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Keith_Richards [23]
2 years ago
6

In practice, if a voltmeter was connected across any combination of the terminals, the potential difference would be less than w

hat is calculated. State why this is so and explain how the difference is kept to a minimum in modern transformers.
Physics
1 answer:
VladimirAG [237]2 years ago
8 0

Answer:

This difference is kept to a minimum because the resistance in transformers is a few tens of ohms and the resistance of modern voltmeters is of the order of MΩ.

Explanation:

A voltmeter is built by a galvanometer and a resistance in series, this set is connected in parallel to the resistance where the voltage is to be measured, therefore the voltage is divided between the voltmeter and the element to be measured, consequently the measured voltage It is less than the calculated one, since for them the resistance of the voltmeter is assumed infinite.

This difference is kept to a minimum because the resistance in transformers is a few tens of ohms and the resistance of modern voltmeters is of the order of MΩ.

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Answer:

so his far point according to this pair of glass is 200 cm

Explanation:

power of old pair of corrective glasses is given as

P = -0.5 dioptre

now we have

f = \frac{1}{P}

f = -2 m

f = -200 cm

now we know that for normal vision the maximum distance of vision is for infinite distance

so by lens formula we have

\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}

\frac{1}{d_i} + 0 = \frac{1}{200}

d_i = 200 cm

so his far point according to this pair of glass is 200 cm

7 0
3 years ago
Does the earths magnetic field change with time?
ZanzabumX [31]

Answer:

As per the fossil fuel records, magnetic field reversal does not impact living beings. It will take almost a century for the poles to complete the shift. Meanwhile, the earth is left with almost zero magnetic field.

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3 years ago
Give 2 reasons for fitting heavy commercial vehicles with many tyres​
forsale [732]
Better weight distribution and more stability
4 0
2 years ago
You leave the doctor's office after your annual checkup and recall that you weighed 688 N in her office. You then get into an el
Lapatulllka [165]

Answer:

a=0.5418\ m.s^{-2} upwards

a=1.283\ m.s^{-2} downwards

Explanation:

Given:

weight of the person, w=688\ N

So, the mass of the person:

m=\frac{w}{g}

m=\frac{688}{9.81}

m=70.132\ kg

  • Now if the apparent weight in the elevator, w_a= 726\ N

<u>Then the difference between the two weights is :</u>

\Delta w=w_a-w

\Delta w=726-688

\Delta w=38\ N is the force that acts on the body which generates the acceleration.

Now the corresponding acceleration:

a=\frac{\Delta w}{m}

a=\frac{38}{70.132}

a=0.5418\ m.s^{-2} upwards, because the normal reaction that due to the weight of the body is increased here.

  • Now if the apparent weight in the elevator, w_a= 598\ N

<u>Then the difference between the two weights is :</u>

\Delta w=w-w_a

\Delta w=688-598

\Delta w=90\ N is the force that acts on the body which generates the acceleration.

Now the corresponding acceleration:

a=\frac{\Delta w}{m}

a=\frac{90}{70.132}

a=1.283\ m.s^{-2} downwards, because the normal reaction that due to the weight of the body is decreased here.

7 0
3 years ago
Read 2 more answers
A series circuit has a capacitor of 0.25 × 10⁻⁶ F, a resistor of 5 × 10³ Ω, and an inductor of 1H. The initial charge on the cap
viktelen [127]

Answer:

q = (3 + e^{-4000 t} - 4 e^{-1000 t})\times 10^{-6}

at t = 0.001 we have

q = 1.55 \times 10^{-6} C

at t = 0.01

q = 2.99 \times 10^{-6} C

at t = infinity

q = 3 \times 10^{-6} C

Explanation:

As we know that they are in series so the voltage across all three will be sum of all individual voltages

so it is given as

V_r + V_L + V_c = V_{net}

now we will have

iR + L\frac{di}{dt} + \frac{q}{C} = 12 V

now we have

1\frac{d^2q}{dt^2} + (5 \times 10^3) \frac{dq}{dt} + \frac{q}{0.25 \times 10^{-6}} = 12

So we will have

q = 3\times 10^{-6} + c_1 e^{-4000 t} + c_2 e^{-1000 t}

at t = 0 we have

q = 0

0 = 3\times 10^{-6} + c_1  + c_2

also we know that

at t = 0 i = 0

0 = -4000 c_1 - 1000c_2

c_2 = -4c_1

c_1 = 1 \times 10^{-6}

c_2 = -4 \times 10^{-6}

so we have

q = (3 + e^{-4000 t} - 4 e^{-1000 t})\times 10^{-6}

at t = 0.001 we have

q = 1.55 \times 10^{-6} C

at t = 0.01

q = 2.99 \times 10^{-6} C

at t = infinity

q = 3 \times 10^{-6} C

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3 years ago
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