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Zinaida [17]
3 years ago
6

A 1-kg iron frying pan is placed on a stove. The pan increases from 20°C to 250°C. If the same amount of heat is added to a pan

with a greater specific heat, what can you predict about the temperature of this second pan
Physics
1 answer:
Nesterboy [21]3 years ago
7 0

Answer;

The temperature change for the second pan will be lower compared to the temperature change of the first pan

Explanation;

-The quantity of heat is given by multiplying mass by specific heat and by temperature change.

That is; Q = mcΔT

This means; the quantity of heat depends on the mass, specific heat capacity of a substance and also the change in temperature.

-Maintaining the same quantity of heat, with another pan of the same mass and greater specific heat capacity would mean that the change in temperature would be much less lower.

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A toroidal solenoid with mean radius r and cross-sectional area A is wound uniformly with N1 turns. A second toroidal solenoid w
lilavasa [31]

Answer:

Mutual inductance, M=2.28\times 10^{-5}\ H

Explanation:

(a) A toroidal solenoid with mean radius r and cross-sectional area A is wound uniformly with N₁ turns. A second thyroidal solenoid with N₂ turns is wound uniformly on top of the first, so that the two solenoids have the same cross-sectional area and mean radius.

Mutual inductance is given by :

M=\dfrac{\mu_oN_1N_2A}{2\pi r}

(b) It is given that,

N_1=550

N_2=290

Radius, r = 10.6 cm = 0.106 m

Area of toroid, A=0.76\ cm^2=7.6\times 10^{-5}\ m^2

Mutual inductance, M=\dfrac{4\pi \times 10^{-7}\times 550\times 290\times 7.6\times 10^{-5}}{2\pi \times 0.106}

M=0.0000228\ H

or

M=2.28\times 10^{-5}\ H

So, the value of mutual inductance of the toroidal solenoid is 2.28\times 10^{-5}\ H. Hence, this is the required solution.

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3 years ago
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find the distance traveled in km when the average speed is 60 km per hour and the time taken is 3 and 1/2 hours​
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Answer:

210 km

Explanation:

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g A bowling ball with a mass of 3.86 kg and a radius of 0.161 m starts from rest at a height of 2.5 m and rolls down a 48.4 o sl
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Answer:

v=1.5m/s

Explanation:

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mgh=\frac{mv^2}{2}+\frac{2mr^2\omega^2}{2*5}=\frac{mv^2}{2}+\frac{mv^2}{5}=\frac{7mv^2}{10}

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v=\sqrt{\frac{10gh}{7}}=\sqrt{\frac{10(9.8m/s^2)(0.161m)}{7}}=1.5m/s

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