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vovangra [49]
3 years ago
14

The product of charge through, and potential across, an electrical device is:

Physics
2 answers:
melomori [17]3 years ago
5 0
The option is Work.

The product of charge and potential is equal to the energy. Adn, as we know work is related to energy as the capacity to do work.  

Alos, because, Potential is given as, V = E/q
or E = Vq
Thus, t<span>he product of charge through, and potential across, an electrical device is:work
</span>  
gregori [183]3 years ago
3 0

Answer:

The product of the charge and voltage is the work .

Explanation:

According to Ohm's law

The voltage is equal to the product of the current and resistance. Its unit is volt.

V= I\times R

Power :

Power is the defined as the electric energy per unit time. its unit is J/s or watt.

P=\dfrac{E}{t}

According to Ohm's law,

The power is equal to the square of voltage divided by the resistance.

P=\dfrac{V^2}{R}

The power equal to the product of resistance and square of current.

P = R\times I^2

The product of current and voltage is equal to the power.

P=I\times V

Current :

The current is equal to the voltage divided by the resistance.

The ampere is the unit of current.

Work : Work is related to the energy.

The Voltage is equal to the energy divided by the charge.

V=\dfrac{E}{q}

The product of charge and potential is called energy.

E=q\times V

Hence, The product of the charge and voltage is the work .  

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Grace rides her bike at a constant speed of 6 miles per hour. How far can she travel in 2 1/2 hours?
aliina [53]

Answer:

15 miles

Explanation:

6 miles per hour

2 1/2 hours

6 x 2 = 12

6 x 1/2 = 3

12 + 3 = 15

7 0
3 years ago
Two ropes are attached to either side of a 100.0 kg wagon as shown below. The rope on the right is pulled at an angle 40.0° rela
NikAS [45]

The acceleration of the wagon is found by applying Newton's Second Law of motion.

1. The responses for question 1 are;

  • x-component of the tension in the rope on the right is approximately <u>91.93 N</u>
  • y-component of the tension in the rope on the right is approximately <u>71.135 N</u>
  • x-component of the tension in the rope on the left is -80.0 N
  • y-component of the tension in the rope on the left is 0

2. The net force in the x-direction is approximately <u>11.93 N</u>

3. The net acceleration of the wagon in the horizontal direction is approximately <u>0.1193 m/s²</u>.

Reasons:

The given parameters are;

Mass of the wagon, m = 100.0 kg

Angle of inclination to the horizontal of the rope to the right, θ = 40.0°

Tension in the rope on the right = 120.0 N

Direction in which the rope on the left is pulled = To the west

Tension in the rope on the left = 80.0 N

1. The <em>x</em> and <em>y</em> component of the tension in the rope on the right are;

x-component = 120.0 N × cos(40.0°) ≈ <u>91.93 N</u>

y-component = 120.0 N × sin(40.0°) ≈ <u>77.135 N</u>

The <em>x</em> and <em>y</em> component of the tension in the rope on the left are;

x-component = 80.0 N × cos(180°) = <u>-80.0 N</u>

y-component = 80.0 N × sin(180°) = <u>0.0 N</u>

2. The net force in the horizontal direction, Fₓ, is found as follows;

Fₓ = The x-component of the rope on the left + The x-component of the rope on the right

Which gives;

Fₓ = 91.93 N - 80.0 N = <u>11.93 N</u>

3. The net acceleration of the block is given as follows;

According to Newton's Second Law of motion, we have;

Force in the horizontal direction, Fₓ = Mass of wagon, m × Acceleration of the wagon in the horizontal direction, aₓ

Fₓ = m × aₓ

Therefore;

\displaystyle a_x = \frac{F_x}{m}  \approx \frac{11.93 \, N}{100.0 \, kg} = \mathbf{0.1193 \ m/s^2}

  • The acceleration of the wagon in the horizontal direction, aₓ ≈ <u>0.1193 m/s²</u>.

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