Answer:
the final energy of the system is 35.5 kJ.
Explanation:
Given;
initial energy of the system, E₁ = 10 kJ
heat transferred to the system, q₁ 30 kJ
Heat lost to the surrounding, q₂ = 5kJ
heat gained by the system, Q = q₁ - q₂ = 30 kJ - 5kJ = 25 kJ
work done on the system, W = 500 J = 0.5 kJ
Apply first law of thermodynamic,
ΔU = Q + W
where;
ΔU is change in internal energy
Q is the heat gained by the system
W is work done on the system
ΔU = 25kJ + 0.5 kJ
ΔU = 25.5 kJ
The final energy of the system is calculated as;
E₂ = E₁ + ΔU
E₂ = 10 kJ + 25.5 kJ
E₂ = 35.5 kJ.
Therefore, the final energy of the system is 35.5 kJ.
Incomplete question.The Complete question is here
A flat uniform circular disk (radius = 2.00 m, mass = 1.00 ✕ 102 kg) is initially stationary. The disk is free to rotate in the horizontal plane about a friction less axis perpendicular to the center of the disk. A 40.0-kg person, standing 1.25 m from the axis, begins to run on the disk in a circular path and has a tangential speed of 2.00 m/s relative to the ground.
a.) Find the resulting angular speed of the disk (in rad/s) and describe the direction of the rotation.
b.) Determine the time it takes for a spot marking the starting point to pass again beneath the runner's feet.
Answer:
(a)ω = 1 rad/s
(b)t = 2.41 s
Explanation:
(a) initial angular momentum = final angular momentum
0 = L for disk + L............... for runner
0 = Iω² - mv²r ...................they're opposite in direction
0 = (MR²/2)(ω²) - mv²r
................where is ω is angular speed which is required in part (a) of question
0 = [(1.00×10²kg)(2.00 m)² / 2](ω²) - (40.0 kg)(2.00 m/s)²(1.25 m)
0=200ω²-200
200=200ω²
ω = 1 rad/s
b.)
lets assume the "starting point" is a point marked on the disk.
The person's angular speed is
v/r = (2.00 m/s) / (1.25 m) = 1.6 rad/s
As the person and the disk are moving in opposite directions, the person will run part of a revolution and the turning disk would complete the whole revolution.
(angle) + (angle disk turns) = 2π
(1.6 rad/s)(t) + ωt = 2π
t[1.6 rad/s + 1 rad/s] = 2π
t = 2.41 s
Answer:
The magnitude of the magnetic force acting on the conductor is 0.75 Newton
Explanation:
The parameters given in the question are;
The length of the straight segment of wire, L = 25 cm = 0.25 m
The current carried in the wire, I = 5 A
The orientation of the wire with the magnetic field = Perpendicular
The strength of the magnetic field in which the wire is located, B = 0.60 T
The magnetic force, 'F', is given by the following formula;
F = 
·L×
= I·L·B·sin(θ)
Where;
= The current flowing, I
L = The length of the wire
= The magnetic field strength, B
θ = The angle of inclination of the conductor to the magnetic field
Where I = 5 A, L = 0.25 m, B = 0.60 T, and θ = 90°, we get;
F = 5 A × 0.25 m × 0.60 T × sin(90°) = 0.75 N
Therefore
The magnitude of the magnetic force, F = 0.75 N.
The distance between Jupiter and the sun is 5.2 AU.
According to Kepler's third law, the square of the period of revolution of planets is proportional to the cube of their mean distances from the sun. From this; T^2 = r^3.
Now, we are told that the orbital period (T) is 11. 9 Earth years. We have to make the distance the subject of the formula.
r =T^2/3
r = (11.9)^2/3
r = 5.2 AU
Learn more: brainly.com/question/15207516
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