The pressure needed : 800 kPa
<h3>Further explanation</h3>
Given
V₁ = 0.4 m³
P₁ = 100 kPa
T₁ = 20 + 273 = 293 K
V₂ = 0.05 m³
T₂ = T₁ = 293 K
Required
The final pressure(P₂)
Solution
Boyle's Law
At a fixed temperature, the gas volume is inversely proportional to the pressure applied
Input the value :
P₂=P₁V₁/V₂
P₂=100 x 0.4 / 0.05
P₂=800 kPa
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The total energy TE = mgh + 1/2 mU^2; where h = 20 m, g = 9.81 m/sec^2, and U = 10 mps. When the ball reaches max height H, all that TE will be potential energy PE = mgH = TE.
So there you are. TE = mgh + 1/2 mU^2 = mgH = TE from the conservation of energy. Solve for H.
1) H = (gh + 1/2 U^2)/g = h + U^2/2g = ? meters where everything on the RHS is given. You can do the math.
2) As the ball drops from H to h, it picks up KE as the potential energy mgH is converted when the potential energy is diminished to mgh, where h < H. So PE - pe = ke = mg(H - h) = 1/2 mv^2 so solve for v = sqrt(2g(H - h)) and, again, everything is given. You can do the math.
3) Same deal as 2) except now its V = sqrt(2gH) because all the PE = mgH = 1/2 mV^2 = KE when it is about to hit the ground. You can do the math.
Significant ear drainage or a scarred tympanic membrane can lead to inaccurate results on a tympanic temperature reading.
<h3>Why inaccurate tympanic temperature reading?</h3>
Significant ear drainage or a scarred tympanic membrane can lead to inaccurate results on a tympanic temperature reading.
Although an ear infection or the presence of an ear infection will not significantly affect a tympanic thermometer reading.
If the client has been sleeping on one side, take the temperature on the other side as heat may be increased on the side against the pillow.
Recent consumption of a cold beverage will not affect tympanic temperature.
To learn more about temperature readings, refer https://brainly.ph/question/20039492
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Answer:
a) ω=n
b) V = R*n
c) a = 0
d) 
Explanation:
Angular velocity is given by the derivative of θ:
ω = n
Linear Velocity will be:
V = ω*R = n*R
Tangential acceleration will be the derivative of the linear velocity. Since velovity is constant:
a = 0
Radial acceleration is given by:
- The equivalent capacitance between point a and b is 5.87μf.
- The charge at 20μf is 93.92 μC.
- The charge at 6μf is 67.8 μC.
- The charge at C and 3μf is 26.12 μC.
<h3>
Sum of capacitance of C and 3μF</h3>
The sum of the capacitance is calculated as follows;
1/Ct = 1/C + 1/3
1/Ct = 1/10μf + 1/3μf
1/Ct = (3μf + 10μf)/30μf²
1/Ct = 13μf/30μf²
Ct = 30μf²/13μf
Ct = 2.31μf
<h3>Total capacitance in parallel arrangement</h3>
The total capacitance in parallel arrangement is calculated as follows;
Ct = 2.31μf + 6μf = 8.31μf
<h3>Equivalent capacitance between point a and b</h3>
1/Ct = 1/8.31μf + 1/20μf
1/Ct = 0.1703
Ct = 5.87μf
<h3>Charge flowing in each capacitor</h3>
Maximum voltage is delivered in 20μf, q = CV
<u>charge for 20μf:</u>
q = (5.87 x 16)μC
q = 93.92 μC
<h3>Equivalent capacitance for C, 3μf and 6μf</h3>
Ct = 2.31μf + 6μf = 8.31uf
<u>charge for 6μf:</u>
q = (6/8.31) x 93.92μC
q = 67.8μC
<h3>Total charge for C and 3μf</h3>
q = 93.92μC - 67.8μC = 26.12 μC
charge for C = charge 3μf = 26.12 μC
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