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tiny-mole [99]
2 years ago
12

A current of 5. 0 amperes is passing through a piece of wire. Determine how long it takes for 30 coulombs of charge to pass thro

ugh this wire
Physics
1 answer:
Ainat [17]2 years ago
3 0

The time taken for the charge to pass through the cross-section of the wire is 6.0s.

Given the data in the question;

Current; I = 5.0A

Charge; Q = 30C

Time; t =  \ ?

<h3>Current</h3>

Current is simply the rate of flow of charged particles i.e electrons caused by EMF or voltage.

If a charge passes through the cross-section of a conductor in a given time, the current I is expressed as;

I = \frac{Q}{t}

Where Q is the charge and t is time elapsed.

We substitute our given values into the expression above to determine the time it took the charge to pass through this wire.

I = \frac{Q}{t}\\ \\5.0A = \frac{30C}{t} \\\\t = \frac{30C}{5.0A} \\\\t = 6.0s

Therefore, the time taken for the charge to pass through the cross-section of the wire is 6.0s.

Learn more about current: brainly.com/question/3192435

You might be interested in
A 15 m ladder with a mass of 51 kg is leaning against a frictionless wall, which makes an angle of 60 degrees with the horizonta
timofeeve [1]

Answer:

Explanation:

Given that, .

Mass of ladder is 51kg

Then, it weight is

WL = mg = 51 × 9.81 = 500.31N

This weight will act at the midpoint of the ladder

Length of ladder is 15m

The ladder makes an angle 55°C with the horizontal

An object whose mass is 81kg is at 4m from the bottom of the ladder

Then, weight of object

Wo = mg = 81 × 9.81 = 794.61 N

Using newton second law

Check attachment

Ng is normal force on the ground

Ff is the horizontal frictional force

Nw Is the normal force on the wall

ΣFy = 0

Ng = Wo + WL

Ng = 794.61 + 500.31

Ng = 1294.92 N

Also

ΣFx = 0

Ff — Nw = 0

Then,

Ff = Nw

Now taking moment about point A.

Check attachment

using the principle of equilibrium

Sum of clockwise moment equals to sum of anti-clockwise moment

Also note that the Normal force on the wall is not perpendicular to the ladder, so we will resolve that and also the weights of ladder and weight of object

Clockwise = Anticlockwise

Wo•Cos60 × 4 + WL•Cos60 × 7.5 = Nw•Sin60 × 15

794.61Cos60 × 4 + 500.31Cos60 × 7.5 = Nw × Sin60 × 15

1589.22 + 1876.163 = 12.99•Nw

3465.383 = 12.99•Nw

Nw = 3465.383 / 12.99

Nw = 266.77 N

Since, Nw = Ff

Then, Ff = 266.77N

the horizontal force exerted by the ground on the ladder is 266.77 N

8 0
3 years ago
Read 2 more answers
The sleigh is being pulled with a force of 800N and has a mass of 200 Kg.
kiruha [24]
The answer is 4 m/s/s
4 0
3 years ago
Read 2 more answers
A. A land speed car can decelerate at 9.8m/s. How long does it take the car to come to a complete stop from a run of 885 km/hr (
Nimfa-mama [501]

Answer:

A. 25.08 s

B. 3082.53 m

C. 3×10⁵ m/s²

Explanation:

A. Determination of the time.

This can be obtained as illustrated below:

Acceleration (a) = –9.8 m/s²

Initial velocity (u) = 245.8 m/s

Final velocity (v) = 0 m/s

Time (t) =.?

v = u + at

0 = 245.8 + (–9.8 × t)

0 = 245.8 – 9.8t

Collect like terms

0 – 245.8 = – 9.8t

– 245.8 = – 9.8t

Divide both side by –9.8

t = –245.8 / –9.8

t = 25.08 s

Therefore, it will take 25.08 s for the car to come to a complete stop.

B. Determination of the distance travelled by the car.

Acceleration (a) = –9.8 m/s²

Initial velocity (u) = 245.8 m/s

Final velocity (v) = 0 m/s

Distance (s) =?

v² = u² + 2as

0² = 245.8² + (2 × –9.8 × s)

0 = 60417.64 – 19.6s

Collect like terms

0 – 60417.64 = – 19.6s

– 60417.64 = – 19.6s

Divide both side by –19.6

s = –60417.64 / –19.6

s = 3082.53 m

Thus, the car travelled a distance of 3082.53 m before stopping completely.

C. Determination of the acceleration of the object.

Initial velocity (u) = 0 m/s

Final velocity (v) = 600 m/s

Distance (s) = 0.6 m

Acceleration (a) =?

v² = u² + 2as

600² = 0² + (2 × a × 0.6)

360000 = 0 + 1.2a

360000 = 1.2a

Divide both side by 1.2

a = 360000 / 1.2

a = 300000 = 3×10⁵ m/s²

7 0
3 years ago
You have two windlasses (winches) named X and Y. Windlass X drum diameter is 50cm and windlass Y drum diameter is 75cm. Both are
Blababa [14]

You have two windlasses  named X and Y. Windlass X drum diameter is 50cm and windlass Y drum diameter is 75cm. Both are hoisting a 50kg weight, Hence, X will lift faster and easier than Y

<h3>What is windlasses (winches) will lift faster?</h3>

Generally, the equation for the Torque is  mathematically given as

Torque(X)=weight x radius of drum

T=50x9.81x0.25

T=122.625 Nm

For, Torque (Y)

T=50x9.81x0.375

T=183.93 Nm

Generally, the equation for the Equation of Power is  mathematically given as

P= Torque * Angular velocity

Angular velocity (w)= 2 x pi x N/60

As a consequence, torque has an inverse relationship to speed. Because of this, the torque will decrease as the speed rises.

According to the previous statement, windlass X has a speed of lift that is at its maximum, but windlass Y has a speed that is very low.

Additionally, power is exactly proportional to torque, which means that the highest possible torque required the highest possible amount of energy and vice versa.

Therefore, X should raise at the highest possible speed while using the least amount of energy.

In conclusion, X will lift faster and easier than Y

Read more about windlasses

brainly.com/question/14289204

#SPJ1

4 0
1 year ago
A solid sphere has a radius of 0.200 m and a mass of 150.0 kg. how much work is required to get the sphere rolling with an angul
Allisa [31]

Here in this case we can use work energy theorem

As per work energy theorem

Work done by all forces = Change in kinetic Energy of the object

Total kinetic energy of the solid sphere is ZERO initially as it is given at rest.

Final total kinetic energy is sum of rotational kinetic energy and translational kinetic energy

KE = \frac{1}{2}Iw^2 +\frac{1}{2} mv^2

also we know that

I = \frac{2}{5}mR^2

w= \frac{v}{R}

Now kinetic energy is given by

KE = \frac{1}{2}(\frac{2}{5}mR^2)w^2 +\frac{1}{2} m(Rw)^2

KE = \frac{1}{5}mR^2w^2 +\frac{1}{2} mR^2w^2

KE = \frac{7}{10}mR^2w^2

KE = \frac{7}{10}*150*(0.200)^2(50)^2

KE = 10500 J

Now by work energy theorem

Work done = 10500 - 0 = 10500 J

So in the above case work done on sphere is 10500 J

7 0
2 years ago
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