Answer:
In free fall, mass is not relevant and there's no air resistance, so the acceleration the object is experimenting will be equal to the gravity exerted. If the object is falling on our planet, the value of gravity is approximately 9.81ms2 .
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 .
Answer: for insulation of heat
Explanation:
Windows in cold countries have double glazing windows to provide a barrier against the outside temperature by creating a buffer zone between two glasses.
The air or any other gas-filled between the glasses act as an insulator and offer great resistance to outside temperature thereby maintaining the inside temperature intact.
The acorn was at a height of <u>4.15 m</u> from the ground before it drops.
The acorn takes a time t to fall through a distance h₁, which is the length of the scale. When the acorn reaches the top of the scale, its velocity is u.
Calculate the speed of the acorn at the top of the scale, using the equation of motion,
Since the acorn falls freely under gravity, its acceleration is equal to the acceleration due to gravity g.
Substitute 2.27 m for s (=h₁), 0.301 s for t and 9.8 m/s² for a (=g).
If the acorn starts from rest and reaches a speed of 6.067 m/s at the top of the scale, it would have fallen a distance h₂ to achieve this speed.
Use the equation of motion,
Substitute 6.067 m/s for v, 0 m/s for u, 9.8 m/s² for a (=g) and h₂ for s.
The height h above the ground at which the acorn was is given by,
The acorn was at a height <u>4.15m</u> from the ground before dropping down.
Assuming constant acceleration, the goalie slows the ball from 18 m/s to rest in 0.035 s, so that the acceleration felt by the ball is
