All categories

3 years ago

A 1.5 m wire carries a 9A current when a potential difference of 84 v is applied. what is the resistance of the wire

2 answers:

8 0

Ohm's law states that I=V/R (Current=volts divided by resistance). Since we're looking for resistance, we'll rewrite it as R=V/I. Then just plug in the numbers; R=84/9, R= 9 1/3 or 28/3. The resistance of the wire is 9.33... or 9 1/3 ohm's, depending on how you wanna write it.

Ann [662]3 years ago

3 0

Answer : The resistance of the wire is, 9.33 ohms

Solution : Given,

Electric current = 9 A

Potential difference = 84 V

Using ohm's law,

$V=I\times R$

or,

$R=\frac{V}{I}$

where,

R = resistance of wire

V = potential difference

I = electric current

Now put all the given values in the above formula, we get the resistance of the wire.

$R=\frac{84V}{9A}=9.33\Omega$

Therefore, the resistance of the wire is, 9.33 ohms

You might be interested in

As you approach your vehicle and perform checks it is not necessary to always

Greeley [361]

1.) Have your keys in hand before approaching or entering your car.

2. Be alert to other pedestrians and drivers.

3.) Search for signs of movement between, beneath and around objects to both sides of your vehicle

4.) check the spare tire for proper inflation

4 0

2 years ago

A vector has an x component of 25.0 units and a y

sammy [17]

Here is the answer to your question

8 0

3 years ago

How can you justify that force is transferred to lift and throw soil using shovel?<br>

Natali [406]

Forces are needed to lift, turn, move, open, close, push, pull, and so on. When you throw a ball, you are using force to make the ball move through the air. More than one force can act on an object at the same time.

4 0

2 years ago

A baseball pitcher throws the ball towards the batter at 90 mph. His bat connects with the ball for a line drive, after which th

forsale [732]

Answer:

$F=-18412.9N$ , where the minus indicates the direction is opposite to that of the throw.

Explanation:

Since MKS stands for meter-kilogram-second and we know that:

$1\ hour = 3600\ seconds$

$1\ mile = 1600\ meters$

$1000g = 1kg$

We can write that:

$\frac{1\ hour}{3600\ seconds}=1$

$\frac{1600\ meters}{1\ mile}=1$

$\frac{1kg}{1000g}=1$

These are conversion factors, equal to 1, so multiplying our results by them won't change their value, only their units.

So we have that:

$90 mph=90 \frac{miles}{hour}(\frac{1\ hour}{3600\ seconds})(\frac{1600\ meters}{1\ mile})=40m/s$

$110 mph=110 \frac{miles}{hour}(\frac{1\ hour}{3600\ seconds})(\frac{1600\ meters}{1\ mile})=48.89m/s$

$145 g=145 g(\frac{1kg}{1000g})=0.145kg$

Newton's 2nd Law tells us that F=ma, and the definition of acceleration is $a=\frac{\Delta v}{\Delta t}$ , so we have:

$F=m\frac{\Delta v}{\Delta t}=m\frac{v_f-v_i}{t}$

Taking the throw direction as the positive one, for our values we have:

$F=m\frac{v_f-v_i}{t}=(0.145kg)\frac{(-48.89m/s)-(+40m/s)}{0.0007s}=-18412.9N$

4 0

3 years ago

You have a battery marked " 6.00 V 6.00 V ." When you draw a current of 0.383 A 0.383 A from it, the potential difference betwee

Archy [21]

Answer:

V = 4.81 V

Explanation:

As the potential difference between the battery terminals, is less than the rated value of the battery, this means that there is some loss in the internal resistance of the battery.
We can calculate this loss, applying Ohm's law to the internal resistance, as follows:

$V_{rint} = I* r_{int}$

The value of the potential difference between the terminals of the battery, is just the voltage of the battery, minus the loss in the internal resistance, as follows:

$V = V_{b} - V_{rint} = 5.03 V = 6.0 V - 0.383 A* r_{int}$

We can solve for rint, as follows:

$r_{int} = \frac{V_{b}-V}{I} =\frac{6.0V-5.03V}{0.383A} = 2.53 \Omega$

When the circuit draws from battery a current I of 0.469A, we can find the potential difference between the terminals of the battery, as follows:

$V = V_{b} - V_{rint} = 6.0 V - 0.469 A* 2.53 \Omega= 6.0 V - 1.19 V = 4.81 V$

As the current draw is larger, the loss in the internal resistance will be larger too, so the potential difference between the terminals of the battery will be lower.

5 0

2 years ago