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3 years ago

A circuit has a current of 3.6 A and a resistance of 5.0 Ω. What is the voltage applied to the circuit?

Physics

2 answers:

Inessa [10]3 years ago

7 0

The key formula ===> Voltage = (current) x (resistance)

Plug in the numbers given ===> Voltage = (3.6 A) x (5.0 ohms)

Dooda multiplication ===> Voltage = 18 volts

seraphim [82]3 years ago

5 0

Answer: The correct answer is 18 V.

Explanation:

The mathematical expression from Ohm's law is as follows:

V=IR

Here, V is the voltage, R is the resistance and I is the current.

It is given in the problem that A circuit has a current of 3.6 A and a resistance of 5.0 Ω.

Put R= 5 ohm and I= 3.6 A in the above expression.

V= (3.6)(5)

V=18 V.

Therefore, the value of voltage is 18 V.

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Read 2 more answers

What force must the deltoid muscle provide to keep the arm in this position?

ruslelena [56]

Answer:

Deltoid Force, $F_{d} = \frac {r_{a}mgsin\alpha_{a}}{r_{d}sin\alpha_{d}}$

Additional Information:

Some numerical information are missing from the question. However, I will derive the formula to calculate the force of the deltoid muscle. All you need to do is insert the necessary information and calculate.

Explanation:

The deltoid muscle is the one keeping the hand arm in position. We have two torques that apply to the rotating of the arm.

1. The torque about the point in the shoulder for the deltoid muscle, $T_{Deltoid}$

2. The torque of the arm, $T_{arm}$

Assuming the arm is just being stretched and there is no rotation going on,

$T_{Deltoid}$ = 0

$T_{arm}$ = 0

⇒ $T_{Deltoid} = T_{arm}$

$r_{d}F_{d}sin\alpha_{d} = r_{a}F_{a}sin\alpha_{a}$

Where,

$r_{d}$ is radius of the deltoid

$F_{d}$ is the force of the deltiod

$\alpha_{d}$ is the angle of the deltiod

$r_{a}$ is the radius of the arm

$F_{a}$ is the force of the arm , $F_{a} = mg$ which is the mass of the arm and acceleration due to gravity

$\alpha_{a}$ is the angle of the arm

The force of the deltoid muscle is,

$F_{d} = \frac {r_{a}F_{a}sin\alpha_{a}}{r_{d}sin\alpha_{d}}$

but $F_{a} = mg$ ,

∴ $F_{d} = \frac {r_{a}mgsin\alpha_{a}}{r_{d}sin\alpha_{d}}$

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3 years ago