Problem 1. Cylinders and a pendulum A uniform solid cylinder of radius r and mass m can roll inside a hollow cylinder of radius
R > r without slipping. A pendulum of length l = R-r and mass M = m/2 is attached to the center of the smaller cylinder. Find the normal frequencies and normal modes of this Y.
1 answer:
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Answer:
1.98 atm
Explanation:
Given that:
Temperature = 28.0 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (28 + 273.15) K = 301.15 K
n = 1
V = 0.500 L
Using ideal gas equation as:
PV=nRT
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L atm/ K mol
Applying the equation as:
P × 0.500 L = 1 ×0.0821 L atm/ K mol × 301.15 K
⇒P (ideal) = 49.45 atm
Using Van der Waal's equation
R = 0.0821 L atm/ K mol
Where, a and b are constants.
For Ar, given that:
So, a = 1.345 atm L² / mol²
b = 0.03219 L / mol
So,
⇒P (real) = 47.47 atm
Difference in pressure = 49.45 atm - 47.47 atm = 1.98 atm
Answer:
On moon time period will become 2.45 times of the time period on earth
Explanation:
Time period of simple pendulum is equal to ....eqn 1 here l is length of the pendulum and g is acceleration due to gravity on earth
As when we go to moon, acceleration due to gravity on moon is times os acceleration due to gravity on earth
So time period of pendulum on moon is equal to
--------eqn 2
Dividing eqn 2 by eqn 1
So on moon time period will become 2.45 times of the time period on earth
Answer:
Explanation:
M = 1.989 x 10^30 kg
R = 6.96 x 10^8 m
G = 6.67 x 10^-11 Nm²/kg²
Let the velocity is v.
v = 6.17 x 10^5 m/s