weight = mg acts
downwards <span>
normal force = N acts upwards.
and force F acts at an angle θ below the horizontal.
(Let us assume that the woman pushes from the left, so F is
acted towards the right, which is below the horizontal)
so that, Frictional force, f=us*N acts towards the left
Now we balance the forces along x and y directions:
y direction: N = mg + F sinΘ
x direction: us * N = F cosΘ
We let the value of µs be equal to a value such that any F
will not be able to move the crate. Then, if we increase F by an amount F',
then the force pushing the crate towards the right also increases by F' cosΘ. Additionally,
the frictional force f must raise by exactly this amount.
Since f can’t exceed us*N, so the normal force must increase
by F' cosΘ/us.
Also, from the y direction equation, the normal force exceeds
by F' sin Θ.
<span>These two values must be the same, therefore:
<span>us = cot θ</span></span></span>
Answer:
Gravitational
Tension
Normal
Friction.
Explanation:
The forces acting on the sled are:
Tension: the tension from the rope, this is the force that "moves" the sled.
Friction: kinetic friction between the sled and the ground as the sled moves.
There are another two forces that also act on the sled, but that "has no effect"
Gravitational force: This force pulls the sled down, against the floor.
Normal force: This force "opposes" to the gravitational one, so they cancel each other.
These two forces cancel each other, so they have no direct impact on the movement of the sled. BUT, the friction force depends on the weight of the moving object, and the weight of the moving object depends on the gravitational force, so we need gravitational force in order to have friction force.
Then we can conclude that the forces acting on the sled are:
Gravitational
Tension
Normal
Friction.
After one meter, 3.4% of the light is gone ... either soaked up in the fiber
material or escaped from it. So only (100 - 3.4) = 96.6% of the light
remains, to go on to the next meter.
After the second meter, 96.6% of what entered it emerges from it, and
that's 96.6% of 96.6% of the original signal that entered the beginning
of the fiber.
==> After 2 meters, the intensity has dwindled to (0.966)² of its original level.
It's that exponent of ' 2 ' that corresponds to the number of meters that the light
has traveled through.
==> After 'x' meters of fiber, the remaininglight intensity is (0.966) ^x-power
of its original value.
If you shine 1,500 lumens into the front of the fiber, then after 'x' meters of
cable, you'll have
<em>(1,500) · (0.966)^x</em>
lumens of light remaining.
=========================================
The genius engineers in the fiber design industry would not handle it this way.
When they look up the 'attenuation' of the cable in the fiber manufacturer's
catalog, it would say "15dB per 100 meters".
What does that mean ? Break it down: 15dB in 100 meters is <u>0.15dB per meter</u>.
Now, watch this:
Up at the top, the problem told us that the loss in 1 meter is 3.4% . We applied
super high mathematics to that and calculated that 96.6% remains, or 0.966.
Look at this ==> 10 log(0.966) = <em><u>-0.15</u> </em> <== loss per meter, in dB .
Armed with this information, the engineer ... calculating the loss in 'x' meters of
fiber cable, doesn't have to mess with raising numbers to powers. All he has to
do is say ...
-- 0.15 dB loss per meter
-- 'x' meters of cable
-- 0.15x dB of loss.
If 'x' happens to be, say, 72 meters, then the loss is (72) (0.15) = 10.8 dB .
and 10 ^ (-10.8/10) = 10 ^ -1.08 = 0.083 = <em>8.3%</em> <== <u>That's</u> how much light
he'll have left after 72 meters, and all he had to do was a simple multiplication.
Sorry. Didn't mean to ramble on. But I do stuff like this every day.
The two points on a periodic wave in a medium are said to be in phase if they have the same amplitude and are moving in the same direction.
Option 4.
Explanation:
A periodic wave is termed for waves which flow in a repetition pattern in a given time scale. Periodic wave can also be termed as a transverse wave. So a transverse wave have various crests and troughs. The two successive crests and two successive troughs are said to be in phase with each other.
Thus, for a periodic wave in a medium, the in phase can be obtained in two points which have the same amplitude and are moving in the same direction.
As amplitude is a scalar quantity and so direction should be taken into consideration for making the points related to successive crests only in phase with themselves. Also this also relates the points related to successive troughs to be in phase with each other. But a crest and a trough will not be in phase with each other.
Thus, option 4, that is the two points on a periodic wave in a medium are said to be in phase if they have the same amplitude and are moving in the same direction.
