Answer:
The frequency of the oscillations in terms of fo will be f2=fo/3
E xplanation:
T= 
=1:3
⇒f2=fo\3
Here frequency f is inversely poportional to square root of mass m.
so the value of remainder of frequency f2 and fo is equal to 1:3.
⇒
= 
⇒
= 1:3
⇒f2=
We know, weight = mass * gravity
10 = m * 9.8
m = 10/9.8 = 1.02 Kg
Now, Let, the gravity of that planet = g'
g' = m/r² [m,r = mass & radius of that planet ]
g' = M/10 / (1/2R)² [M, R = mass & radius of Earth ]
g' = 4M / 10R²
g' = 2/5 * M/R²
g' = 2/5 * g
g' = 2/5 * 9.8
g' = 3.92
Weight on that planet = planet's gravity * mass
W' = 3.92 * 1.02
W' = 4 N
In short, Your Answer would be 4 Newtons
Hope this helps!
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>
That can be considered alive.
