Richardson pulls a toy 3.0 m across the floor by a string, applying a force of0.50 N. During the first meter, the string is para
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The question is missing some parts. Here is the complete question.
An ideal gas is contained in a vessel at 300K. The temperature of the gas is then increased to 900K.
(i) By what factor does the average kinetic energy of the molecules change, (a) a factor of 9, (b) a factor of 3, (c) a factor of , (d) a factor of 1, or (e) a factor of
?
Using the same choices in part (i), by what factor does each of the following change: (ii) the rms molecular speed of the molecules, (iii) the average momentum change that one molecule undergoes in a colision with one particular wall, (iv) the rate of collisions of molecules with walls, and (v) the pressure of the gas.
Answer: (i) (b) a factor of 3;
(ii) (c) a factor of ;
(iii) (c) a factor of ;
(iv) (c) a factor of ;
(v) (e) a factor of 3;
Explanation: (i) Kinetic energy for ideal gas is calculated as:
where
n is mols
R is constant of gas
T is temperature in Kelvin
As you can see, kinetic energy and temperature are directly proportional: when tem perature increases, so does energy.
So, as temperature of an ideal gas increased 3 times, kinetic energy will increase 3 times.
For temperature and energy, the factor of change is 3.
(ii) Rms is root mean square velocity and is defined as
Calculating velocity for each temperature:
For 300K:
For 900K:
Comparing both veolcities:
For rms, factor of change is
(iii) Average momentum change of molecule depends upon velocity:
q = m.v
Since velocity has a factor of and velocity and momentum are proportional, average momentum change increase by a factor of
(iv) Collisions increase with increase in velocity, which increases with increase of temperature. So, rate of collisions also increase by a factor of .
(v) According to the Pressure-Temperature Law, also known as Gay-Lussac's Law, when the volume of an ideal gas is kept constant, pressure and temperature are directly proportional. So, when temperature increases by a factor of 3, Pressure also increases by a factor of 3.
Answer:
Option B, 93 cm
Explanation:
An diagram of the seed's motion is attached to this solution.
This is very close to a projectile motion question. And the quantity to be calculated, how far along the grant a seed released would travel is called the Range.
And this would be obtained from the equations of motion,
First of, the height of the plant is related to some quantities of the motion with this relation.
H = u(y) t + 0.5g(t^2)
U(y) = initial vertical component of velocity = 0 m/s, H = height at which motion began, = 20cm = 0.2 m
That means t = √(2H/g)
The horizontal distance covered, R,
R = u(x) t + 0.5g(t^2) = u(x) t (the second part of the equation goes to zero as the vertical component of the acceleration of this motion is 0)
(substituting the t = √(2H/g) derived from above
R = u(x) √(2H/g)
Where u(x) = the initial horizontal component of the bomb's velocity = maximum initial speed, that is, 4.6 m/s, H = vertical height at which the seed was released = 20 cm = 0.2 m, g = acceleration due to gravity = 9.8 m/s2
R = 4.6 √(2×0.2/9.8) = 0.929 m = 0.93 m = 93 cm. Option B.
QED!
gravitational potential energy is given by formula
here we need to compare the gravitational potential energy of stone 2 with respect to stone 1
so we will say
given that
now we have