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Zolol [24]
2 years ago
6

A copper wire carries 1 amp of electric current. what kind of charge does the electron flow create in the wire?

Physics
1 answer:
torisob [31]2 years ago
7 0
A current is just the development of electrons. Since there is no electric field from a (steady) current, this implies the wire is still electrically unbiased, ie uncharged. Accordingly the appropriate response is no charge. I trust the appropriate response will help you.
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A distance of 2.00 mm separates two objects of equal mass. If the gravitational force between them is 0.0104 N, find the mass of
aleksklad [387]

Given the distance r = 2/1000 m, the force between them F = 0.0104 N, the mass of the two object can be calculated using formula:

F = G(m1m2)/r^2 since the mass are equal F = G (m^2)/r^2

And where G = is the gravitational constant (6.67E-11 m3 s-2 kg-1)

The mass of the two objects are 24.96 kg

6 0
2 years ago
A seasoned mini golfer is trying to make par on a tricky number five hole. The golfer can complete the hole by hitting the ball
liberstina [14]

Answer:5.17 m/s

Explanation:

Given

let u be the speed at cliff initial point

range over cliff is 1.45 m

and range of projectile is given by

R=\frac{u^2\sin 2\theta }{g}

1.45=\frac{u^2\sin 90}{9.8}

u=3.77 m/s

Conserving Energy

E_{bottom}=E_{initial\ point\ at\ cliff}

Kinetic energy=Kinetic energy +Potential energy gained

Let v be the initial velocity

\frac{mv^2}{2}=mgh+\frac{mu^2}{2}

v^2=u^2+2gh

v=\sqrt{u^2+2gh}

v=\sqrt{3.77^2+2\time 9.8\times 0.64}

v=\sqrt{26.75}=5.17 m/s

5 0
2 years ago
A 4.0-mF capacitor initially charged to 50 V and a 6.0-mF capacitor charged to 30 V are connected to each other with the positiv
Juli2301 [7.4K]

Answer:

<em>The final charge on the 6.0 mF capacitor would be 12 mC</em>

Explanation:

The initial charge on 4 mF capacitor  = 4 mf  x 50 V = 200 mC

The initial Charge on 6 mF capacitor  = 6 mf x 30 V =180 mC

Since the negative ends are joined together  the total charge on both capacity would be;

q = q_{1} -q_{2}

q = 200 - 180

q = 20 mC

In order to find the final charge on the 6.0 mF capacitor we have to find the combined voltage

q = (4 x V) + (6 x V)

20 = 10 V

V = 2 V

For the final charge on 6.0 mF;

q = CV

q = 6.0 mF x 2 V

q =  12 mC

Therefore the final charge on the 6.0 mF capacitor would be 12 mC

5 0
2 years ago
Read 2 more answers
Suppose that air resistance cannot be ignored. For the position at which the person has jumped from the platform and the cord re
Kamila [148]

Answer:

The stretch cord stores potential energy as a result of stretching but due to kinetic energy, it will move back to its original state. Since air resistance is not being ignored in this case, it will experience a slight delay in stretching at first.

Explanation:

In case, where air resistance is being ignored the stretch cord will stretch as it normally does.              

  • Air resistance is a force that any object experiences as a result of its motion through the air.  

There are various factors that affect air resistance like speed, the density of air, area, the shape of an object etc. Meanwhile, the density of air changes with temperature or altitude. <em>Hence this force is not constant but is thought to be constant during short time frames.  </em>

6 0
2 years ago
A spaceship of mass 8600 kg is returning to Earth with its engine turned off. Consider only the gravitational field of Earth. Le
Katyanochek1 [597]

Answer:

\Delta KE = 4.20\times 10^{13}\ J

Explanation:

given,

mass of spaceship(m) = 8600 Kg

Mass of earth = 5.972 x 10²⁴ Kg

position of movement of space ship

R₁ = 7300 Km

R₂ = 6700 Km

the kinetic energy of the spaceship increases by = ?

Increase in Kinetic energy = decrease in potential energy

    \Delta KE = GMm (\dfrac{1}{R_2}-\dfrac{1}{R_1})

    \Delta KE = GMm (\dfrac{R_1-R_2}{R_2R_1})

    \Delta KE = 6.67 \times 10^{-11}\times 5.972 \times 10^{24}\times 8600 (\dfrac{7300 - 6700}{7300 \times 6700})

    \Delta KE = 6.67 \times 10^{-11}\times 5.972 \times 10^{24}\times 8600 (\dfrac{600}{48910000})

    \Delta KE = 4.20\times 10^{13}\ J

5 0
2 years ago
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