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4 years ago

Who developed a unit of electric potential and electromotive force

Physics

1 answer:

ehidna [41]4 years ago

4 0

Answer:

Alessandro Volta

Explanation:

Alessandro Volta composed a letter to his amigo, Sir Joseph Banks at the Royal Society of London, in March 1800. In it, Volta drew an image of his voltaic heap, an exacting heap of 24 copper and zinc plates, in addition to cardboard circles absorbed salt. Volta at that point associated copper wires to both the top and base to utilize it as power source.
The voltaic heap and wires would at that point, in principle, charge something helpful, similar to a light or a little engine, however since neither existed in 1800, Volta put the two wires on the tip of his tongue. His tongue, wet with salivation, was a decent conductor and finished the circuit. Volta gave himself a decent stun, accordingly demonstrating he had made another sort of battery, or electrochemical vitality source.
While researchers presently don't prescribe putting live wires on your tongue, Volta was respected for his commitments to science and we named the volt (V), or the unit of measure for electric potential, in his respect.

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A police car is traveling north on a straight road at a constant 16.0 m/s. An SUV traveling north at 30.0 m/s passes the police

Nastasia [14]

Answer:

It will take 15.55s for the police car to pass the SUV

Explanation:

We first have to establish that both the police car and the SUV will travel the same distance in the same amount of time. The police car is moving at constant velocity and the SUV is experiencing a deceleration. Thus we will use two distance fromulas (for constant and accelerated motions) with the same variable for t and x:

1. $x=x_{0}+vt$

2. $x=x_{0}+v_{0}t+\frac{at^{2}}{2}$

Since both cars will travel the same distance x, we can equal both formulas and solve for t:

$vt = v_{0}t+\frac{at^2}{2}\\\\ 16\frac{m}{s}t =30\frac{m}{s}t-\frac{1.8\frac{m}{s^{2}} t^{2}}{2}$

We simplify the fraction present and rearrange for our formula so that it equals 0:

$0.9\frac{m}{s^{2}} t^{2}-14\frac{m}{s}t=0 \\\\ t(0.9\frac{m}{s^{2}}t-14\frac{m}{s})=0$

In the very last step we factored a common factor t. There is two possible solutions to the equation at $t=0$ and:

$0.9\frac{m}{s^{2}}t-14\frac{m}{s}=0 \\\\ 0.9\frac{m}{s^{2}}t =14\frac{m}{s} \\\\ t =\frac{14\frac{m}{s}}{0.9\frac{m}{s^{2}}}=15.56s$

What this means is that during the displacement of the police car and SUV, there will be two moments in time where they will be next to each other; at $t=0 s$ (when the SUV passed the police car) and $t=15.56s$ (when the police car catches up to the SUV)

8 0

3 years ago

PLEASEE HELP!!! LOTS OF POINTS AND BRAINLIEST!!!!!!!!

Dennis_Churaev [7]

A clock and a battery

3 0

3 years ago

Read 2 more answers

The momentum of a 5-kilogram object moving at 6 meters per second is-

goldfiish [28.3K]

I think the answer is 30 but I’m not sure

3 0

2 years ago

Mass and Weight: What is the mass of an object that experiences a gravitational force of 685 N near Earth's surface where g

11Alexandr11 [23.1K]

Answer:

m = 69.9 kg

Explanation:

The mass and the weight of an object are two different quantities. Mass is basically the amount of matter that is present in a body. It remains same everywhere in the universe and measured in kilograms.

Weight is basically a force. It is the force by which earth attracts everything towards itself. The weight of an object changes from planet to planet, with the change in value of the gravitational acceleration (g).

Therefore, the relation between mass and weight of an object is given by the following formula:

W = mg

m = W/g

where,

m = mass = ?

W = Weight = 685 N

g = 9.8 m/s²

Therefore,

m = (685 N)/(9.8 m/s²)

4 0

3 years ago

I know how the voltage increasing/not enough/stays the same and how long the battery lasts with series and parallel connections.

kirill [66]

Answer:

It comes out the positive side of the battery and goes in to the negative side of the battery

Explanation:

There are already electrons in wires in a circuit before you add the battery. By adding the battery, you're giving the electrons the energy it needs to move along the circuit.

In a series circuit, the circuit is one continuous loop so there is only one path for the electrons to go - out of the positive side of the battery and around the circuit then goes back into the negative side of the battery.

However, with a parallel circuit, there are two or more ways the electrons can go so they take the path of least resistance. The electrons still go out the positive side of a battery but along the circuit, the electrons will go through the path of least resistance ( I tend to think of it like a net with holes in it - the lower the resistance the bigger the holes for the electrons to go through so more can fit in a set amount of time ) but the electrons still go out of the positive side and in through the negative

3 0

2 years ago