Answer:
Magnitude the net torque about its axis of rotation is 2.41 Nm
Solution:
As per the question:
The radius of the wrapped rope around the drum, r = 1.33 m
Force applied to the right side of the drum, F = 4.35 N
The radius of the rope wrapped around the core, r' = 0.51 m
Force on the cylinder in the downward direction, F' = 6.62 N
Now, the magnitude of the net torque is given by:
where
= Torque due to Force, F
= Torque due to Force, F'
Now,
The net torque comes out to be negative, this shows that rotation of cylinder is in the clockwise direction from its stationary position.
Now, the magnitude of the net torque:
Answer:
v = 98.75 km/h
Explanation:
Given,
The distance driver travels towards the east, d₁ = 135 km
The time period of the travel, t₁ = 1.5 h
The halting time, tₓ = 46 minutes
The distance driver travels towards the east, d₂ = 215 km
The time period of the travel, t₁ = 2 h
The average speed of the vehicle before stopping
v₁ = d₁/t₁
= 135/1.5
= 90 km/h
The average speed of vehicle after stopping
v₂ = d₂/t₂
= 215/2
= 107.5 km/h
The total average velocity of the driver
v = (v₁ +v₂) /2
= (90 + 107.5)/2
= 98.75 km/h
Hence, the average velocity of the driver, v = 98.75 km/h
The main formula is given by Eb/nucleon = Eb/ mass of nucleid
as for <span>52He, the mass is 5
so by applying Einstein's formula Eb=DmC², Eb=</span><span>binding energy
</span><span>52He-----------> 2 x 11p + 3 x10n is the equation bilan
</span>so Dm=2 mp + (5-2)mn-mnucleus, mp=mass of proton=1.67 10^-27 kg
mn=mass of neutron=<span>1.67 10^-27 kg
</span><span>m nucleus= 5
Dm= 2x</span>1.67 10^-27 kg+ 3x<span>1.67 10^-27 kg-5= - 4.9 J
Eb= </span> - <span>4.9 J x c²= -4.9 x 9 .10^16= - 45 10^16 J
so the answer is Eb /nucleon = Eb/5= -9.10^16 J, but 1eV=1.6 . 10^-19 J
so </span><span>-9.10^16 J/ 1.6 10^-19= -5.625 10^35 eV
the final answer is </span><span>Eb /nucleon </span><span>= -5.625 x10^35 eV</span>
