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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

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Suppose that the height of the incline is h = 14.7 m. Find the speed at the bottom for each of the following objects. 1.solid sp
tensa zangetsu [6.8K]

Answer:

1. 14.4 m/s  2. 13.2 m/s 3. 12.0 m/s 4. 13.9 m/s

Explanation:  

Assuming no friction present, the different objects roll without slipping, so there is a constant relationship between linear and angular velocity, as follows:

ω= v/r

If no friction exists, the change in total kinetic energy must be equal in magnitude to the change in the gravitational potential energy:

∆K = -∆U

 ½ *m*v² + ½* I* ω²  = m*g*h

Simplifying and replacing the value of the angular velocity:

½ * v² + ½ I *(v/r)² = g*h (1)

In order to answer the question, we just need to replace h by the value given, and I (moment of inertia) for the value for each different object, as follows:

  •  Solid Sphere I = 2/5* m *r²

                Replacing in (1):

                ½ * v² + ½ (2/5 *m*r²) *(v/r)² = g*h

                Replacing by the value given for h, and solving for v:

                v = √(10/7*9.8 m/s2*14.7 m)  = 14. 4 m/s

  • Spherical shell I=2/3*m*r²

                Replacing in (1):

                ½ * v² + ½ (2/3 *m*r²) *(v/r)² = g*h

                Replacing by the value given for h, and solving for v:

                v = √(6/5*9.8 m/s2*14.7 m)  = 13.2 m/s

  • Hoop   I= m*r²

                Replacing in (1):

                ½ * v² + ½ (m*r²) *(v/r)² = g*h

               Replacing by the value given for h, and solving for v:

               v = √(9.8 m/s2*14.7 m)  = 12.0 m/s

  • Cylinder I = 1/2 * m* r²

                 Replacing in (1):

                ½ * v² + ½ (1/2 *m*r²) *(v/r)²= g*h

                 Replacing by the value given for h, and solving for v:

                v = 2*√(1/3*9.8 m/s2*14.7 m)  = 13.9 m/s

5 0
3 years ago
A 35 kg box rests on the back of a truck. The coefficient of static friction bet?005 (part 1 of 2)A 35 kg box rests on the back
elena-14-01-66 [18.8K]

Answer with Explanation:

We are given that

Mass of box=35 kg

Coefficient of static friction between box and truck bed=0.202

Acceleration due to gravity=9.8 m/s^2

a.We have to find the force by which the box accelerates forward.

Force by which box accelerates=\mu mg=0.202\times 9.8\times 35

Force by which box accelerates=62.286 N

b.We have to find the maximum acceleration can the truck have before the box slides.

Force =friction force

ma=\mu mg

a=\mu g=9.8\times 0.202=1.9796 m/s^2

Hence, the truck can have maximum acceleration before the box slide=1.9796 m/s^2

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Bruno the bat flies at a speed of 0.5 m/s in circle of radius 1 m. What is his acceleration?
beks73 [17]

Answer:

Acceleration is 0.25m/s^2

Explanation:

Given the following :

Speed = 0.5m/s

Radius(r) of circle = 1m

Acceleration round a circular path is given as :

a = v^2 / r

Where

a = acceleration of the body

v = speed / velocity

r = radius

Therefore,

a = v^2 / r

a = (0.5)^2 / 1

a = 0.25m/s^2

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