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andrezito [222]
3 years ago
8

The power in an electric circuit varies inversely with the resistance. If the power is 2,200 watts when the resistance is 25 ohm

s, find the resistance when the power is 1,100 watts.
Physics
1 answer:
valentinak56 [21]3 years ago
6 0

Answer:

The value of resistance when power is 1100 watts = R_{2} = 50 ohms

Explanation:

Power P_{1} = 2200 Watts

Resistance R_{1} = 25 ohms

Power P_{2} = 1100 Watts

Resistance R_{2} = we have to calculate

Given that the power in an electric circuit varies inversely with the resistance

⇒ P ∝ \frac{1}{R}

⇒ \frac{P_{2} }{P_{1} } = \frac{R_{1} }{R_{2} }

⇒ \frac{1100}{2200} = \frac{25}{R_{2} }

⇒ R_{2} = 50 ohms

This is the value of resistance when power is 1100 watts.

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an empty density bottle weighed 23.5 and 48.4g when filled with water .The empty bottle was partially filled with sand until and
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Density of Sand is 2.653g/cm^{3}.

Explanation:

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Weight of bottle when completely filled water=48.4gm

So amount of water required to fill the bottle=Weight of bottle filled with water-W

Amount of water required to fill the bottle(w_{max})=48.4gm-23.5gm

w_{max}=24.9g

Since we know density of water d_{w} =1g/cm^{3} and w_{max}

We can calculate volume of empty space in the bottle(V).

w_{max}=d_{w}\timesV

V=\frac{w_{max} }{d_{w} }

V=\frac{24.9}{1}

V=24.9 g/cm^{3}

Now bottle is partially filled with sand,and weight of bottle is (w_{s})36.5gm

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Amount of sand added (m_{s})=36.5-Weight of the bottle

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After filling the bottle with water again,the weight of the bottle becomes (W_{2}=56.5g)

Therefore,

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As the density of water = 1g/cm^{3}

Amount of water (in grams )=Volume of water occupied

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Therfore volume of water added to the sand filled bottle(V_{w})=20cm^{3}

As we know the total volume of the water bottle(V),

Volume of the sand occupied in the water bottle=V-V_{w}

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V_{s}=4.9g

We know,

Density=Mass/Volume

Therefore,

density of sand = \frac{m_{s} }{V_{s} }

density of sand =\frac{13}{4.9}

density of sand = 2.653g/cm^{3}

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