Ask question

Ask question

All categories

3 years ago

10

A current of 9 A flows through an electric device with a resistance of 43 Ω. What must be the applied voltage in this particular

circuit?

1 answer:

lubasha [3.4K]3 years ago

8 0

Voltage, V = IR

Where I is current in Ampere, R is Resistance in Ohms.

V = 9A * 43 Ω

V = 387 V

You might be interested in

Is this object showing acceleration for the first 2 seconds? explain your answer.

iVinArrow [24]

Yes. The line is increasing. The flat line at the top of the graph is where there is not acceleration and the decreasing line is deceleration.

3 0

3 years ago

A mass resting on a horizontal, frictionless surface is attached to one end of a spring; the other end is fixed to a wall. It ta

My name is Ann [436]

Answer:

a). $K=528.92 \frac{N}{m}$

b). $m=4.84kg$

Explanation:

a).

The work of the spring is find by the formula:

$w_s =\frac{1}{2}*k*x^2$

So knowing the work can find the constant K'

$3.2J =\frac{1}{2}*k*(0.11m)^2$

Solve for K'

$K=\frac{2*W_s}{x^2}=\frac{2*3.2J}{0.11m^2}$

$K=528.92 \frac{N}{m}$

b).

The force of the spring realice a motion so using the force and knowing the accelerations can find the mass

$F=m*a$

$m=\frac{F}{a}=\frac{K*x}{a}$

$m=\frac{528.9*0.11m}{12m/s^2}$

$m=4.84kg$

8 0

3 years ago

Suppose the maximum power delivered by a car's engine results in a force of 16000 N on the car by the road. In the absence of an

joja [24]

Answer:

Approximately $9.7\; \rm m \cdot s^{-2}$ .

Explanation:

Assuming that there is no other force on this vehicle, the $16000\; \rm N$ force from the road would be the only force on this vehicle. The net force would then be equal to this $16000\; \rm N\!$ force. The size of the net force would be $16000\; \rm N\!\!$ .

Let $m$ denote the mass of this vehicle and let $\Sigma F$ denote the net force on this vehicle.

By Newton's Second Law of motion, the acceleration of this vehicle would be proportional to the net force on this vehicle. In other words, the acceleration of this vehicle, $a$ , would be:

$\begin{aligned}a &= \frac{\Sigma F}{m}\end{aligned}$ .

For this vehicle, $\Sigma F = 16000\; \rm N$ whereas $m = 1650\; \rm kg$ . The acceleration of this vehicle would be:

$\begin{aligned}a &= \frac{16000\; \rm N}{1650\; \rm kg} \\ &= \frac{16000\; \rm kg \cdot m\cdot s^{-2}}{1650\; \rm kg}\\ &\approx 9.7 \; \rm m \cdot s^{-2}\end{aligned}$ .

8 0

3 years ago

5. A sled with no initial velocity accelerates at a rate of 5.0 m/s2 down a hill. How

ad-work [718]

Answer:

60 m/s

Explanation:

$a = \frac{vf - vi}{t}$

$5.0 = \frac{vf - 0}{12}$

$5.0 \times 12 = vf$

$60 = vf$

6 0

3 years ago

How to round numbers and find significant digits in physics?? <br> Simple answer? Please???

Mrrafil [7]

Answer:

Heya how have you been doing

3 0

3 years ago