Here Change in Kinetic Energy
= Work Done by Friction
Therefore, substituting the
given values to the equation, we get
0.5 * m * (vFinal^2 -
vInitial^2) = µ m g * d
Therefore
0.5*( 5.90^2 - Vfinal^2 ) =
0.100*9.8*2.10
Therefore
vfinal = 5.54 m/sec
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Answer:
The final angular speed is 16.1 rad/s
Explanation:
Given;
initial moment of inertia, I₁ = 2.56 kg.m²
final moment of inertia, I₂ = 0.40 kg.m²
initial angular speed, ω₁ = 0.4 rev/s = 2.514 rad/s
Apply the principle of conservation of angular momentum;
I₁ω₁ = I₂ω₂
where;
ω₂ is the final angular speed
ω₂ = (I₁ω₁) / (I₂)
ω₂ = (2.56 x 2.514) / (0.4)
ω₂ = 16.1 rad/s
Therefore, the final angular speed is 16.1 rad/s
Explanation:
Below is an attachment containing the solution.
Answer:
45C in a minute is
45/60 C in a second
.75 C in a second is 3/4 of an ampere.
Explanation:
Answer:
At 3.86K
Explanation:
The following data are obtained from a straight line graph of C/T plotted against T2, where C is the measured heat capacity and T is the temperature:
gradient = 0.0469 mJ mol−1 K−4 vertical intercept = 0.7 mJ mol−1 K−2
Since the graph of C/T against T2 is a straight line, the are related by the straight line equation: C /T =γ+AT². Multiplying by T, we get C =γT +AT³ The electronic contribution is linear in T, so it would be given by the first term: Ce =γT. The lattice (phonon) contribution is proportional to T³, so it would be the second term: Cph =AT³. When they become equal, we can solve these 2 equations for T. This gives: T = √γ A .
We can find γ and A from the graph. Returning to the straight line equation C /T =γ+AT². we can see that γ would be the vertical intercept, and A would be the gradient. These 2 values are given. Substituting, we f ind: T =
√0.7/ 0.0469 = 3.86K.
