How far should flammable material be kept from electrical panels?

Acid rain is formed when chemicals in the air get into rain and up the acidity levels.

tester [92]

False! Because the acid rain it becomes a chemicals it does not air get into the rain and up the acidity levels.
Hope it helped you.

-Charlie

:)

5 0

3 years ago

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If you wish to take a picture of a bullet traveling at 500 m/s, then a very brief flash of light produced by an RC discharge thr

Agata [3.3K]

Answer:

Resistance in the flash tube, $R=3.97\times 10^{-3}\ \Omega$

Explanation:

It is given that,

Speed of the bullet, v = 500 m/s

Distance between one RC constant, d = 1 mm = 0.001 m

Capacitance, $C=503\ \mu F=503\times 10^{-6}\ F$

The time constant of RC circuit is given by :

$\tau=RC$

R is the resistance in the flash tube

$R=\dfrac{\tau}{C}$ ..........(1)

Speed of the bullet is given by total distance divided by total time taken as :

$v=\dfrac{d}{\tau}$

$\tau=\dfrac{d}{v}$

$\tau=\dfrac{0.001}{500}$

$\tau=0.000002\ s$

Equation (1) becomes :

$R=\dfrac{0.000002}{503\times 10^{-6}}$

$R=3.97\times 10^{-3}\ \Omega$

So, the resistance in the flash tube is $3.97\times 10^{-3}\ \Omega$ . Hence, this is the required solution.

8 0

2 years ago

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A refrigerator having a power rating of 350W operates for 12 hours a day. calculate the cost of electrical energy to operate it

rodikova [14]

Answer:

See the answers below.

Explanation:

The cost of energy can be calculated by multiplying each given value, a dimensional analysis must be taken into account in order to calculate the total value of the cost in Rs.

$Cost=0.350[kW]*12[\frac{hr}{1day}]*30[days]*4.5[\frac{Rs}{kW*hr} ]=567[Rs]$

The fuse can be calculated by knowing the amperage.

$P=V*I$

where:

P = power = 350 [W]

V = voltage = 240 [V]

I = amperage [amp]

Now clearing I from the equation above:

$I=P/V\\I=350/240\\I=1.458[amp]$

The fuse should be larger than the current of the circuit, i.e. about 2 [amp]

3 0

2 years ago

Chapter 36, Problem 007 Light of wavelength 586 nm is incident on a narrow slit. The angle between the first diffraction minimum

svp [43]

Answer:

$d =4.77\times 10^{-5}\ m$

Explanation:

given

wavelength of light λ = 479 nm

= 479 x 10⁻⁹ m

the angle

θ = 1.15 / 2 = 0.575°

using

condition for diffraction minimum ,

d sinθ = m λ

for first minimum m = 1

d sinθ = λ

therefore ,

slit width

$d =\dfrac{\lambda}{sin\theta}$

$d =\dfrac{479\times 10^{-9}}{sin 0.575^0}$

$d =\dfrac{479\times 10^{-9}}{0.01}$

$d =4.77\times 10^{-5}\ m$

hence, the width of the slit is equal to $d =4.77\times 10^{-5}\ m$

3 0

2 years ago

You are asked to design a cylindrical steel rod 50.0 cm long, with a circular cross section, that will conduct 170.0 J/s from a

zepelin [54]

Answer:

You are asked to design a cylindrical steel rod 50.0 cm long, with a circular cross section, that will conduct 170.0 J/s from a furnace at 350.0 ∘C to a container of boiling water under 1 atmosphere.

Explanation:

Given Values:

L = 50 cm = 0.5 m

H = 170 j/s

To find the diameter of the rod, we have to find the area of the rod using the following formula.

Here Tc = 100.0° C

k = 50.2

H = k × A × $\frac{[T_{H -}T_{C} ] }{L}$