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Luda [366]
3 years ago
13

Can someone help me with this real quick?

Physics
1 answer:
ch4aika [34]3 years ago
4 0

Answer:

7,546 J

Explanation:

recall that Potential energy is given by

P.E = mgΔh

where m = 70kg (given)

g = 9.8 m/s² (acceleration due to gravity)

Δh = change in height

= distance from top of building to top of car

= height of building - height of car

= (5+8) - 2

= 11m

substituting all these into the equation:

P.E = mgΔh

= 70 x 9.8 x 11

= 7,546 J

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A 90. 0-kg ice hockey player hits a 0. 150-kg puck, giving the puck a velocity of 45. 0 m/s. If both are initially at rest and i
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The distance traveled by the hockey player is 0.025 m.

<h3>The principle of conservation of linear momentum;</h3>
  • The principle of conservation of linear momentum states that, the total momentum of an isolated system is always conserved.

The final velocity of the hockey play is calculated by applying the principle of conservation of linear momentum;

m_1v_1 = m_2 v_2\\\\v_1 = \frac{m_2 v_2}{m_1} \\\\v_1 = \frac{0.150 \times 45}{90} \\\\v_1 = 0.075 \ m/s

The time taken for the puck to reach 15 m is calculated as follows;

t = \frac{d}{v} \\\\t = \frac{15\ m}{45 \ m/s} \\\\t = 0.33 \ s

The distance traveled by the hockey player at the calculated time is;

d = vt\\\\d = 0.075 \ m/s \ \times 0.33 \ s\\\\d = 0.025 \ m

Learn more about conservation of linear momentum here: brainly.com/question/7538238

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What describes the particles in a liquid
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mark brainliest :))

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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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