What is the mean free path for the molecules in an ideal gas when the pressure is 100 kPa and the temperature is 300 K given tha
t the collision cross-section for the molecules of that gas is 2.0 � 10-20 m2? The Boltzmann constant is 1.38 � 10-23J/K, Avogadro's number is 6.02 � 1023 molecules/mole, and the ideal gas constant is R = 8.314 J/mol � K = 0.0821 L � atm/mol � K.
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Answer:
F=
Explanation:
The magnitude of force required to pull the lid off the box by air pressure.
We know that Pressure, P= Force(F)/Area(A)
Force, F= P×A
Given: A=
P=.
Therefore, F=.
F=
To solve this problem we will use the definition of the kinematic equations of centrifugal motion, using the constants of the gravitational acceleration of the moon and the radius of this star.
Centrifugal acceleration is determined by
Where,
v = Velocity
r = Radius
From the given data of the moon we know that gravity there is equivalent to
While the radius of the moon is given by
If we rearrange the function to find the speed we will have to
The speed for this to happen is 1.7km/s
Answer:
Part a)
a = -9.81 m/s/s
Part b)
v = 0
Part c)
v = 9.81 m/s
Part d)
Explanation:
Part a)
During the motion of ball it will have only gravitational force on the ball
so here the acceleration of the ball is only due to gravity
so it is given as
Part b)
As we know that ball is moving against the gravity
so here the velocity of ball will keep on decreasing as the ball moves upwards
so at the highest point of the motion of the ball the speed of ball reduce to zero
Part c)
We know that the total time taken by the ball to come back to the initial position is T = 2 s
so in this time displacement of the ball will be zero
Part d)
at the maximum height position we know that the final speed will be zero
so we will have
here we have