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baherus [9]
3 years ago
12

Give two ways in which the water vapour changes as it passes down the glass tube in the condenser

Physics
1 answer:
SCORPION-xisa [38]3 years ago
6 0
<span>It changes from a liquid into a solid which means it is changing states. It also is changing temperature due to the condenser increasing or decreasing the temperature in order for the vapor to be transformed in liquid.</span>
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____ is a type of circuit were there is a direct connection between two points in a circuit that aren't supposed to be directly
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Answer:

Short

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A proton moves through a magnetic field at 20.3 % of the speed of light. At a location where the field has a magnitude of 0.0062
galina1969 [7]

Answer:

Force on the proton will be .73\times 10^{-14}N

Explanation:

We have given speed of proton is 20.3% of speed of light

Speed of light c=3\times 10^8m/sec

So speed of proton v=\frac{3\times 10^8\times 23}{100}=6.9\times 10^7m/sec

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Angle between velocity and magnetic field \Theta =137^{\circ}

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4 0
3 years ago
When the amount of water in a river increases so much that the river overflows its channel, a flood occurs.True or false
suter [353]

That's true, and there are also other mechanisms that can also create floods.

4 0
3 years ago
I only need help on #15! Thanks!
Vanyuwa [196]

I would have to say that 'B' and 'D' are both correct.

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7 0
3 years ago
The addition of 9.0×105 J is required to convert a block of ice at -10 ∘C to water at 11 ∘C. i need help this is due in less tha
Law Incorporation [45]

The mass of the block of ice is 2.24 kg

Explanation:

The amount of heat needed for the whole process consists of three different amounts of heat:

Q_1: the amount of heat needed to raise the temperature of the block of ice from -10^{\circ}C to 0^{\circ}C

Q_2: the amount of heat needed to melt the block of ice at melting point

Q_3: the amount of heat needed to raise the temperature of the water from 0^{\circ}C to 11^{\circ}C

The total amount of heat needed can be written as

Q=Q_1+Q_2+Q_3=mC_i\Delta T_1 + m\lambda_f + mC_w\Delta T_3

where we have:

Q=9.0 \cdot 10^5 J (total amount of heat required)

m is the mass of the block of ice

C_i = 2108 J/kg^{\circ}C is the specific heat of ice

\lambda_f=3.34\cdot 10^5 J/kg is the latent heat of fusion of ice

C_w=4186 J/kg^{\circ}C is the specific heat capacity of water

\Delta T_1 = 0-(-10)=10^{\circ}C is the change in temperature in the 1st process

\Delta T_3 = 11-0=11^{\circ}C is the change in temperature in the 3rd process

Solving the equation for m, we find the mass of the block of ice:

m=\frac{Q}{C_i\Delta T_1 + \lambda_f+C_w\Delta T_3}=\frac{9.0\cdot 10^5}{(2108)(10)+3.34\cdot 10^5+(4186)(11)}=2.24 kg

Learn more about specific heat capacity:

brainly.com/question/3032746

brainly.com/question/4759369

#LearnwithBrainly

8 0
3 years ago
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