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2 years ago

Input Voltage = 220V

Physics

1 answer:

muminat2 years ago

4 0

Answer:

Correct option is

450 V and 15 A

⇒T

200N

3000W

=15A

Efficciencyofthetransformer

η=

out

⇒

100

3000

100×6

90×5000

=450V

Hence,

option (A) is correct answer.

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Rahim is watching his favorite football team on television. In order to work, the television must be plugged into an electrical

Rasek [7]

I think the answer is B

8 0

3 years ago

Read 2 more answers

. If a pendulum-driven clock gains 5.00 s/day, what fractional change in pendulum length must be made for it to keep perfect tim

inn [45]

Answer:

The appropriate response will be "Length must be increased by 0.012%".

Explanation:

The given values is:

ΔT = 5 s/day

Now,

⇒ $\frac{\Delta T}{T} =\frac{5}{24\times 60\times 60}$

On multiplying both sides by "100", we get

⇒ $\frac{\Delta T}{T}\times 100 =\frac{500}{24\times 60\times 60}$

⇒ $\frac{\Delta T}{T}\times 100=0.005787$ (%)

∵ $T=2\pi\sqrt{\frac{l}{g} }$

On substituting the values, we get

⇒ $\frac{\Delta T}{T}$ % = $\frac{1}{2}\times \frac{\Delta l}{l}$ %

On applying cross multiplication, we get

⇒ $\frac{\Delta l}{l}$ % = $2\times \frac{\Delta T}{T}$ %

⇒ = $2\times 0.05787$

⇒ = $0.011574$

⇒ = $0.012$ %

6 0

3 years ago

During an adiabatic process an object does 100 J of work and its temperature decreases by 5 K. During another process it does 25

Liono4ka [1.6K]

Answer:

The heat capacity for the second process is 15 J/K.

Explanation:

Given that,

Work = 100 J

Change temperature = 5 k

For adiabatic process,

The heat energy always same.

$dQ=0$

$dU=-dW$

We need to calculate the number of moles and specific heat

Using formula of heat

$dU=nC_{v}dT$

$nC_{v}=\dfrac{dU}{dT}$

Put the value into the formula

$nC_{v}=\dfrac{-100}{5}$

$nC_{v}=-20\ J/K$

We need to calculate the heat

Using formula of heat

$dQ=nC_{v}(dT_{1})+dW_{1}$

Put the value into the formula

$dQ=-20\times5+25$

$dQ=-75\ J$

We need to calculate the heat capacity for the second process

Using formula of heat

$dQ=nC_{v}(dT_{1})$

Put the value into the formula

$-75=nC_{v}\times(-5)$

$nC_{v}=\dfrac{-75}{-5}$

$nC_{v}=15\ J/K$

Hence, The heat capacity for the second process is 15 J/K.

5 0

3 years ago

car was moving in a straight road of length 320 km it covered 240 km with an average velocity 75 km/hr then it ran out of fuel a

Stella [2.4K]

The average velocity of the car for the whole journey is 69.57 km/h.

The given parameters:

Length of the road, L = 320 km
Distance covered = 240 km at 75 km/h
time spent refueling, t₂ = 0.6 hr
Final velocity, = 100 km/hr

The time spent by the before refueling is calculated as follows;

$t = \frac{d}{v} \\\\t_1 = \frac{240}{75} \\\\t_1 = 3.2 \ hours$

The time spent by the car for the remaining journey;

$t_3 = \frac{320 - 240}{100} \\\\t_3 = 0.8 \ hr$

The total time of the journey is calculated as follows;

$t = t_1 + t_2 + t_3\\\\t = 3.2 \ hr \ + \ 0.6 \ hr \ + \ 0.8 \ hr\\\\t = 4.6 \ hours$

The average velocity of the car for the whole journey is calculated as follows;

$v = \frac{total \ distance }{total \ time} \\\\v = \frac{320}{4.6} \\\\v = 69.57 \ km/h$

Learn more about average velocity here: brainly.com/question/6504879

6 0

3 years ago

A 0.111 kg hockey puck moving at 55 m/s is caught by a 80. kg goalie at rest. with what speed does the goalie slide on the frict

madreJ [45]

Answer:

0.076 m/s

Explanation:

Momentum is conserved:

m v = (m + M) V

(0.111 kg) (55 m/s) = (0.111 kg + 80. kg) V

V = 0.076 m/s

After catching the puck, the goalie slides at 0.076 m/s.

8 0

3 years ago