Answer: c living in a camber in an under water habitat
Explanation:
The minimum value of the coefficient of static friction between the block and the slope is 0.53.
<h3>Minimum coefficient of static friction</h3>
Apply Newton's second law of motion;
F - μFs = 0
μFs = F
where;
- μ is coefficient of static friction
- Fs is frictional force
- F is applied force
μ = F/Fs
μ = F/(mgcosθ)
μ = (250)/(50 x 9.8 x cos15)
μ = 0.53
Thus, the minimum value of the coefficient of static friction between the block and the slope is 0.53.
Learn more about coefficient of friction here: brainly.com/question/20241845
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Answer:
<u> </u><u>»</u><u> </u><u>Image</u><u> </u><u>distance</u><u> </u><u>:</u>

- v is image distance
- u is object distance, u is 10 cm
- f is focal length, f is 5 cm

<u> </u><u>»</u><u> </u><u>Magnification</u><u> </u><u>:</u>
• Let's derive this formula from the lens formula:

» Multiply throughout by fv

• But we know that, v/u is M

- v is image distance, v is 10 cm
- f is focal length, f is 5 cm
- M is magnification.

<u> </u><u>»</u><u> </u><u>Nature</u><u> </u><u>of</u><u> </u><u>Image</u><u> </u><u>:</u>
- Image is magnified
- Image is erect or upright
- Image is inverted
- Image distance is identical to object distance.
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 .
Answer:
A) 
B) F = 1632.65 N
Explanation:
Given details
outside air speed is given as 
since inside air is atmospheric , 
a) By using bernoulli equation between outside and inside of flight


![\Delta P = \frac{1}{2} \rho[ v_2^2 -v_1^2]](https://tex.z-dn.net/?f=%5CDelta%20P%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5Crho%5B%20v_2%5E2%20-v_1%5E2%5D)
![\Delta P = \frac{1}{2} 1.29 [ 150^2 - 0^2]](https://tex.z-dn.net/?f=%5CDelta%20P%20%3D%20%5Cfrac%7B1%7D%7B2%7D%201.29%20%5B%20150%5E2%20-%200%5E2%5D)

b) force exerted on window
Area of window 
We know that force is given as


F = 1632.65 N