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.
Answer:
I think kilogram.
Explanation:
If you refer to science book you can find.
Answer:
I = 0.75 A
Explanation:
The question is not to calculate the resistance, but to calculate the current. (The resistance is already given by the value of 2 Ohm).
U = I * R
I = U / R
with U = 1.5 V and R = 2 Ohm
I = 1.5 / 2
I = 0.75 A
Answer:
Explanation:
Use the one-dimensional equation
which says that the final velocity of an object is equal to the object's initial velocity plus its acceleration times time. We are looking for time. That means the equation looks like this:
0 = 30 + (-4)t and
-30 = -4t so
t = 7.5 sec
Answer:
0.739
Explanation:
If we treat the four tire as single body then
W ( weight of the tyre ) = mass × acceleration due to gravity (g)
the body has a tangential acceleration = dv/dt = 5.22 m/s², also the body has centripetal acceleration to the center = v² / r
where v is speed 25.6 m/s and r is the radius of the circle
centripetal acceleration = (25.6 m/s)² / 130 = 5.041 m/s²
net acceleration of the body = √ (tangential acceleration² + centripetal acceleration²) = √ (5.22² + 5.041²) = 7.2567 m/s²
coefficient of static friction between the tires and the road = frictional force / force of normal
frictional force = m × net acceleration / m×g
where force of normal = weight of the body in opposite direction
coefficient of static friction = (7.2567 × m) / (9.81 × m)
coefficient of static friction = 0.739
