Answer:
n this question, we are asked to find the probability that
R1 is normally distributed with mean 65 and standard deviation 10
R2 is normally distributed with mean 75 and standard deviation 5
Both resistor are connected in series.
We need to find P(R2>R1)
the we can re write as,
P(R2>R1) = P(R2-R1>R1-R1)
P(R2>R1) = P(R2-R1>0)
P(R2>R1) = P(R>0)
Where;
R = R2 - R1
Since both and are independent random variable and normally distributed, we can do the linear combinations of mean and standard deviations.
u = u2-u1
u = 75 - 65 = 10ohm
sd = √sd1² + sd2²
sd = √10²+5²
sd = √100+25 = 11.18ohm
Now we will calculate the z-score, to find P( R>0 )
Z = ( X -u)/sd
the z score of 0 is
z = 0 - 10/11.18
z= - 0.89
Answer:
P = 4.745 kips
Explanation:
Given
ΔL = 0.01 in
E = 29000 KSI
D = 1/2 in
LAB = LAC = L = 12 in
We get the area as follows
A = π*D²/4 = π*(1/2 in)²/4 = (π/16) in²
Then we use the formula
ΔL = P*L/(A*E)
For AB:
ΔL(AB) = PAB*L/(A*E) = PAB*12 in/((π/16) in²*29*10⁶ PSI)
⇒ ΔL(AB) = (2.107*10⁻⁶ in/lbf)*PAB
For AC:
ΔL(AC) = PAC*L/(A*E) = PAC*12 in/((π/16) in²*29*10⁶ PSI)
⇒ ΔL(AC) = (2.107*10⁻⁶ in/lbf)*PAC
Now, we use the condition
ΔL = ΔL(AB)ₓ + ΔL(AC)ₓ = ΔL(AB)*Cos 30° + ΔL(AC)*Cos 30° = 0.01 in
⇒ ΔL = (2.107*10⁻⁶ in/lbf)*PAB*Cos 30°+(2.107*10⁻⁶ in/lbf)*PAC*Cos 30°= 0.01 in
Knowing that PAB*Cos 30°+PAC*Cos 30° = P
we have
(2.107*10⁻⁶ in/lbf)*P = 0.01 in
⇒ P = 4745.11 lb = 4.745 kips
The pic shown can help to understand the question.
Question:
1. Some are known as "nun" buoys
2. They are labeled with odd numbers
3. If it is lighted, the light color is green
4. Some are known as "can" buoys
Answer:
The correct option is;
1. Some are known as "nun" buoys
Explanation:
Based on the lateral system, on the starboard side, one can find the red even numbered marks while the odd-numbered, green, marks are located on the port side of a channel such that the buoy numbers increase as a vessel travels upstream.
The red buoys are cones shaped in appearance and have triangular reflective sign markings embossed and they are of different types included in the order of lower water depth
1. NUN buoy
2. Lighted buoy
3. Light
4. Day beacon
Therefore, the correct option is 1. Some are known as "NUN" buoys.
Answer:
Explanation:
In a particular application involving airflow over a heated surface, the boundary layer temperature distribution, T(y), may be approximated as:
[ T(y) - Ts / T∞ - Ts ] = 1 - e^( -Pr (U∞y / v) )
where y is the distance normal to the surface and the Prandtl number, Pr = Cpu/k = 0.7, is a dimensionless fluid property. a.) If T∞ = 380 K, Ts = 320 K, and U∞/v = 3600 m-1, what is the surface heat flux? Is this into or out of the wall? (~-5000 W/m2 , ?). b.) Plot the temperature distribution for y = 0 to y = 0.002 m. Set the axes ranges from 380 to 320 for temperature and from 0 to 0.002 m for y. Be sure to evaluate properties at the film temperature.
The use of smart cranes doesn't enable transportation companies to: B. Manually load or unload crates onto a train with several workers.
<h3>What is a
smart crane?</h3>
A smart crane can be defined as a mechanical equipment that is designed and developed to automatically load, unload, or control the movement of heavy equipment (objects) from one point to another, especially through the use of a projecting arm or beam.
<h3>The use of smart cranes.</h3>
Generally, various transportation companies use a smart crane to achieve and perform the following tasks:
- An ability to hire fewer workers to transport crates onto trains or haulage vehicles.
- An ability to control the movement of heavy objects.
- An ability to incorporate technological advancements into their routine work or daily practices.
Hence, the use of smart cranes by transportation companies completely abolishes the need to manually load or unload crates onto a train with several workers.
Read more on smart cranes here: brainly.com/question/25845985
