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Feliz [49]
3 years ago
10

A 0.5 kg ball rolls down a frictionless inclined plane. The height of the inclined plane is 0.5m. What is the kinetic energy of

the ball when it reaches the bottom of the inclined plane?
Chemistry
1 answer:
joja [24]3 years ago
8 0

Answer:

2.45Joules

Explanation:

since kinetic energy is K.E.=1/2mv^{2}

lost in potential energy is gain of kinetic energy

mgh=1/2mv^2

9.8*2*0.5=v^2

v^{2}=9.8

v=\sqrt{9.8} =3.13m/s

now k.e. = 0.5*m*v^{2} =0.5*0.5*9.8 =2.45J

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Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/m
koban [17]

Here is the full question:

Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/min, and the circulated air is then pumped out at the same rate. If there is an initial concentration of 0.2% carbon dioxide, determine the subsequent amount in the room at any time.

What is the concentration at 10 minutes? (Round your answer to three decimal places.

Answer:

0.046 %

Explanation:

The rate-in;

R_{in} = \frac{0.04}{100}*2000

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The rate-out

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We can say that:

\frac{dA}{dt}= 0.8-\frac{A}{3}

where;

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A(0)= 0.002 × 6000

A(0)= 12

\frac{dA}{dt} +\frac{A}{3} =0.8

Integration of the above linear equation =

e^{\int\limits \frac {1}{3}dt } = e^{\frac{1}{3}t

so we have:

e^{\frac{1}{3}t}\frac{dA}{dt}} +\frac{1}{3}e^{\frac{1}{3}t}A = 0.8e^{\frac{1}{3}t

\frac{d}{dt}[e^{\frac{1}{3}t}A] = 0.8e^{\frac{1}{3}t

Ae^{\frac{1}{3}t} =2.4e\frac{1}{3}t +C

∴ A(t) = 2.4 +Ce^{-\frac{1}{3}t

Since A(0) = 12

Then;

12 =2.4 + Ce^{-\frac{1}{3}}(0)

C= 12-2.4

C =9.6

Hence;

A(t) = 2.4 +9.6e^{-\frac{t}{3}}

A(0) = 2.4 +9.6e^{-\frac{10}{3}}

A(t) = 2.74

∴ the concentration at 10 minutes is ;

=  \frac{2.74}{6000}*100%

= 0.0456667 %

= 0.046% to three decimal places

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