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Vanyuwa [196]
3 years ago
15

A railroad car of mass M moving at a speed v1 collides and couples with two coupled railroad cars, each of the same mass M and m

oving in the same direction at a speed v2.
(a) What is the speed vf of the three coupled cars after the collision in terms of v1 and v2? vf =

(b) How much kinetic energy is lost in the collision? Answer in terms of M, v1, and v2. ?KE =
Physics
1 answer:
Tatiana [17]3 years ago
7 0

Answer:

Explanation:

Given

mass of each railroad is M

first railroad is moving with velocity v_1 and other two's with v_2

conserving Momentum

Mv_1+2Mv_2=3M v_f

where v_f is the final velocity of combined system

v_f=\frac{v_1+2v_2}{3}

(b)Initial Kinetic Energy

=\frac{1}{2}Mv_1^2+\frac{1}{2}2Mv_2^2

=\frac{1}{2}Mv_1^2+Mv_2^2

Final Kinetic Energy

=\frac{1}{2}3Mv_f^2=\frac{1}{6}M(v_1^2+4v_2^{2}+4v_{1}v_2)

Loss in Kinetic Energy

=\frac{1}{2}Mv_1^2+Mv_2^2-\frac{1}{6}M(v_1^2+4v_2^{2}+4v_{1}v_2)

=M(v_1-v_2^2)^2

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Awnser these pls they r for middle schoolers lol
soldi70 [24.7K]

Answer:

Find the set of value of x Find the set of value of x which satisfy the inequality 2r2- 5x 2 18which satisfy the inequality 2r2- Find the set of value of x which satisfy the inequality 2r2- 5x 2 185x 2 18

8 0
3 years ago
A runner moves west with an initial velocity of 4 m/s. She accelerates 3 m/s2 for 30 seconds. The runner's final velocity is
Firlakuza [10]

Answer:

v = 94m/s

Explanation:

Using the first equation of motion

v = u + at

u = 4m/s , a = 3m/s² , t = 30s , v = ?

v = u + at

v = 4 + 3 × 30

v = 4 + 90

v = 94m/s

I hope this was helpful, please mark as brainliest

3 0
3 years ago
A car has an acceleration of -5 m/s^2. Describe the car’s motion
AlladinOne [14]

Explanation:

because the acceleration is negative, this indicates a deceleration (or slowing down) . Hence we can say that:

The car is decelerating (slowing down), i.e its velocity is decreasing, at a constant rate of 5m/s².

7 0
3 years ago
ANswer the question and give example: Can friction ever perform positive work?
Harman [31]
Yes. Think of block sitting on top of a bigger block. If the bottom block moves, it will drag the top block with it. Since the force of friction on the small block and its displacement are in the same direction, the "work" is positive. The static friction is a passive force, It is not a source of energy; it transmits the force placed on the bottom block. (And the "work" done by the friction on the bottom block is exactly the negative of the work done on the top block.) 
4 0
3 years ago
Read 2 more answers
Fiber-optic cables are used widely for internet wiring, data transmission, and surgeries. When light passes through a fiber-opti
Gwar [14]
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.
5 0
3 years ago
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