(1) Changing Fahrenheit to Celsius:
The formula used to convert from Fahrenheit to Celsius is as follows:
C = <span>(F - 32) * 5/9
</span>We are given that F=200, substitute in the above formula to get the corresponding temperature in Celsius as follows:
C = (200-32) * (5/9) = 93.333334 degrees Celsius
(2) Changing the Fahrenheit to kelvin:
The formula used to convert from Fahrenheit to kelvin is as follows:
K = <span>(F - 32) * 5/9 + 273.15
</span>We are given that F = 200. substitute in the above formula to get the corresponding temperature in kelvin as follows:
K = (200-32)*(5/9) + 273.15 = 366.483334 degrees kelvin
Answer:
Explanation:
Theorem of Binomial Distribution will apply here.
n = 29 , p = .67 , q = 0.33
mean = np = 29 x .67 = 19.43
Standard Deviation = √npq
= √29 x .67 x .33
= √6.4
= 2.53
=
The related concept to solve this exercise is given in the expressions that the magnetic field has both as a function of the number of loops, current and length, as well as inductance and permeability. The first expression could be given as,
The magnetic field H is given as,
Here,
n = Number of turns of the coil
I = Current that flows in the coil
l = Length of the coil
From the above equation, the number of turns of the coil is,
The magnetic field is again given by,
Where the minimum inductance produced by the solenoid coil is B.
We have to obtain n, that
Replacing with our values we have that,
Therefore the number of turn required is 28Truns
John weighs 200 pounds.
In order to lift himself up to a higher place, he has to exert force of 200 lbs.
The stairs to the balcony are 20-ft high.
In order to lift himself to the balcony, John has to do
(20 ft) x (200 pounds) = 4,000 foot-pounds of work.
If he does it in 6.2 seconds, his RATE of doing work is
(4,000 foot-pounds) / (6.2 seconds) = 645.2 foot-pounds per second.
The rate of doing work is called "power".
(If we were working in the metric system (with SI units),
the force would be in "newtons", the distance would be in "meters",
1 newton-meter of work would be 1 "joule" of work, and
1 joule of work per second would be 1 "watt".
Too bad we're not working with metric units.)
So back to our problem.
John has to do 4,000 foot-pounds of work to lift himself up to the balcony,
and he's able to do it at the rate of 645.2 foot-pounds per second.
Well, 550 foot-pounds per second is called 1 "horsepower".
So as John runs up the steps to the balcony, he's doing the work
at the rate of
(645.2 foot-pounds/second) / (550 ft-lbs/sec per HP)
= 1.173 Horsepower. GO JOHN !
(I'll betcha he needs a shower after he does THAT 3 times.)
_______________________________________________
Oh my gosh ! Look at #26 ! There are the metric units I was talking about.
Do you need #26 ?
I'll give you the answers, but I won't go through the explanation,
because I'm doing all this for only 5 points.
a). 5
b). 750 Joules
c). 800 Joules
d). 93.75%
You're welcome.
And #27 is 0.667 m/s .
