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3 years ago

What is the current when the voltage is 18 volts and the resistance is 6 ohms?

Physics

2 answers:

juin [17]3 years ago

6 0

Voltage = current(I) * resistance (R)
V = 18
R = 6

18 = I * 6
I = 18/6 = 3 Amps or D

sergey [27]3 years ago

4 0

The answer to this question is D.) 3 amps

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The amount of energy that must be absorbed or lost to raise or lower the temperature of 1 g of liquid water by 1°c _____.

Fiesta28 [93]

Answer:

4.2 J

Explanation:

Specific heat capacity: This is defined as the amount of a heat required to rise a unit mass of a substance through a temperature of 1 K

From specific heat capacity,

Q = cmΔt.............................. Equation 1

Where Q = amount of energy absorbed or lost, c = specific heat capacity of water, m = mass of water, Δt = Temperature rise.

Given: m = 1 g = 0.001 kg, Δt = 1 °C

Constant : c = 4200 J/kg.°C

Substitute into equation 1

Q = 0.001×4200(1)

Q = 4.2 J.

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4 years ago

The rotational inertia of a thin rod about one end is 1/3 ML2. What is the rotational inertia of the same rod about a point loca

zlopas [31]

Answer:

The value is $I = 0.0932 ML ^2$

Explanation:

From the question we are told that

The rotational inertia about one end is $I_R = \frac{1}{3} ML^2$

The location of the axis of rotation considered is $d = 0.4 L$

Generally the mass of the portion of the rod from the axis of rotation considered to the end of the rod is $0.4 M$

Generally the length of the rod from the its beginning to the axis of rotation consider is

$k = 1 - 0.4 L = 0.6L$

Generally the mass of the portion of the rod from the its beginning to the axis of rotation consider is

$m = 1- 0.4 M = 0.6 M$

Generally the rotational inertia about the axis of rotation consider for the first portion of the rod is

$I_{R1} = \frac{1}{3} (0.6 M )(0.6L)^2$

$I_{R1} = \frac{1}{3} (0.6 M )L^2 0.6^2$

Generally the rotational inertia about the axis of rotation consider for the second portion of the rod is

$I_{R2} = \frac{1}{3} (0.6 M )(0.6L)^2$

=> $I_{R2} = \frac{1}{3} (0.4 M )(0.4L)^2$

=> $I_{R2} = \frac{1}{3} (0.4 M )L^2 0.4^2$

Generally by the principle of superposition that rotational inertia of the rod at the considered axis of rotation is

$I = \frac{1}{3} (0.6 M )L^2 0.6^2 + \frac{1}{3} (0.4 M )L^2 0.4^2$

=> $I = \frac{1}{3} ML ^2 [0.6 * (0.6)^2 + 0.4 * (0.4)^2 ]$

=> $I = 0.0932 ML ^2$

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