Answer:
2.857 gm
Explanation:
<u>Step 1: Energy from uranium fission
</u>
The ratio of energy reeled from uranium to coal= 2.5 million times=2.5*10⁶
<u>step2: mass of uranium 235 required
</u>
To get the same energy that of 1 ton coal, the mass reuied will be = (1/2.5) * 10⁻⁶ ton=0.4 * 10⁻³ Kg
where we have taken 1 ton = about 1000Kg
<u>step3: mass of uranium
</u>
mass of u-235 = 70% of natural Uranium=0.7 Mu
So Mu= (1/0.7 )* mass of U-235=(0.4 * 10⁻³ Kg)/0.7=0.571 gm
<u>Step 4: Mass of ore
</u>
mass of Mu = 20% of ore=0.2 M
So, mass of ore= (1/0.2 )* mass of MU-=(0.571)/0.2 gm=2.857 gm
Answer:
Work done is 12.3 J
Explanation:
We have,
Mass of puck, m = 0.35 kg
Force of friction acting on the puck when it slides is 0.15 N
Distance travelled by the puck is 82 m.
It is required to find the work done on the puck. Finally the puck comes to rest and the force of friction is acting on it. It means the applied force is 0.15 N. Work done is given by
The work done on the puck is 12.3 J.
Explanation:
a)
θ = 4.91 + 9.7t + 2.06t² when t = 0
θ = 4.91 rad
θ = 4.91 + 9.7t + 2.06t²
ω = dθ/dt = 9.7 + 2.06t, when t =0
ω = dθ/dt = 9.7 + 0
ω = 9.7 rad/s
α = d²θ/dt² = 2.06
α= 2.06 rad/s²
b) please use same method above for t = 2.94 s
