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zvonat [6]
3 years ago
14

a television set having apower rating of 120w and electric lawnmower of power rating 1kw are both connected to a 250v supply if

3A 5A and 10A fuses are available state which is the most appropriate for each appliance​
Engineering
1 answer:
Soloha48 [4]3 years ago
8 0

Answer:

<u>Television </u>

Power = 120 w

p.d = 240 v

I = p / v

I = 120 / 240 = 0.5

Fuse can be used 3A

<u>lawnmower</u>

Power = 1000w

p.d = 240 v

I = 1000/ 240 = 4.16

Fuse can be used = 5A

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4 years ago
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Air modeled as an ideal gas enters a combustion chamber at 20 lbf/in.2
motikmotik

Answer:

The answer is "112.97 \ \frac{ft}{s}"

Explanation:

Air flowing into thep_1 = 20 \ \frac{lbf}{in^2}

Flow rate of the mass m  = 230.556 \frac{lbm}{s}

inlet temperature T_1 = 700^{\circ} F

PipelineA= 5 \times 4 \ ft

Its air is modelled as an ideal gas Apply the ideum gas rule to the air to calcule the basic volume v:

\to \bar{R} = 1545 \ ft \frac{lbf}{lbmol ^{\circ} R}\\\\ \to M= 28.97 \frac{lb}{\bmol}\\\\ \to pv=RT \\\\\to v= \frac{\frac{\bar{R}}{M}T}{p}

      = \frac{\frac{1545}{28.97}(70^{\circ}F+459.67)}{20} \times \frac{1}{144}\\\\=9.8 \frac{ft3}{lb}

V= \frac{mv}{A}

   = \frac{230.556 \frac{lbm}{s} \times 9.8 \frac{ft^3}{lb}}{5 \times 4 \ ft^2}\\\\= 112.97 \frac{ft}{s}

8 0
3 years ago
Overshoot for an underdamped system is defined as the maximum displacement of the system at the end of its first half cycle. Wha
Kisachek [45]

Answer:

\zeta =0.69026

Explanation:

We have given no more than 5% overshoot

So maximum overshoot = 5 % =0.05

Maximum overshoot is given by M_P=e^\frac{-\pi \zeta }{\sqrt{1-\zeta ^2}}

Here \zeta is the damping ratio

We know that cos\Theta =\zeta

So \sqrt{1-\zeta ^2}=sin\Theta

So 0.05=e^{-\pi cot\Theta }

ln0.05={-\pi cot\Theta }

0.954={ cot\Theta }

\Theta =46.3485

cos\Theta =0.69026

\zeta =0.69026

4 0
3 years ago
A plane, opaque, surface M has the following properties: gray, diffuse, absorptivity = 0.7, surface area = 0.5 m2 , temperature
BaLLatris [955]

Answer:

The rate of energy absorbed per unit time is 3500W.

Explanation:

From the question, we were given the following parameters;

Plane, opaque, gray, diffuse surface

â = 0.7

Surface area, A = 0.5m²

Incoming radiant energy, G = 10000w/m²

T = 500°C

Rate of energy absorbed is âAG;

âAG = 0.7 × 0.5 × 10000

âAG = 3500W.

The energy absorbed is measured in watts and denoted by the symbol W.

7 0
4 years ago
A step-down transformer (turns ratio = 1:7) is used with an electric train to reduce the voltage from the wall receptacle to a v
SOVA2 [1]

Answer:

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Explanation:

6 0
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