Required value of initial speed of the bullet be ( 4M/m)√(gL).
Given parameters:
Mass of the bullet =m.
Mass of the bob of the pendulum = M.
speed of the bullet before collision = v
Speed of the bullet after collision = v/2.
Length of the pendulum stiff rod = L.
Let speed transmitted to the pendulum be u.
Using principle of conservation of momentum:
mv = Mu + mv/2
⇒ Mu = mv/2
⇒ u = (m/M)v/2
We know that: to make the bob over the top of the trajectory without falling backward in its circular path, required speed be = √(4gL). [ where g = acceleration due to gravity]
To be minimum initial speed the bullet must have in order for the pendulum bob to just barely swing through a complete vertical circle:
u = √(4gL)
⇒ (m/M)v/2 = √(4gL)
⇒ v =( 4M/m)√(gL).
Hence, minimum required speed of the bullet be ( 4M/m)√(gL).
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Answer:
The velocity when the ball hits the ground is obtained using v2. 2 = v1. 2 + 2 g Dy with v1=0 and Dy=h. Thus solving for v2 yields 17.1 m/s v2 = 2 g h =.
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Given:
m = 0.240 kg = 240 g, the mass of O₂
V = 3.10 L = 3.10 x 10⁻³ m³, the volume
Because the molar mass of oxygen is 16, the number of moles of O₂ is
n = (240 g)/(2*16 g/mol) = 7.5 mol
As an ideal gas,
p*V = nRT
or
V = (nRT)/p
where R = 8.314 J/(mol-K)
When
p = 0.910 atm = (0.910 atm) * (101325Pa/atm) = 92205.75 Pa
T = 27 °C = (27 + 273) K = 300 K
then the volume is

V = (0.2029 m³)*(10³ L/m³) = 202.9 L
Answer: 203 liters
Answer:

Explanation:
For n-=1 state hydrogen energy level is split into three componets in the presence of external magnetic field. The energies are,
,
,

Here, E is the energy in the absence of electric field.
And
are the highest and the lowest energies.
The difference of these energies

is known as Bohr's magneton.
B=2.5 T,
Therefore,

Now,

Therefore, the energy difference between highest and lowest energy levels in presence of magnetic field is 