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olga2289 [7]
3 years ago
14

A 1.00-kg object is attached by a thread of negligible mass, which passes over a pulley of negligible mass, to a 2.00-kg object.

The objects are positioned so that they are the same height from the floor and then released from rest. What are the speeds of the objects when they are separated vertically by 1.00 m?
Physics
1 answer:
Anna [14]3 years ago
3 0

Answer:

a = 3.27 m/s²

v = 2.56 m/s

Explanation:

given,

mass A = 1 kg

mass B = 2 kg

vertical distance between them = 1 m

F_d = mg

F_d = 2 \times 9.8

F_d = 19.6\ N

F_u = mg

F_u = 1 \times 9.8

F_u = 9.8\ N

F_{net} = 19.6 - 9.8

F_{net}=9.8\ N

F = (m_1+m_2)a

9.8 = (2+1)a

a = 3.27 m/s²

The speed of the system at that moment is:

v² = u² + 2×a×s

v² = 0² + 2× 3.27 × 1

v ² = 6.54

v = 2.56 m/s

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Let us assume we need to find its velocity.

We know that,

v=\dfrac{dr}{dt}\\\\=\dfrac{d}{dt}(3.0 t^2 i + 5.0j+ 6.0 tk) \\\\=(6ti+6k)\ m/s

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3 years ago
A 1500 kg car traveling 5.0 m/s collides head on with a 3000 kg truck traveling
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  • v_2=7m/s

Using law of conservation of momentum

\\ \sf\Rrightarrow m_1v_1-m_2v_2=(m1+m2)v_3

\\ \sf\Rrightarrow 1500(5)-3000(7)=(1500+3000)v_3

\\ \sf\Rrightarrow 7500-21000=4500v_3

\\ \sf\Rrightarrow -13500=4500v_3

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6 0
2 years ago
Explain the difference between the terms "rotation" and "revolution"
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Answer:

Explanation:

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Two buses are driving along parallel freeways that are 5mi apart, one heading east and the other heading west. Assuming that eac
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Answer:

101.54m/h

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l=\sqrt{13^2-5^2}=12m\\\\\frac{dl}{dt}=-(55m/h+55m/h})\\\\=-110m/h#The speed toward each other.

\frac{dh}{dt}=0, \ \ \ \ h=constant\\\\h^2+l^2=b^2\\\\2h\frac{dh}{dt}+2l\frac{dl}{dt}=2b\frac{db}{dt}\\\\2\times5\times0+2\times12\times(-110)=2\times13\frac{db}{dt}\\\\\frac{db}{dt}=-101.54m/h

Hence, the rate at which the distance between the buses is changing when they are 13mi apart is 101.54m/h

4 0
2 years ago
Please do number 25! Explain how you got your answer with detail to get Brainliest! Thank you!
Ad libitum [116K]
John weighs 200 pounds.
In order to lift himself up to a higher place, he has to exert force of 200 lbs.

The stairs to the balcony are 20-ft high.
In order to lift himself to the balcony, John has to do
(20 ft) x (200 pounds)  =  4,000 foot-pounds of work.

If he does it in 6.2 seconds, his RATE of doing work is
(4,000 foot-pounds) / (6.2 seconds)  =  645.2 foot-pounds per second.

The rate of doing work is called "power".

(If we were working in the metric system (with SI units),
the force would be in "newtons", the distance would be in "meters",
1 newton-meter of work would be 1 "joule" of work, and
1 joule of work per second would be 1 "watt".
Too bad we're not working with metric units.)

So back to our problem.

John has to do 4,000 foot-pounds of work to lift himself up to the balcony,
and he's able to do it at the rate of 645.2 foot-pounds per second.

Well, 550 foot-pounds per second is called 1 "horsepower".

So as John runs up the steps to the balcony, he's doing the work
at the rate of

           (645.2 foot-pounds/second) / (550 ft-lbs/sec per HP)

=  1.173 Horsepower.  GO JOHN !

(I'll betcha he needs a shower after he does THAT 3 times.)
_______________________________________________

Oh my gosh !  Look at #26 !  There are the metric units I was talking about.

Do you need #26 ?

I'll give you the answers, but I won't go through the explanation,
because I'm doing all this for only 5 points.

a).  5
b).  750 Joules
c).  800 Joules
d).  93.75%

You're welcome.

And #27 is 0.667 m/s .
7 0
3 years ago
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