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Juli2301 [7.4K]
3 years ago
7

How many calories are absorbed by a pot of water with a mass of 500g in order to raise the temperature from 20c to 30c?

Physics
1 answer:
LenaWriter [7]3 years ago
8 0
Calories would denote the amount of heat.

Givens are:
Mass = 500 g
t0 = 20C
tf = 30C
C = 1 cal/gC

Formula:
Q=MCt

500g (1cal/gC) 10C= 5000 cal

Thank you for your question. Please don't hesitate to ask in Brainly your queries. 
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Answer:

Explanation:

b) Gravity reduces the initial upward velocity to zero in a time of

t = v/g = 40/10 = 4 s

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or

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Let’s say I am in a bumper car and have a velocity of 14 m/s, driving in the positive x-direction. I and my bumped car have a ma
AlekseyPX

Answer:

160 kg

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Explanation:

m_1 = Mass of first car = 120 kg

m_2 = Mass of second car

u_1 = Initial Velocity of first car = 14 m/s

u_2 = Initial Velocity of second car = 0 m/s

v_1 = Final Velocity of first car = -2 m/s

v_2 = Final Velocity of second car

For perfectly elastic collision

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}\\\Rightarrow m_2v_2=m_{1}u_{1}+m_{2}u_{2}-m_{1}v_{1}\\\Rightarrow m_2v_2=120\times 14+m_2\times 0-(120\times -2)\\\Rightarrow m_2v_2=1920\\\Rightarrow m_2=\frac{1920}{v_2}

Applying in the next equation

v_2=\frac{2m_1}{m_1+m_2}u_{1}+\frac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\frac{2\times 120}{120+\frac{1920}{v_2}}\times 14+\frac{m_2-m_1}{m_1+m_2}\times 0\\\Rightarrow \left(120+\frac{1920}{v_2}\right)v_2=3360\\\Rightarrow 120v_2+1920=3360\\\Rightarrow v_2=\frac{3360-1920}{120}\\\Rightarrow v_2=12\ m/s

m_2=\frac{1920}{v_2}\\\Rightarrow m_2=\frac{1920}{12}\\\Rightarrow m_2=160\ kg

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Answer:

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Answer:

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= 3600joules

D).the system works under isothermal condition so no heat was transferred

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