Large bodies of water<span> such as oceans, seas, and large lakes </span>affect<span> the </span>climate<span> of an area. </span>Water<span> heats and cools more slowly than land. Thus, in the summer, the </span>coastal<span> regions </span>will<span> stay cooler and in winter warmer. A more moderate </span>climate<span> with a smaller temperature range </span>is<span> created.</span>
Answer:
T = 1.766(M-m) Nm where M and m are the 2 masses of the objects
Explanation:
Let m and M be the masses of the 2 objects and M > m so the system would produce torque and rotational motion on the pulley. Force of gravity that exert on each of the mass are mg and Mg. Since Mg > mg, the net force on the system is Mg - mg or g(M - m) toward the heavier mass.
Ignore friction and string mass, and let g = 9.81 m/s2, the net torque on the pulley is the product of net force and arm distance to the pivot point, which is pulley radius r = 0.18 m
T = Fr = g(M - m)0.18 = 0.18*9.81(M - m) = 1.766(M-m) Nm
Answer:
a) 0 J
b) W = nRTln(Vf/Vi)
c) ΔQ = nRTln(Vf/Vi)
d) ΔQ = W
Explanation:
a) To find the change in the internal energy you use the 1st law of thermodynamics:
Q: heat transfer
W: work done by the gas
The gas is compressed isothermally, then, there is no change in the internal energy and you have
ΔU = 0 J
b) The work is done by the gas, not over the gas.
The work is given by the following formula:
n: moles
R: ideal gas constant
T: constant temperature
Vf: final volume
Vi: initial volume
Vf < Vi, then W < 0 and the work is done on the gas
c) The gas has been compressed. Thus, its temperature increases and heat has been transferred to the gas.
The amount of heat is equal to the work done W
d)
Answer:
Systems always tend toward a state of decreasing order unless more energy is provided into the system to counteract this tendency.
Answer:
6 days.
Explanation:
From radioactivity, The expression for half life is given as,
R/R' = 2⁽ᵃ/ᵇ)................... Equation 1
Where R = original mass of the radioactive substance, R' = Remaining mass of the radioactive substance after decay, a = Total time taken to decay, b = half life.
Given: R = 80 g, R' = 10 g, b = 2 days.
Substitute into equation 1
80/10 = 2⁽ᵃ/²⁾
8 = 2⁽ᵃ/²⁾
2³ = 2⁽ᵃ/²)
Equating the base and solving for a
3 = a/2
a = 2×3
a = 6 days.
