Answer:
A) reduced air pressure on the ball.
Explanation:
Answer:
Acceleration (a) = 40 m/s²
Explanation:
Given:
Initial velocity (u) = 6 m/s
Final velocity (v) = 4.4 m/s
Time taken (t) = 0.04sec
Find:
Acceleration (a) = ?
Computation:
We know that,
⇒ v = u + at
⇒ a = (v - u) / t
⇒ Acceleration (a) = (4.4 - 6) / 0.04
⇒ Acceleration (a) = (-1.6) / 0.04
Acceleration (a) = 40 m/s²
The velocity of the ferry relative to the current is 4.5 m/s.
<h3>Relative velocity</h3>
- Relative velocity is the velocity of a body as observed from the reference point of another body either stationary or in motion.
Since the river is flowing parallel to the shore and the ferry is moving perpendicular to the shore, their velocities are at right angles to each other.
The two velocities form a right angled-triangle of sides 2, 4 and a hypotenuse which gives the relative velocity of the ferry to the current.
Using Pythagoras rule:
- Let c be the hypotenuse
- a = velocity of the ferry, and
- b = the velocity of the current, and
c² = 4² + 2²
c² = 16 + 4
c = 20
c = √20
c = 4.47 m/s
c ≈ 4.5 m/s
Therefore, the velocity of the ferry relative to the current is 4.5 m/s.
Learn more about relative velocity and Pythagoras rule at: brainly.com/question/25617868
The first thing you should know for this case is that work is defined as the product of force by the distance traveled in the direction of force.
We have then:
W = Fd
The distance varies, so we must integrate:
from 0 to 20:
W = ∫F (x) dx
W = ∫32xdx
W = 32∫xdx
W = 32 (x ^ 2/2) = (16) (20 ^ 2) = 6400 ft * lbs
answer:
6400 ft * lbs is work done pulling the rope up 20 ft
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.