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ira [324]
3 years ago
10

Which is the correct unit for electric power

Physics
1 answer:
Bumek [7]3 years ago
3 0
The unit for power is watts
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In practice, if a voltmeter was connected across any combination of the terminals, the potential difference would be less than w
VladimirAG [237]

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.

8 0
3 years ago
Why can gases be compressed
Keith_Richards [23]
Gases can be compressed, because they just take up the space surrounding them. The attractive forces between the particles in a gas are very weak, so the particles are free to move in random direction. They just move along until they collide, either with the walls of the container or with each other. Moreover, gases can be compressed because the particles are far apart and they have space to move into.

5 0
3 years ago
Which of the following statements is true?
likoan [24]

Answer:

d.Energy as heat transferred into an object is determined by the amount of work done on the object.

Explanation:

4 0
3 years ago
Read 2 more answers
Steam at 100°C is condensed into a 38.0 g aluminum calorimeter cup containing 280 g of water at 25.0°C. Determine the amount of
KonstantinChe [14]

Answer:

7.2g

Explanation:

From the expression of latent heat of steam, we have

Heat supplied by steam = Heat gain water + Heat gain by calorimeter

mathematically,

m_{s}c_{w} \alpha _{w} + m_{s}l=m_{w}c_{w} \alpha _{w} +m_{c}c_{c} \alpha c_{c}

L=specific latent heat of water(steam)=2268J/g

c_{w}=specific heat capacity=4.2J/gK

c_{c}=specific heat capacity of calorimeter =0.9J/gk

m_{w}=280g

m_{c}=38g

α=change in temperature

\alpha _{c}=(40-25)=15

\alpha _{w}=(40-25)=15

\alpha _{s}=(100-40)=60

Note: the temperature of the calorimeter is the temperature of it content.

From the equation, we can make m_{s} the subject of formula

m_{s}=\frac{m_{w}c_{w} \alpha +m_{c}c_{c}\alpha}{c_{w}\alpha +l}

Hence

m_{s}=\frac{(280*4.2*15) +(38*0.9*15)}{(4.2*60) +2268} \\m_{s}=\frac{18153}{2520}\\ m_{s}=7.2g

Hence the amount of steam needed is 7.2g

6 0
3 years ago
A halfback on an apparent breakaway for a touchdown is tackled from behind. If the halfback has a mass of 98 kg and was moving a
uranmaximum [27]

Answer:

The mutual speed immediately after the touchdown-saving tackle is 4.80 m/s

Explanation:

Given that,

Mass of halfback = 98 kg

Speed of halfback= 4.2 m/s

Mass of corner back = 85 kg

Speed of corner back = 5.5 m/s

We need to calculate their mutual speed immediately after the touchdown-saving tackle

Using conservation of momentum

m_{h}v_{h}+m_{c}v_{c}=m_{h+c}v_{h+c}

Where, m_{h}= mass of halfback

m_{c}=mass of corner back

v_{h}= velocity of halfback

v_{c}= velocity of corner back

Put the value into the formula

98\times4.2+85\times5.5=(98+85)\times v

v=\dfrac{98\times4.2+85\times5.5}{98+85}

v=4.80\ m/s

Hence, The mutual speed immediately after the touchdown-saving tackle is 4.80 m/s

3 0
3 years ago
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