To solve this problem, let us recall that the formula for
gases assuming ideal behaviour is given as:
rms = sqrt (3 R T / M)
where
R = gas constant = 8.314 Pa m^3 / mol K
T = temperature
M = molar mass
Now we get the ratios of rms of Argon (1) to hydrogen (2):
rms1 / rms2 = sqrt (3 R T1 / M1) / sqrt (3 R T2 / M2)
or
rms1 / rms2 = sqrt ((T1 / M1) / (T2 / M2))
rms1 / rms2 = sqrt (T1 M2 / T2 M1)
Since T1 = 4 T2
rms1 / rms2 = sqrt (4 T2 M2 / T2 M1)
rms1 / rms2 = sqrt (4 M2 / M1)
and M2 = 2 while M1 = 40
rms1 / rms2 = sqrt (4 * 2 / 40)
rms1 / rms2 = 0.447
Therefore the ratio of rms is:
<span>rms_Argon / rms_Hydrogen = 0.45</span>
As Potential energy =mgh
m= 0.95kg
h=3 meter
g = 9.8 m/sec^2. ( acceleration due to gravity)
So P.E =(0.95)(9.8)(3)kgm^2/s^2
P.E =27.93 joules
Answer:
This means that the kinetic energy of second object is 48times that of the first object
Explanation:
Kinetic energy is the energy possessed by a body by virtue of its motion e.g motion of an accelerating car. Mathematically,
Kinetic energy = 1/2mv² where;
m is the mass of the object
v is the velocity of the object
If Object 1 of mass m moves with speed v in the positive direction, its kinetic energy will be expressed as;
K1 = 1/2mv²
For Object 2 of mass 3m moving with speed 4v in the negative x-direction, its kinetic energy can be expressed as;
K2 = 1/2(3m)(4v)²
K2 = 1/2(3m)(16v²)
K2 = (3m)(8v²)
K2 = 24mv²
To compare the kinetic energy of both bodies, we will take the ratio of K2:K1 to have;
K2/K1 = 24mv²/(1/2)mv²
K2/K1 = 24/(1/2)
K2/K1 = 48
K2 = 48K1
This means that the kinetic energy of second object is 48times that of the first object and moving in the negative x direction since the body of mass 3m initially moves in the negative x direction.
Answer:
424.26 m/s
Explanation:
Given that Two air craft P and Q are flying at the same speed 300m/s. The direction along which P is flying is at right angles to the direction along which Q is flying. Find the magnitude of velocity of the air craft P relative to air craft Q
The relative speed will be calculated by using pythagorean theorem
Relative speed = sqrt(300^2 + 300^2)
Relative speed = sqrt( 180000 )
Relative speed = 424.26 m/s
Therefore, the magnitude of velocity of the air craft P relative to air craft Q is 424.26 m/s