Engineers are investigating the properties of a material for use as a wrapping product. Three identical

What’s the voltage of a battery in a circuit with resistance of 3 ohms and current of 5 amps?

alexira [117]

Answer: The correct answer is-15 Volts.

Explanation-

Voltage of a battery can be defined as the difference in electric potential that lies between the positive and negative terminals of a battery.

It can be calculated using Ohm's law, which states that the electric potential difference between two points on a circuit is equal to the product of the current that flows between the two points (I) and the total resistance that sis present between the two points. It can be mathematically depicted as-

ΔV = I • R

Putting the value of 'I' and 'R', we get-

ΔV = 5 X 3

= 15 V

8 0

3 years ago

HELP NOWWW Studies based on which of the following were used to describe the current model of the atom?

alexgriva [62]

Well I know that ernest rutherford did the gold foil experiment where he fired alpha particles at gold foil. This experiment founded the nucleus but I don't know if the current model of the atom is based on this.

Cathode ray tube experiments sounds like its to do with electrolysis so i dont think it can be that.

6 0

3 years ago

Read 2 more answers

The current theory of the structure of the

Mariana [72]

Answers:

a) $2.82(10)^{21} kg$

b) $1410 J$

c) $36.62 m/s$

Explanation:

<h3>a) Mass of the continent</h3>

Density $\rho$ is defined as a relation between mass $m$ and volume $V$ :

$\rho=\frac{m}{V}$ (1)

Where:

$\rho=2720 kg/m^{3}$ is the average density of the continent

$m$ is the mass of the continent

$V$ is the volume of the continent, which can be estimated is we assume it as a a slab of rock 5300 km on a side and 37 km deep:

$V=(length)(width)(depth)=(5300 km)(5300 km)(37 km)=1,030,330,000 km^{3} \frac{(1000 m)^{3}}{1 km^{3}}=1.03933(10)^{18} m^{3}$

Finding the mass:

$m=\rho V$ (2)

$m=(2720 kg/m^{3})(1.03933(10)^{18} m^{3})$ (3)

$m=2.82(10)^{21} kg$ (4) This is the mass of the continent

<h3>b) Kinetic energy of the continent</h3>

Kinetic energy $K$ is given by the following equation:

$K=\frac{1}{2}mv^{2}$ (5)

Where:

$m=2.82(10)^{21} kg$ is the mass of the continent

$v=4.8 \frac{cm}{year} \frac{1 m}{100 cm} \frac{1 year}{365 days} \frac{1 day}{24 hours} \frac{1 hour}{3600 s}=1(10)^{-9} m/s$ is the velocity of the continent

$K=\frac{1}{2}(2.82(10)^{21} kg)(1(10)^{-9} m/s)^{2}$ (6)

$K=1410 J$ (7) This is the kinetic energy of the continent

<h3>c) Speed of the jogger</h3>

If we have a jogger with mass $m=77 kg$ and the same kinetic energy as that of the continent $1413 J$ , we can find its velocity by isolating $v$ from (5):