3. Assuming a 6V battery to power an LED. The current limiting resistor is 195 ohms. Using the standard maximum current of the c
1 answer:
You might be interested in
Answer:
V1=5<u>ft3</u>
<u>V2=2ft3</u>
n=1.377
Explanation:
PART A:
the volume of each state is obtained by multiplying the mass by the specific volume in each state
V=volume
v=especific volume
m=mass
V=mv
state 1
V1=m.v1
V1=4lb*1.25ft3/lb=5<u>ft3</u>
state 2
V2=m.v2
V2=4lb*0.5ft3/lb= <u> 2ft3</u>
PART B:
since the PV ^ n is constant we can equal the equations of state 1 and state 2
P1V1^n=P2V2^n
P1/P2=(V2/V1)^n
ln(P1/P2)=n . ln (V2/V1)
n=ln(P1/P2)/ ln (V2/V1)
n=ln(15/53)/ ln (2/5)
n=1.377
Answer:
- <u>0.59</u>
Explanation:
The <em>batch</em> is <em>rejected</em> if any of the <em>random samples are found defective</em>, or, what is the same, it will be accepted only if all 15 samples are good.
The probability that none be defective is the same probability that all the samples are good. Thus, start to calculate the probability that the batch is accepted.
The probability that the first sample is good is 214 /225, because there are 225 - 11 = 214 good samples in 225 doses.
The probability that the second samples is good too is 213/224, because there is 1 less good sample, in the 224 remaining samples.
By the same process, you conclude that the consecutive probabilities of selecting a good sample are: 212/223, 211/222, 210/221, . . . up to 199/211.
The joint probability of all the samples are good is the product of each probability:
The result is: 0.41278 ≈ 0.41
The conclusion is that the probability that all the samples are good and the batch is accepted is 0.41.
Therefore, <em>the probability that the batch is rejected</em> is 1 - 0.41 = 0.59.
Answer:
The inside temperature, is approximately 248 °C.
Explanation:
The parameters given are;
Temperature of the air = 20°C
Carbon steel surface temperature 250°C
Area of surface = 50 cm × 75 cm = 0.5 × 0.75 = 0.375 m²
Convection heat transfer coefficient = 25 W/(m²·K)
Heat lost by radiation = 300 W
Assumption,
Air temperature = 20 °C
Hot plate temperature = 250 °C
Thermal conductivity K = 65.2 W/(m·K)
Steady state heat transfer process
One dimensional heat conduction
We have;
Newton's law of cooling;
q = h×A×( -
) + Heat loss by radiation
= 25×0.325×(250 - 20) + 300
= 2456.25 W
The rate of energy transfer per second is given by the following relation;
Thermal conductivity K = 65.2 W/(m·K)
Therefore;
The inside temperature, = 247.99 °C ≈ 248 °C.