The gravitational potential energy of an object depends on three things. Its mass, its height above the surface of the earth and the pull of gravity (which is assumed to always be 9.8 m/s².
The Formula for finding the GPE is : m x g x h where m = mass, g = gravitational acceleration and h is height from earth's surface.
Using this formula we can find that :
GPE= 75 x 9.8 x 300 = 220500J (where J is the SI unit for GPE and stands for Joules.
Answer:
Half life of S = 3.76secs
Explanation:
The concept of half life in radioactivity is applied. Half life is the time taken for a radioactive material to decay to half of its initial size.
For part 1 - How much signal will be degraded in 1secs = 1/3.9 = 0.2564
for part 2 - How much signal will be degraded in 1secs = 1/104 = 0.009615
Simply say = 1/3.9 + 1/104 = 0.266015
So both part 1 and part 2 took 1/0.266015 = 3.76secs is the half life of S when both pathways are active
A vehicle's power steering system consists of components that generate and transmit power to the wheels.
We can solve this problem using <span>Hagen–Poiseuille equation. Derivation of this equation is a bit complicated so I will just write down the equation.
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This equation gives you the pressure drop <span>in an </span>incompressible<span> and </span>Newtonian<span> fluid in </span>laminar flow<span> flowing through a long cylindrical pipe of the constant cross section.
L is the length of the cylinder, Q is the volumetric flow rate, R is the radius of the pipe, and
is dynamic viscosity.
Dynamic viscosity of water at 20 Celsius is 0.001 PaS.
Now we can calculate the pressure drop:
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Answer: Example 1: Consider a crate being pulled along a frictionless floor (while such a floor is very hard to find, this will still help us understand the concept and we can return to this situation later, after considering friction, and solve it more realistically).
Consider a crate being pulled along a horizontal, frictionless floor. A rope is tied around it and a man pulls on the rope with a force of T. T is the tension in the rope. What happens to the crate?
Before we can apply Newton's Second Law,
F = m a
we must find the net force -- the vector sum of all the forces -- acting on the object. In addition to the force T exerted by the rope, what other forces act on the object?
As discussed in class, in Mechanics, we can restrict our attention to "contact" forces and "gravity". That means gravity pulls down on this crate with a force equal to its weight, w. But the floor supports the crate. The floor responds by pushing up on the crate with a force we call the normal force. "Normal" means "perpendicular". We will call this force n; you may also encounter it labeled N or FN.
Explanation:
