All categories

3 years ago

A charge of 90 C passes through a wire in 1 hour 15 minutes . what is the current in the wire

2 answers:

dsp733 years ago

7 0

Current = C / seconds.

1 hour = 3600 seconds
15 minutes = 900 seconds

1 hour 15 minutes = 4500 seconds

90 C / 4500 sec = 0.02 amperes = 20 milliamperes

timurjin [86]3 years ago

5 0

We calculate current from the formula:
$I= \frac{q}{t}$ , where q is a electric charge transferred over time t
Time should be converted to seconds:
1h 15 min= 75min= 4500s
I= $\frac{90C}{4500s}=0,02A$ Result is in unit-Ampere

You might be interested in

Fill in the blanks below: Urgently need help!!!

miskamm [114]

Answer:

In a circuit , VOLTAGE can be said to be the "source" or the "push of electrons". This push then creates what is known as a  CURRENT , which is the flow of electric charge through the circuit. This flow can the slowed down or restricted by RESISTOR , and this is also what can be harnessed in order to use electric ENERGY .

Explanation:

Voltage:

It is the 'push' that causes charges to move in a wire or other electrical conductor, also it is a Source input to the electric circuit.

Measured in Volts.

Current:

An electric current is the rate of flow of electric charge from a point or through a region.

Measured in Ampere.

Resistor:

Resistor is used to resist the flow of charge or to resist the current called as Resistance.

Measured in Ohms.

Electric Energy:

Electrical energy is a form of energy resulting from the flow of electric charge.

Measured in Joules.

In a circuit , voltage can be said to be the "source" or the "push of electrons". This push then creates what is known as a current, which is the flow of electric charge through the circuit. This flow can the slowed down or restricted by resistor, and this is also what can be harnessed in order to use electric energy.

8 0

3 years ago

2. A powerful experimental sewing machine is powered by a mass-spring system. This

Alexus [3.1K]

We have that the Number of stitches per sec and he mass of oscillation motion is mathematically given as

a) Nt=25stitches per sec

b) m=2.033e-5kg

<h3>Number of stitches per sec and he mass of oscillation motion</h3>

Question Parameters:

This sewing machine is capable of stitching 1,500 stiches in one minute.

If the sewing machine has a spring constant of 0.5 N/m,

Generally the equation for the Number of stitches per sec is mathematically given as

Nt=N/t

Therefore

Nt=1500/60

Nt=25stitches per sec

Generally the equation for the Time t is mathematically given as

$T=2\pi\sqrt{\frac{m}{k}}$

Therefore

$0.04=2\pi\sqrt{\frac{m}{0.5}}\\\\m=\frac{0.5*0.04^2}{4\pi^2}$

m=2.033e-5kg

For more information on Mass visit

brainly.com/question/15959704

7 0

2 years ago

A 240 g toy car is placed on a narrow 60-cm-diameter track with wheel grooves that keep the car going in a circle. The 1.0 kg tr

lesya [120]

Answer:

The track's angular velocity is W2 = 4.15 in rpm

Explanation:

Momentum angular can be find

I = m*r^2

P = I*W

So to use the conservation

P1 + P2 = 0

I1*W1 + I2*W2 = 0

Solve to w2 to find the angular velocity

0.240kg*0.30m^2*0.79m/s=-1kg*0.30m^2*W2

W2 = 0.435 rad/s

W2 = 4.15 rpm

8 0

3 years ago

How often is water added to the Earth system?

rosijanka [135]

Water is never added to earth system. Water forever remains in the water cycle on earth, so it goes from the ground, to the air, to the rain, to the sea, and round and round continuously. This cycle means that there does not need to be new water added to the earth, because it recycles any water that already exists of its own accord.

4 0

4 years ago

A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.15

Oksanka [162]

1) 5.5 N

When the ball is at the bottom of the circle, the equation of the forces is the following:

$T-mg = m\frac{v^2}{R}$

where

T is the tension in the string, which points upward

mg is the weight of the string, which points downward, with

m = 0.158 kg being the mass of the ball

g = 9.8 m/s^2 being the acceleration due to gravity

$m \frac{v^2}{R}$ is the centripetal force, which points upward, with

v = 5.22 m/s being the speed of the ball

R = 1.1 m being the radius of the circular trajectory

Substituting numbers and re-arranging the formula, we find T:

$T=mg+m\frac{v^2}{R}=(0.158 kg)(9.8 m/s^2)+(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=5.5 N$

2) 3.9 N

When the ball is at the side of the circle, the only force acting along the centripetal direction is the tension in the string, therefore the equation of the forces becomes:

$T=m\frac{v^2}{R}$

And by substituting the numerical values, we find

$T=(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=3.9 N$

3) 2.3 N

When the ball is at the top of the circle, both the tension and the weight of the ball point downward, in the same direction of the centripetal force. Therefore, the equation of the force is

$T+mg=m\frac{v^2}{R}$

And substituting the numerical values and re-arranging it, we find

$T=m\frac{v^2}{R}-mg=(0.158 kg)\frac{5.22 m/s)^2}{1.1 m}-(0.158 kg)(9.8 m/s^2)=2.3 N$

4) 3.3 m/s

The minimum velocity for the ball to keep the circular motion occurs when the centripetal force is equal to the weight of the ball, and the tension in the string is zero; therefore:

$T=0\\mg = m\frac{v^2}{R}$

and re-arranging the equation, we find

$v=\sqrt{gR}=\sqrt{(9.8 m/s^2)(1.1 m)}=3.3 m/s$

7 0

3 years ago