Ask question

Ask question

All categories

2 years ago

13

A 1-w, 350-ω resistor is connected to 24 v. Is this resistor operating within its power rating?

1 answer:

seraphim [82]2 years ago

8 0

Answer:

No.

Explanation:

$P_r$ = Power rating = 1 W

R = Resistance = $350\ \Omega$

V = Voltage = $24\ \text{V}$

Power is given by

$P=\dfrac{V^2}{R}\\\Rightarrow P=\dfrac{24^2}{350}\\\Rightarrow P=1.65\ \text{W}$

$1.65\ \text{W}>1\ \text{W}$

So

$P>P_r$

Hence, the resistor is not operating within its power rating.

You might be interested in

A timing light checks the ignition timing in relation to the ____ position.

kogti [31]

Answer:

The timing light is connected to the ignition circuit and used to illuminate the timing marks on the engine's crankshaft pulley or flywheel, with the engine running. The apparent position of the marks, frozen by the stroboscopic effect, indicates the current timing of the spark in relation to piston position.

Explanation:

:)

8 0

2 years ago

I'm supposed to make an accurate 1/4"=1' floor plan. How do you do that?

seropon [69]

Using the measuring tool

8 0

3 years ago

Read 2 more answers

The purpose of pasteurizing milk is to A. Kill pathogens B. Break down milk fat C. Add vitamins and minerals D. Prevent spoilage

motikmotik

Answer: A Kill pathogens

7 0

2 years ago

stimate the maximum efficiency of an automobile engine that has a compression ratio of 5:1.0. Assume the engine operates accordi

Fed [463]

Answer:

Efficiency based on Otto cycle.

Effotto = 47.47%

Explanation:

Efficiency based on Otto cycle.

effotto = 1 – (V2 / V1)^γ-1

effotto = 1 – (1 / 5)^1.4 - 1

effotto = 47.47%

5 0

3 years ago

A steel pipe of 400-mm outer diameter is fabricated from 10-mm-thick plate by welding along a helix that forms an angle of 20° w

Verdich [7]

Explanation:

Outer di ameter $d_{0}=400 \mathrm{mm}[tex] Thickness of the cylinder [tex]t=10 \mathrm{mm}$

$\therefore[tex] Inner diam eter [tex]d_{i}=d_{0}-2 t=400-2 \times 10$

$d_{1}=380 \mathrm{mm}$

Given loading on the cylinder $P=300 \mathrm{kN}$ Helix an gle of the weld form $\theta=20^{\circ}$

(i) Normal stress on the plane at angle $\theta=20^{\circ}$ is

$\sigma=\frac{P \cos ^{2} \theta}{A_{0}}$

$\text { Where } A_{0}=\frac{\pi}{4}\left(d_{0}^{2}-d_{1}^{2}\right)$

$\quad=\frac{\pi}{4}\left(400^{2}-380^{2}\right)$

$=12252.21 \mathrm{mm}^{2}$

$=12.25221 \times 10^{-9} \mathrm{m}^{2}$

$\sigma=\frac{-300 \times 10^{2} \times \cos ^{2} 20}{12.25221 \times 10^{-1}}$

$=-21.6 \mathrm{MPa}$

(ii) Shear stress along an angle of $\theta=20^{\circ}$ is $\tau=\frac{P}{A_{0}} \cos \theta \sin \theta$

$=\frac{-300 \times 10^{-1} \times \cos 20 \times \sin 20}{12.25221 \times 10^{-3}}$

$=-7.86 \mathrm{MPa}$

3 0

3 years ago