All categories

2 years ago

A circuit supplied with 110V carries 5 Amps current. Calculate the Resistance value of the circuit?

Physics

1 answer:

IgorC [24]2 years ago

6 0

Answer:

The value is $R = 22 \ \Omega$

Explanation:

From the question we are told that

The voltage is $V = 110 \ V$

The current is $I = 5 \ A$

Generally the resistance value is mathematically represented as

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

=> $R = \frac{110}{5}$

=> $R = 22 \ \Omega$

You might be interested in

*please refer to photo* An electric field of magnitude 5.25 ✕ 10^5N/C points due south at a certain location. Find the magnitude

kvv77 [185]

Answer:

Approximately $3.86\; {\rm N}$ (given that the magnitude of this charge is $-7.35\; {\rm \mu C}$ .)

Explanation:

If a charge of magnitude $q$ is placed in an electric field of magnitude $E$ , the magnitude of the electrostatic force on that charge would be $F = E\, q$ .

The magnitude of this charge is $q = 7.35\; {\rm \mu C}$ . Apply the unit conversion $1\; {\rm \mu C} = 10^{-6}\; {\rm C}$ :

$\begin{aligned} q &= 7.35\; {\mu C} \times \frac{10^{-6}\; {\rm C}}{1\; {\mu C}} = 7.35\times 10^{-6}\; {\rm C}\end{aligned}$ .

An electric field of magnitude $E = 5.25\times 10^{5}\; {\rm N \cdot C^{-1}}$ would exert on this charge a force with a magnitude of:

$\begin{aligned}F &= E\, q \\ &= 5.25 \times 10^{5}\; {\rm N \cdot C^{-1}} \times (-7.35\times 10^{-6}\; {\rm C}) \\ &\approx 3.86\; {\rm N}\end{aligned}$ .

Note that the electric charge in this question is negative. Hence, electrostatic force on this charge would be opposite in direction to the the electric field. Since the electric field points due south, the electrostatic force on this charge would point due north.

4 0

1 year ago

Which is the best example of potential energy?

svetoff [14.1K]

First one, holding a basketball in the air. Potential energy is the energy it has mostly from gravity. The further you go from the center of mass, the more energy.

8 0

2 years ago

Read 2 more answers

A professor drives off with his car (mass 870 kg), but forgot to take his coffee mug (mass 0.47 kg) off the roof. The coefficien

SSSSS [86.1K]

Answer: $6.86 m/s^2$

Explanation:

Given

mas of car=870 kg

coffee mug mass=0.47 kg

coefficient of static friction between mug and roof $\mu _s= 0.7$

Coefficient of kinetic Friction $\mu _k=0.5$

maximum car acceleration is $\mu \times g$

here coefficient of static friction comes in to action because mug is placed over car . If mug is moving relative to car then \mu _k is come into effect

$a_{max}=0.7\times 9.8=6.86 m/s^2$

8 0

2 years ago

A cheetah can accelerate from rest at the rate of 4m /s

lesya [120]

Acceleration of cheetah (a) = 4m/s²
time = 10s
initial velocity(u) = 0
final velocity = v
distance travelled = s

v = u +at = 0 + 10×4 = 40m/s
s = (v²-u²)/2a = 40²/(2×4) = 1600/8 = 200m

6 0

2 years ago

The value of g is greater at the poles than at the equator why

babunello [35]

Answer:

because the gravitational pull is maximum at the poles and decreases as it comes down toward the equator.

6 0

2 years ago