Answer:
θ = 21.8º
Explanation:
We can solve this exercise using Newton's conditions for equilibrium, in the attached we can see a diagram of the forces.
The most used coordinate system is an axis parallel to the plane (x axis) and an axis perpendicular to the plane (y axis), let's write Newton's equations on this axes
Y Axis
N- 
= 0
N =
X axis
fr - Wₓ = 0 (1)
Let's use trigonometry to find the normal ones
sin θ = Wₓ / W
cos θ = / W
Wx = W sin θ
= W cos θ
The friction force has the formula
fr = μ N
fr = μ (W cos θ)
We substitute in 1
μ mg cosθ = mg sin θ
μ cos θ = sin θ
tan θ = μ
θ = tan⁻¹ μ
calculate
θ = tan⁻¹ 0.40
θ = 21.8º
The answer is "the same than the mercury in the bigger tube".
If one barometer tube has twice the cross-sectional area of another, mercury in the smaller tube will rise the same than the mercury in the bigger tube.
The mercury will rise to the point where the column of mercury has the same weight as the force exerted by the atmosphere.
The force exerted by the atmosphere is pressure * cross-sectional area
Anf the weight of the column of mercury, W, will be:
W = m* g
where m = density * volume, and volume = cross-sectional area * height
=> W = density * cross-sectional area * height
Then, you make W = F and get:
density * cross-sectional area * height = P * cross-sectional area
The term cress-sectional area appears on both sides so it gets cancelled, and the height of the column of mercury does not depend on the cross-sectional area of the barometer.
Answer:
B.
Explanation:
I'm not sure if it's the right answer but I think it is. I hope I helped out in some way.
When a positive rod is placed to the right of sphere B, and the spheres are separated, the reason behind this is the same charge on the sphere as the rod i.e. the right of the sphere also had a positive charge. Thus, the same positive charges could not reside on the right side surface of the sphere due to which it separation happens.
Read more on Brainly.com -
brainly.com/question/4135790#readmore
Answer: 2.83 minutes
Explanation:
It is understood that trains are approaching. That is, they have speeds of equal magnitude but opposite. When train A travels x meters northbound, then train B travels the same distance southbound.
Therefore trains approach at a speed of:
Then:
Where x is the distance between the trains
So the time in which both trains meet is:
This is:
<em />
<em>How long will it be before they reach one another ?</em>
<h3>2.83 minutes</h3>
