Work is obtained by multiplying the force and the object's displacement. The force and displacement and force should be in the same direction in order to have work.
W = F x d
d = W / F
Substituting the known values,
d = 352 J / 45 N = 7.82 m
Thus, the displacement of the student is 7.82 m.
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:
L = 5,955 m
Explanation:
For this exercise we must use the relation
R = ρ L / A
where R is the resistance that indicates that it is 1 Ω, the resistivity is taken from the tables ρ = 2.82 10⁻⁸ Ω m, L is the length of the wire and A is the cross section.
As it indicates to us in volume of aluminum to use we divide the two terms by the length
R / L = ρ L / (A L)
the volume of a body is its area times its length, therefore
R / L = ρ L / V
R = ρ L² / V
we clear the length of the wire
L = √ R V /ρ
we reduce the volume to SI units
v = 1 cm³ (1m / 10² cm)³ = 1 10⁻⁶ m
let's calculate
L = √ (1 1 10⁻⁶ / 2.82 10⁻⁸)
L = √ (0.3546 10²)
L = 5,955 m
The answer is b because it’s something being stretched and it
