Newton's<span> first </span>law of motion<span> has been frequently stated throughout this lesson. An</span>object<span> at rest stays at rest and an </span>object<span> in </span>motion<span> stays in </span>motion<span> with the same speed and in the same direction unless </span>acted<span> upon by an </span>unbalanced force<span>.</span>
The work done in the isothermal process is 10 joule.
We need to know about the isotherm process to solve this problem. The isotherm process can be described as a process where the initial temperature system will be the same as the final temperature. Hence, the internal energy change will be zero.
ΔU = 0
Hence,
ΔU = Q - W
0 = Q - W
Q = W
It means that the heat transferred is the same as the work done.
From the question above, we know that the heat transferred is 10 joule. Thus, the work done in the isothermal process is 10 joule.
Find more on isothermal at: brainly.com/question/17097259
#SPJ4
Question:
The operations manager for a well-drilling company must recommend whether to build a new facility, expand his existing one, or do nothing. He estimates that long-run profits (in $000) will vary with the amount of precipitation (rainfall) as follows:
Alternative Precipitation
Low Normal High
Do nothing -100 100 300
Expand 350 500 200
Build new 750 300 0
If he feels the chances of low, normal, and high precipitation are 30 percent, 20 percent, and 50 percent respectively, What is EVPI (Expected value of Perfect Information)?
A. $140,000
B. $170,000
C. $285,000
D. $305,000
E. $475,000
Answer:
D. $170,000
Explanation:
The expected long run profits are for
Low Normal High
Do nothing -100*0.3 100*0.2 300*0.5 = 140
Expand 350*0.3 500*0.2 200*0.5 = 305
Build new 750*0.3 300*0.2 0*0.5 = 285
Therefore the expected long run profits are
$140,000
$305,000
$285,000
Based on his selected option being either to build new or to expand, the most profitable option is to expand
=$305,000
EVPI = EPPI-EMV =$170,000
Answer: 17.83 AU
Explanation:
According to Kepler’s Third Law of Planetary motion <em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”. </em>
(1)
Talking in general, this law states a relation between the <u>orbital period</u> 
of a body (moon, planet, satellite, comet) orbiting a greater body in space with the <u>size</u> 
of its orbit.
However, if is measured in <u>years</u>, and is measured in <u>astronomical units</u> (equivalent to the distance between the Sun and the Earth: 
), equation (1) becomes:
(2)
This means that now both sides of the equation are equal.
Knowing 
and isolating from (2):
![a=\sqrt[3]{T^{2}}=T^{2/3}](https://tex.z-dn.net/?f=a%3D%5Csqrt%5B3%5D%7BT%5E%7B2%7D%7D%3DT%5E%7B2%2F3%7D)
(3)
(4)
Finally:
(5)
Answer:
vB = 15.4 m/s
Explanation:
Principle of conservation of energy:
Because there is no friction the mechanical energy is conserve
ΔE = 0
ΔE : mechanical energy change (J)
K : Kinetic energy (J)
U: Potential energy (J)
K = (1/2)mv²
U = m*g*h
Where :
m: mass (kg)
v : speed (m/s)
h : hight (m)
Ef - Ei = 0
(K+U)final - (K+U)initial =0
(K+U)final = (K+U)initial
((1/2)mv²+m*g*h)final = ((1/2)mv²+m*g*h)initial , We divided by m both sides of the equation:
((1/2)vB² + g*hB = (1/2 )vA²+ g*hA
(1/2) (vB)² + (9.8)*(14.7) = 0 + (9.8)(26.8 )
(1/2) (vB)² = (9.8)(26.8 ) - (9.8)*(14.7)
(vB)² = (2)(9.8)(26.8 - 14.7)
(vB)² = 237.16
vB = 15.4 m/s : speed of the cart at B
