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3 years ago

Who developed a unit of electric potential and electromotive force

Physics

1 answer:

ehidna [41]3 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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_____________________ are structures in cardiac muscle tissue that allow impulses to spread across the heart more rapidly.

Darya [45]

Gap junctions in the intercalated discs allow impulses to be spread across the heart more quickly. This is because gap junctions allow particles/signals to pass through, thus making cells with gap junctions more able to interact.

One more thing—you posted this in the physics section rather than biology.

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3 years ago

Calculate the magnitude of the acceleration due to gravity on the surface of Earth due to the Moon.

Fudgin [204]

Answer:

$g'_h=1.096\times 10^{-5}\ m.s^{-2}$

Explanation:

We know that the gravity on the surface of the moon is,

$g'=\frac{g}{6}$

$g'=1.63\ m.s^{-2}$

<u>Gravity at a height h above the surface of the moon will be given as:</u>

$g'_h=\frac{G.m}{(r+h)^2}$ ..........................(1)

where:

G = universal gravitational constant

m = mass of the moon

r = radius of moon

We have:

$G=6.67\times 10^{-11}\ m^3.s^{-2}.kg^{-1}$

$m=7.35\times 10^{22}\ kg$

$r=1.74\times 10^6\ m$

$h=384.4\times 10^6\ m$ is the distance between the surface of the earth and the moon.

Now put the respective values in eq. (1)

$g'_h=\frac{6.67\times 10^{-11}\times 7.35\times 10^{22}}{(1.74\times 10^6+384.4\times 10^6)^2}$

$g'_h=1.096\times 10^{-5}\ m.s^{-2}$ is the gravity on the moon the earth-surface.

4 0

2 years ago

A 2 nC point charge is at the origin, and a second 5 nC point charge is on the x-axis at x = 8 m. Find the electric field (magni

dimaraw [331]

Answer:

The magnitude of the electric field is 5.75 N/C towards positive x- axis.

Explanation:

Given that,

Point charge at origin = 2 nC

Second charge = 5 nC

Distance at x axis = 8 m

We need to calculate the electric field at the point x = 2 m

Using formula of electric field

$E=\dfrac{1}{4\pi\epsilon_{0}}(\dfrac{q_{1}}{r_{1}^2}+\dfrac{q_{2}}{r_{2}^2})$

Put the value into the formula

$E=9\times10^{9}\times(\dfrac{2\times10^{-9}}{2^2}+\dfrac{5\times10^{-9}}{(8-2)^2})$

$E=5.75\ N/C$

The direction is toward positive x- axis.

Hence, The magnitude of the electric field is 5.75 N/C towards positive x- axis.

7 0

2 years ago

A 60 kg skier starts from rest at the top of a frictionless slope of height of 35 meters. The velocity at the bottom is v. If a

Solnce55 [7]

Answer:

Explanation:

Speed of skier without parachute

= √ 2gh

= √ 2 x 9.8 x 35

= 26.2 m / s

Speed of skier with parachute

net force downwards

mg - 200

= 60 x 9.8 -200

= 388 N

acceleration = 388 / 60

a = 6.47 m / s

v = √ 2ah

= √ 2 x 6.47 x 35

= 21.28 m / s

8 0

2 years ago

Read 2 more answers

S Four point charges each having charge Q are located at the corners of a square having sides of length a. Find expressions for(

Ket [755]

The total electric potential at the center of the square due to the four charges is V = √2Q/πÈa.

<h3>What do you mean by electric potential? </h3>

The amount of work needed to move a unit charge from a reference point to a specific point against an electric field. It's SI unit is volt.

V = kq/r

Where V represents electric potential, K is coulomb constant, q is Charge and r is distance between any two around charge to the point charge.

Electric potential at O due to four charges is given by,

V = 4KQ/ r

where, r = √2a/2 = a/√2

V = 4k × Q√2/a

V = √2Q/πÈa

The total electric potential at the center of the square due to the four charges is V = √2Q/πÈa.

To learn more about electric potential refer to:

brainly.com/question/12645463

#SPJ4

3 0

1 year ago