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oksano4ka [1.4K]
3 years ago
12

What force is produced when a battery causes an electric current to flow trough a circuit made of copper wire

Physics
1 answer:
zzz [600]3 years ago
7 0
You'll hear that force called different things in different places. It
may be called "electromotive force", "EMF", "potential difference",
or "voltage".

It's just a matter of somehow causing the two ends of the wire
to have different electrical potential. When that happens, the
free electrons in the copper suddenly have a burning desire to
travel ... away from the end that's more negative, toward the end
that's more positive, and THAT's an "electric current".
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An empty parallel plate capacitor is connected between the terminals of a 4.4-V battery and charged up. The capacitor is then di
torisob [31]

Answer:

d

Explanation:

idk

7 0
3 years ago
A flying squirrel has a mass of 0. 765 kg. He jumps off the tree and hits the ground with 125 joules of energy. Determine how hi
uranmaximum [27]

Answer:

Height, h = 16.67 m

Explanation:

We have,

Mass of a squirrel is 0.765 kg.

He jumps off the tree and hits the ground with 125 joules of energy.

It is required to find the height up on the tree the squirrel was when it jumped.

The energy possessed by the squirrel is called its gravitational potential energy. It can be given by :

E=mgh

h is height up on the tree the squirrel was when it jumped

h=\dfrac{E}{mg}\\\\h=\dfrac{125\ J}{0.765\ kg\times 9.8\ m/s^2}\\\\h=16.67\ m

So, the squirrel will go to a height of 16.67 m.

6 0
3 years ago
50POINTS! Find the orbital speed of a satellite in a circular orbit 1700km above the surface of the Earth. M_earth 5.97e24kg, r_
ivolga24 [154]
<h2>Answer: 7020.117 m/s</h2>

Explanation:

The velocity of a satellite describing a circular orbit is<u> constant</u> and defined by the following expression:  

V=\sqrt{G\frac{M}{R}} (1)  

Where:  

G=6.674({10}^{-11})\frac{N{m}^{2}}{{kg}^{2}} is the gravity constant

M_{Earth}=5.97{10}^{24}kg the mass of the massive body around which the satellite is orbiting, in this case, the Earth .

R=r_{Earth}+h=8080000m the radius of the orbit (measured from the center of the planet to the satellite).  

This means the radius of the orbit is equal to <u>the sum</u> of the average radius of the Earth r_{Earth} and the altitude of the satellite above the Earth's surface h.

Note this orbital speed, as well as orbital period, does not depend on the mass of the satellite. It depends on the mass of the massive body (the Earth).

Now, rewriting equation (1) with the known values:

V=\sqrt{(6.674({10}^{-11})\frac{N{m}^{2}}{{kg}^{2}})\frac{5.97{10}^{24}kg}{8080000m}}

V=7020.117\frac{m}{s}  

6 0
3 years ago
Read 2 more answers
A centrifuge in a biology laboratory rotates at an angular speed of 3,400 rev/min. When switched off, it rotates 48.0 times befo
IgorLugansk [536]

Answer:

- 210 rad/s²

Explanation:

n = frequency of rotation = 3400/60 = 170/3 per sec.

angular velocity  ω ( 0 ) at time 0  = 2π n = 2π x 170/3

angular velocity at time t = ω(t) = 0

now, ω²( t) = w²(o) + 2α Φ ( α = angular acceleration and Φ = angular displacement) = 2π x 48 rad.

0 = ( 2π x 170/3 )² + 2α x 48 x 2π

α = - (2π x 170 x 170 )/ (3 x 3 x 2 x 48 ) = 210 rad / s²

7 0
3 years ago
What area must the plates of a capacitor be if they have a charge of 5.7uC and an electric field of 3.1 kV/mm between them? O 0.
kirill [66]

Answer:

Area of the plates of a capacitor, A = 0.208 m²

Explanation:

It is given that,

Charge on the parallel plate capacitor, q = 5.7\ \mu C=5.7\times 10^{-6}\ C

Electric field, E = 3.1 kV/mm = 3100000 V/m

The electric field of a parallel plates capacitor is given by :

E=\dfrac{q}{A\epsilon_o}

A=\dfrac{q}{E\epsilon_o}

A=\dfrac{5.7\times 10^{-6}\ C}{3100000\ V/m\times 8.85\times 10^{-12}\ F/m}

A = 0.208 m²

So, the area of the plates of a capacitor is 0.208 m². Hence, this is the required solution.

8 0
3 years ago
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