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diamong [38]
2 years ago
13

A. Compare solids, liquids, and gases in terms of the way their volumes change (or don't change) when placed into different cont

ainers.
b. A sample of gas in a rigid container that cannot change volume. Use the kinetic molecular theory to explain why the pressure in the container will decreases as the temperature rises.

c. Write one or two sentences to describe how heat will flow between two objects of different temperatures that are touching.

d. Write one or two sentences to compare the internal energy of a molecular gas with the internal energy of a monatomic gas.

e. Potential energy from intermolecular forces is included in the internal energies of which states of matter?

f. What is the average velocity of atoms in 1.00 mol of krypton (a monatomic gas) at 347 K? For m, use 0.0838

g. What is the internal energy of 4.00 mol of diatomic nitrogen gas N2, at 455 K?
Physics
1 answer:
svet-max [94.6K]2 years ago
7 0

a) Solids keep shape, liquids take shape of containers but don't spill, gases take container's shape and spill out

b) if you heat gas, speed of its molecules will increase and they'll push the container's walls stronger, so the pressure will increase when the container heated

c) Heat flows from warmer to colder bodies

d) For monatomic gases it's U=1.5nRT only, molecular gas has bonds between atoms so total internal energy increases

e) Of gases

f) v=\sqrt{\frac{3RT}{M}}=321 m/s

g) U=5/2*nRT=37830.85 J

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A toy rocket, launched from the ground, rises vertically with an acceleration of 28 m/s 2 for 9.7 s until its motor stops. Disre
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Answer:

5080.86m

Explanation:

We will divide the problem in parts 1 and 2, and write the equation of accelerated motion with those numbers, taking the upwards direction as positive. For the first part, we have:

y_1=y_{01}+v_{01}t+\frac{a_1t^2}{2}

v_1=v_{01}+a_1t

We must consider that it's launched from the ground (y_{01}=0m) and from rest (v_{01}=0m/s), with an upwards acceleration a_{1}=28m/s^2 that lasts a time t=9.7s.

We calculate then the height achieved in part 1:

y_1=(0m)+(0m/s)t+\frac{(28m/s^2)(9.7s)^2}{2}=1317.26m

And the velocity achieved in part 1:

v_1=(0m/s)+(28m/s^2)(9.7s)=271.6m/s

We do the same for part 2, but now we must consider that the initial height is the one achieved in part 1 (y_{02}=1317.26m) and its initial velocity is the one achieved in part 1 (v_{02}=271.6m/s), now in free fall, which means with a downwards acceleration a_{2}=-9,8m/s^2. For the data we have it's faster to use the formula v_f^2=v_0^2+2ad, where d will be the displacement, or difference between maximum height and starting height of part 2, and the final velocity at maximum height we know must be 0m/s, so we have:

v_{02}^2+2a_2(y_2-y_{02})=v_2^2=0m/s

Then, to get y_2, we do:

2a_2(y_2-y_{02})=-v_{02}^2

y_2-y_{02}=-\frac{v_{02}^2}{2a_2}

y_2=y_{02}-\frac{v_{02}^2}{2a_2}

And we substitute the values:

y_2=y_{02}-\frac{v_{02}^2}{2a_2}=(1317.26m)-\frac{(271.6m/s)^2}{2(-9.8m/s^2)}=5080.86m

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