Answer:
I don't know this answer at all
Explanation:
I don't know about these problems
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 .
Recall this kinematic equation:
a = 
This equation gives the acceleration of the object assuming it IS constant (the velocity changes at a uniform rate).
a is the acceleration.
Vi is the initial velocity.
Vf is the final velocity.
Δt is the amount of elapsed time.
Given values:
Vi = 0 m/s (the car starts at rest).
Vf = 25 m/s.
Δt = 10 s
Substitute the terms in the equation with the given values and solve for a:
a = 
<h3>a = 2.5 m/s²</h3>
Answer:
See the answers below
Explanation:
To solve this problem we must use the following equation of kinematics.
where:
Vf = final velocity [m/s]
Vo = initial velocity [m/s]
a = acceleration [m/s²]
t = time [s]
<u>First case</u>
Vf = 6 [m/s]
Vo = 2 [m/s]
t = 2 [s]
![6=2+a*2\\4=2*a\\a=2[m/s^{2} ]](https://tex.z-dn.net/?f=6%3D2%2Ba%2A2%5C%5C4%3D2%2Aa%5C%5Ca%3D2%5Bm%2Fs%5E%7B2%7D%20%5D)
<u>Second case</u>
Vf = 25 [m/s]
Vo = 5 [m/s]
a = 2 [m/s²]
![25=5+2*t\\t = 10 [s]](https://tex.z-dn.net/?f=25%3D5%2B2%2At%5C%5Ct%20%3D%2010%20%5Bs%5D)
<u>Third case</u>
Vo =4 [m/s]
a = 10 [m/s²]
t = 2 [s]
![v_{f}=4+10*2\\v_{f}=24 [m/s]](https://tex.z-dn.net/?f=v_%7Bf%7D%3D4%2B10%2A2%5C%5Cv_%7Bf%7D%3D24%20%5Bm%2Fs%5D)
<u>Fourth Case</u>
Vf = final velocity [m/s]
Vo = initial velocity [m/s]
a = acceleration [m/s²]
t = time [s]
![v_{f}=5+8*10\\v_{f}=85 [m/s]](https://tex.z-dn.net/?f=v_%7Bf%7D%3D5%2B8%2A10%5C%5Cv_%7Bf%7D%3D85%20%5Bm%2Fs%5D)
<u>Fifth case</u>
Vf = final velocity [m/s]
Vo = initial velocity [m/s]
a = acceleration [m/s²]
t = time [s]
