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stepan [7]
3 years ago
11

A 150 g ceramic serving bowl is warmed 52.0°C when it absorbs 2.2 kcal of heat from a serving of hot food. What is the specific

heat of the china dish?
Physics
1 answer:
allochka39001 [22]3 years ago
8 0

Answer:

1.18010256J/Kg°C

Explanation:

The quantity of heat energy is expressed as:

Q = mct

c is the specific heat capacity

t is change in temperature

Q = 4.184×2.2

Q = 9.2048Joules

m = 150g = 0.15kg

t = 52°C

Substitute and get c

9.2048 = 0.15c(52)

9.2048 = 7.8c

c = 9.2048/7.8

c = 1.18010256

Hence the specific heat capacity of the dish is 1.18010256J/Kg°C

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A proton moves at constant velocity in the direction, through a region in which there is an electric field and a magnetic field.
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Answer:

F_{net}=0

Explanation:

It is given that, a proton moves at constant velocity, through a region in which there is an electric field and a magnetic field such that,

The electric field is, E = 800 V/m

Magnetic field, B = 0.25 T

We know that the net force in the region of magnetic and electric field is given by Lorentz forces. But here, the proton moves with constant velocity. So, the net force acting on it is 0.

i.e.

F_{net}=0

Hence, this is the required solution.

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An electron passes through a point 2.83 cm 2.83 cm from a long straight wire as it moves at 35.5 % 35.5% of the speed of light p
igor_vitrenko [27]

Answer:

The magnitude of electron acceleration is 2.34 \times 10^{15} \frac{m}{s^{2} }

Explanation:

Given:

Distance from the wire to the field point r = 2.83 \times 10^{-2} m

Speed of electron v = 35.5 \%c

Current I = 17.7 A

For finding the acceleration,

First find the magnetic field due to wire,

  B = \frac{\mu _{o}I }{2\pi r }

Where \mu_{o} = 4\pi   \times 10^{-7}

  B = \frac{4\pi \times 10^{-7}  \times 17.7 }{2\pi (2.83 \times 10^{-2} ) }

  B = 12.50 \times 10^{-5} T

The magnetic force exerted on the electron passing through straight wire,

  F = qvB  

  F = 1.6 \times 10^{-19} \times 0.355 \times 3 \times 10^{8} \times 12.50 \times 10^{-5}

  F = 21.3 \times 10^{-16} N

From the newton's second law

  F = ma

Where m = mass of electron = 9.1 \times 10^{-31} kg

So acceleration is given by,

   a = \frac{F}{m}

   a = \frac{21.3 \times 10^{-16} }{9.1 \times 10^{-31} }

   a = 2.34 \times 10^{15} \frac{m}{s^{2} }

Therefore, the magnitude of electron acceleration is 2.34 \times 10^{15} \frac{m}{s^{2} }

7 0
2 years ago
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