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2 years ago

What is the voltage of this circuit.

Physics

1 answer:

liberstina [14]2 years ago

6 0

Answer:

110.4 v

Explanation:

Find the equivalent resistance of the two parallel circuits

R1 and R2 in parallel = R1 * R2 / (R1+ R2) = <u>1 ohm </u>

similarly R4 and R5 = <u>4.77 ohms</u>

Now you can add the three resistances into one (R3=10)

1 ohm + 10 ohm + 4.77 ohm = 15.77 ohms

Now

V = IR

V = 7 amps * 15.77 ohm = 110.4 v

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If ram is 40kg and travels 200m in 30 sec find his power

WARRIOR [948]

Work done

N×200
mg200
40(10)(200)
400(200)
80000J

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80000/30
8000/3
2668W

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2 years ago

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A point P1 is located by the vector A = (3.74)î + (1.64)ĵ and a point P2 is located by the vector B = (1.60)î + (3.66)ĵ. The vec

seropon [69]

Answer:

$\vec{C} = ( \ - 2.14 \ , \ 2.02 \ )$

Explanation:

The vector that point from point P1 to point P2 its found simply by taking the vector at which point P2 its located and subtracting the vector at which point P1 its located:

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So:

$\vec{C} = ( \ 1.60 \ , \ 3.66 \ ) - ( \ 3.74 \ , \ 1.64 \ )$

$\vec{C} = ( \ 1.60 \ - \ 3.74 \ , \ 3.66 \ - \ 1.64 \ )$

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3 years ago

What frequency is received by a person watching an oncoming ambulance moving at 110 km/h and emitting a steady 800-Hz sound from

Bas_tet [7]

Answer:

check photo for solve

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3 years ago

Volume of an block is 5 cm3. If the density of the block is 250 g/cm3, what is the mass of the block ?

9966 [12]

Answer:

1.25kg

Explanation:

Simply multiply volume and density together

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3 years ago

A coin with a diameter 3.00 cm rolls up a 30.0 inclined plane. The coin starts out with an initial angular speed of 60.0 rad/s

sweet [91]

This question is in complete.The question is

A coin with a diameter 3.00 cm rolls up a 30.0° inclined plane. The coin starts out with an initial angular speed of 60.0 rad/s and rolls in a straight line without slipping. If the moment of inertia of the coin is(1/2) MR² , how far will the coin roll up the inclined plane (length along the ramp)? Hint: Conservation of mechanical energy.

Answer:

distance=0.124 m

Explanation:

$mgh=mglSin\alpha =(1/2)Iw_{i}^{2}+(1/2)mv^{2}\\ v=wR\\Solve for L\\L=((1/2)(1/2)0.015^{2}*60^{2}+(1/2)(60*0.015^{2} ))/9.8Sin30\\ L=0.124m$

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3 years ago