Ef= -mgh+3/4mv^2

Since Ef=Ei=0

Mgh=3/4mv^2

gh=3/4v^2

h=0.75v^2/g

plug in givens to get h= 3.57m

5 0
3 years ago
A multimeter in an RL circuit records an rms current of 0.600 A and a 50.0-Hz rms generator voltage of 110 V. A wattmeter shows
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Answer:

(a) The impedance in the circuit is Z=183.33\Omega.

(b)The resistance is R=38.89\Omega.

(c) The inuctance is 0.57 H.

Explanation:

(a)

The expression for the impedance is as follows:

Z=\frac{V_rms}{I_rms}

Here, V_rms is the rms voltage and I_rms is the rms current.

PutV_rms=110 V and I_rms=0.600 A.

Z=\frac{110}{0.600}

Z=183.33\Omega

Therefore, the impedance in the circuit is Z=183.33\Omega.

(b)

The expression for the average power is as follows;

P_{a}=I_{rms}^{2}R

Here, P_{a} is the average power and R is the resistance.

Calculate the resistance by rearranging the above expression.

R=\frac{P_{a}}{I_{rms}^{2}}

Put P_{a}=14W and

R=\frac{14}{{0.600}^{2}}

R=38.89\Omega

Therefore, the resistance is R=38.89\Omega.

(c)

The expression for the impedance is as follows;

Z^{2}=R^{2}+X_{L}^{2}

Here,X_{L} is the inductive reactance.

Put Z=183.33\Omega and R=38.89\Omega.

(183.33)^{2}=(38.89)^{2}+X_{L}^{2}

X_{L}=179.16\Omega

The expression for the inductive reactance in terms of  frequency is as follows;

X_{L}=2\pi fL

Here, L is the inductance.

Calculate the inductance by rearranging the above expression.

L=\frac{X_{L}}{2\pi f}

Put X_{L}=179.16\Omega and f=50Hz.

L=\frac{179.16}{2\pi (50)}

L=0.57 H

Therefore, the inuctance is 0.57 H.

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