To answer the two questions, we need to know two important equations involving centripetal movement:
v = ωr (ω represents angular velocity <u>in radians</u>)
a = 
Let's apply the first equation to question a:
v = ωr
v = ((1800*2π) / 60) * 0.26
Wait. 2π? 0.26? 60? Let's break down why these numbers are written differently. In order to use the equation v = ωr, it is important that the units of ω is in radians. Since one revolution is equivalent to 2π radians, we can easily do the conversion from revolutions to radians by multiplying it by 2π. As for 0.26, note that the question asks for the units to be m/s. Since we need meters, we simply convert 26 cm, our radius, into meters. The revolutions is also given in revs/min, and we need to convert it into revs/sec so that we can get our final units correct. As a result, we divide the rate by 60 to convert minutes into seconds.
Back to the equation:
v = ((1800*2π)/60) * 0.26
v = (1800*2(3.14)/60) * 0.26
v = (11304/60) * 0.26
v = 188.4 * 0.26
v = 48.984
v = 49 (m/s)
Now that we know the linear velocity, we can find the centripetal acceleration:
a =
a = 
a = 9234.6 (m/
)
Wow! That's fast!
<u>We now have our answers for a and b:</u>
a. 49 (m/s)
b. 9.2 * 
(m/)
If you have any questions on how I got to these answers, just ask!
- breezyツ
Answer:
t = 3.516 s
Explanation:
The most useful kinematic formula would be the velocity of the motorcylce as a function of time, which is:
Where v_0 is the initial velocity and a is the acceleration. However the problem states that the motorcyle start at rest therefore v_0 = 0
If we want to know the time it takes to achieve that speed, we first need to convert units from km/h to m/s.
This can be done knowing that
1 km = 1000 m
1 h = 3600 s
Therefore
1 km/h = (1000/3600) m/s = 0.2777... m/s
100 km/h = 27.777... m/s
Now we are looking for the time t, for which v(t) = 27.77 m/s. That is:
27.777 m/s = 7.9 m/s^2 t
Solving for t
t = (27.7777 / 7.9) s = 3.516 s
Those two units can be compared to a 'mile per hour' and a 'mile per hour - hour'.
One is a rate. The other is a quantity, after maintaining a rate for some time.
-- 'Joule' is a unit of energy. It's the amount of work (energy) you do
when you push with a force of 1 newton though a distance of 1 meter.
Lifting 10 pound of beans 3 feet off the floor takes about 40.7 joules of energy.
-- 'Watt' is a <u><em>rate</em></u> of using energy . . . 1 joule per second.
If you lift 10 pounds 3 feet off the floor in 1 second, your <em>power</em> is 40.7 watts.
-- 'Watt-second' is the amount of energy used in one second,
at the rate of 1 joule per second . . . 1 joule.
-- 'Watt-hour' is the amount of energy used in one hour,
at the rate of 1 joule per second . . . 3,600 joules.
-- 'Kilowatt' is a bigger <em>rate</em> of using energy . . . 1,000 joules per second.
-- 'Kilowatt - second' is the amount of energy used in one second,
at the rate of 1,000 joules per second . . . 1,000 joules .
-- 'Kilowatt - hour' is the amount of energy used in one hour,
at the rate of 1,000 joules per second . . . 3,600,000 joules .
Depending on where you live, 3,600,000 joules of energy bought
from the electric company costs something between 5¢ and 25¢.
It is wasted, most likely as light, in this case, or it is lost during the transport of electricity.
333.15 Kelvins are equal to 60 degrees celsius
