Inspections may be_____ or limited to a specific area such as electrical or plumbing

Pepsi [2]

With a phone charger.

7 0

3 years ago

A particular motor rotates at 3000 revolutions per minute. What is its speed in rad/sec, and how many seconds does it takes to m

Leno4ka [110]

Answer:

ω=314.15 rad/s.

0.02 s.

Explanation:

Given that

Motor speed ,N= 3000 revolutions per minute

N= 3000 RPM

The speed of the motor in rad/s given as

$\omega=\dfrac{2\pi N}{60}\ rad/s$

Now by putting the values in the above equation

$\omega=\dfrac{2\pi \times 3000}{60}\ rad/s$

ω=314.15 rad/s

Therefore the speed in rad/s will be 314.15 rad/s.

The speed in rev/sec given as

$\omega=\dfrac{ 3000}{60}\ rad/s$

ω= 50 rev/s

It take 1 sec to cover 50 revolutions

That is why to cover 1 revolution it take

$\dfrac{1}{50}=0.02\ s$

4 0

3 years ago

An asbestos pad is square in cross section, measuring 5 cm on a side at its small end increasing linearly to 10 cm on a side at

polet [3.4K]

Answer:

q = 1.73 W

Explanation:

given data

small end = 5 cm

large end = 10 cm

high = 15 cm

small end is held = 600 K

large end at = 300 K

thermal conductivity of asbestos = 0.173 W/mK

solution

first we will get here side of cross section that is express as

$S = S1 + \frac{S2-S1}{L} x$ ...............1

here x is distance from small end and S1 is side of square at small end

and S2 is side of square of large end and L is length

put here value and we get

S = 5 + $\frac{10-5}{15} x$

S = $\frac{0.15 + x}{3}$ m

and

now we get here Area of section at distance x is

area A = S² ...............2

area A = $(\frac{0.15 + x}{3})^2$ m²

and

now we take here small length dx and temperature difference is dt

so as per fourier law

heat conduction is express as

heat conduction q = $\frac{-k\times A\ dt}{dx}$ ...............3

put here value and we get

heat conduction q = $-k\times (\frac{0.15 + x}{3})^2 \ \frac{dt}{dx}$

it will be express as

$q \times \frac{dx}{(\frac{0.15 + x}{3})^2} = -k (dt)$

now we intergrate it with limit 0 to 0.15 and take temp 600 to 300 K

$q \int\limits^{0.15}_0 {\frac{dx}{(\frac{0.15 + x}{3})^2 } = -0.173 \int\limits^{300}_{600} {dt}$

solve it and we get

q (30) = (0.173) × (600 - 300)

q = 1.73 W

5 0

3 years ago

What is the voltage output (in V) of a transformer used for rechargeable flashlight batteries, if its primary has 515 turns, its

kow [346]

<h2>Answer:</h2>

7532V

<h2>Explanation:</h2>

For a given transformer, the ratio of the number of turns in its primary coil ( $N_{p}$ ) to the number of turns in its secondary coil ( $N_{s}$ ) is equal to the ratio of the input voltage ( $V_{p}$ ) to the output voltage ( $V_{s}$ ) of the transformer. i.e