Answer:
The experiment to determine the coefficient of static friction between a rough block and an incline plane is detailed in the explanation.
Explanation:
A simple experiment is as follows:
Take a weight in which its mass is known and hold it with a roller at the end of the inclined plane with a rope.
Place the weight on the plane inclined horizontally with the ground, forming a zero degree angle.
Gradually lift the weight so that the angle of incline begins to increase.
Observe that the weight will begin to slide. This angle will be the angle theta.
In this way, the coefficient of friction will be calculated as the tangent of the angle theta.
Take a different weight and repeat the procedure to get different values of the coefficient of friction.
By averaging the coefficients of friction, a final value will be obtained.
Answer:
<u>B. East of NYC</u>
Explanation:
We can make this conclusion because for one reason New York City is the Northern Hemisphere, and according to the Coriolis effect <em>when objects in the Northern Hemisphere travel long distances above the ground, </em><em>they will be deflected to the right </em>as a result of earth's rotation.
Hence, going by the map of New York on the Northern Hemisphere, <em>when the (object) flight is deflected to the right, for someone in NYC it will indicate a </em><em>shift to the east of NYC</em><em> </em>since the pilot failed to take the Coriolis Effect into account.
Answer:
False.
Explanation:
Potential of Hydrogen is the full form of PH.
Potential of Hydrogen tells about the acidity or basicity of the solution.
The pH of a solution is a measure of the concentration of hydrogen ions 
ions and hydroxide ions 
in a solution.
So, the given statement is false.
Nothing will happen until the current is switched.
I assume friction is the only force acting on the book as it slides.
(A) By the work-energy theorem, the total work performed on the book as it slides is equal to the change in its kinetic energy:
<em>W</em> = ∆<em>K</em>
<em>W</em> = 1/2 (1.50 kg) (1.25 m/s)² - 1/2 (1.50 kg) (3.21 m/s)²
<em>W</em> ≈ -6.56 J
(B) Using the work-energy theorem again, the speed <em>v</em> of the book at point C is such that
-0.750 J = 1/2 (1.50 kg) <em>v</em> ² - 1/2 (1.50 kg) (1.25 m/s)²
==> <em>v</em> = 0.750 m/s
(C) Take the left side to be positive, then solve again for <em>v</em>.
0.750 J = 1/2 (1.50 kg) <em>v</em> ² - 1/2 (1.50 kg) (1.25 m/s)²
==> <em>v</em> ≈ 1.60 m/s
