The dP/dt of the adiabatic expansion is -42/11 kPa/min
<h3>How to calculate dP/dt in an adiabatic expansion?</h3>
An adiabatic process is a process in which there is no exchange of heat from the system to its surrounding neither during expansion nor during compression
Given b=1.5, P=7 kPa, V=110 cm³, and dV/dt=40 cm³/min
PVᵇ = C
Taking logs of both sides gives:
ln P + b ln V = ln C
Taking partial derivatives gives:
Substitutituting the values b, P, V and dV/dt into the derivative above:
1/7 x dP/dt + 1.5/110 x 40 = 0
1/7 x dP/dt + 6/11 = 0
1/7 x dP/dt = - 6/11
dP/dt = - 6/11 x 7
dP/dt = -42/11 kPa/min
Therefore, the value of dP/dt is -42/11 kPa/min
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3/3...............................
Answer:
dimension are 350 m and 350 m and
area is 122500 m²
Step-by-step explanation:
Given data
perimeter is 700 m
to find out
the largest area and its dimensions
solution
we consider x and y are the sides
so area is xy
perimeter is 2x+ y = 700
so we can say y = 700 -2x
so area will be = x (700 - 2x)
area = 700x - 2x²
we take derivative here and equal to zero
dA/dt = 700x - 2x²
700 - 4x = 0
x = 700/4
x = 175
so will be
y = 700 -2x
y = 700 -2(175)
y = 350
so dimension are 350 m and 350m
area is 122500 m²
"T is a subset of P"
Not true since triangle has three sides but parallelogram has four sides.
"E is a subset of I"
True since equilateral triangles are isosceles triangles with all angles equal.
"S is a subset of T"
True since scalene triangles are still triangle.
"I ⊂ E"
False since there are isosceles triangles those are not equilateral triangles. Namely triangle with angles 20°, 20°, 140°
"T ⊂ E"
False since not all triangles are equilateral. Scalene triangle is one of counterexamples.
"R ⊂ P"
True since rectangles are parallelograms with right angles.
Final answer: <span>E is a subset of I, </span>S is a subset of T, and R ⊂ P.
Hope this helps.
The longer an object, the longer the shadow they cast at a given time of the day and vice versa. This is called direct proportionality. At the same time,
If 15 foot flagpole ---- casts 11 foot shandow
28 foot tree -------- casts ?? foot shadow
The proportion would therefore be:
Length of shadow cast by tree = (28/15)*11 = 20.533 foot.
