The cost of developing thermonuclear power with plasmabe defended because D. It can provide an inexpensive power source.
<h3>How did the
cost of developing t
hermonuclear power defended?</h3>
The cost of developing thermonuclear power defended becvause we can see in the paragraph how it was told that the generation of ths power can be donee through the understanding of the occurrence of plasmain nature,
It should be noted that this thermonuclear power with plasmabe posses the characteristics which make it to exist in the ionosphere, and it can be felt in the flames as well; as in the chemical and nuclearexplosions.
In conclusion the power can be seen as an inexpensive source power because the p[roduction of this power cn be found in most of the thing that can be found around us as discused above.
Therefore, option D is correct.
Read more about cost at:
brainly.com/question/25109150
#SPJ1
The mass of ice melted as a result of friction between the ice and the horizontal surface is 2.78g
<u>Explanation:</u>
Given,
Temperature, T = 0°C
Initial mass, Mi = 62kg
Speed, s = 5.48m/s
Distance, x = 26.8m
Friction is present.
Mass of ice melted = ?
We know,
The amount of energy required for the melting of ice is exactly equal to the initial kinetic energy of the block of ice
and
Therefore, 
KE = 930.94 Joules
Ice melting lateral heat is 334 kJ/kg = 334000 J/kg.
Therefore, the melted mass of the ice = 930.94 / 334000 = 0.00278 kg = 2.78 g.
Thus, The mass of ice melted as a result of friction between the ice and the horizontal surface is 2.78g
Picture #1:
GPE = (mass) x (gravity) x (height)
GPE = (2 kg) x (9.8 m/s²) x (40 m) = 784 joules
KE = (1/2) (mass) (speed²)
KE = (1/2) (2 kg) (5 m/s)²
KE = (1 kg) (25 m²/s²) = 25 joules
Picture #2:
KE = (1/2) (mass) (speed²)
KE = (1/2) (2 kg) (10 m/s)²
KE = (1 kg) (100 m²/s²) = 100 joules
Picture #3:
GPE = (mass) x (gravity) x (height)
GPE = (20 kg) x (9.8 m/s²) x (2 m) = 392 joules
KE = (1/2) (mass) (speed²)
KE = (1/2) (20 kg) (5 m/s)²
KE = (10 kg) (25 m²/s²) = 250 joules
Picture #4:
GPE = (mass) x (gravity) x (height)
98 joules = (1 kg) x (9.8 m/s²) x (height)
Height = (98 joules) / (1 kg x 9.8 m/s²)
Height = 10 meters
Picture #5:
GPE = (mass) x (gravity) x (height)
39,200 Joules = (mass) x (9.8 m/s²) x (20 m)
Mass = (39,200 joules) / (9.8 m/s² x 20 m)
Mass = 200 kg
Answer:
Explanation:
Given that,
Mass of counterweight m= 4kg
Radius of spool cylinder
R = 8cm = 0.08m
Mass of spool
M = 2kg
The system about the axle of the pulley is under the torque applied by the cord. At rest, the tension in the cord is balanced by the counterweight T = mg. If we choose the rotation axle towards a certain ~z, we should have:
Then we have,
τ(net) = R~ × T~
τ(net) = R~•i × mg•j
τ(net) = Rmg• k
τ(net) = 0.08 ×4 × 9.81
τ(net) = 3.139 Nm •k
The magnitude of the net torque is 3.139Nm
b. Taking into account rotation of the pulley and translation of the counterweight, the total angular momentum of the system is:
L~ = R~ × m~v + I~ω
L = mRv + MR v
L = (m + M)Rv
L = (4 + 2) × 0.08
L = 0.48 Kg.m
C. τ =dL/dt
mgR = (M + m)R dv/ dt
mgR = (M + m)R • a
a =mg/(m + M)
a =(4 × 9.81)/(4+2)
a = 6.54 m/s
Force of gravity =mass*graviational acceleration
gravitational acceleration=g=9.81
mass=Density*Volume=.08*7840
force of gravity= .08*7840*9.81
gg
