A car is traveling around a horizontal circular track with radius r = 220 m at a constant speed v = 16 m/s as shown. The angle θ
1 answer:
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Answer:
Explanation:
The equation fo potential energy is PE = mgh, where m is the mass of the ball, g is the pull of gravity (constant at 9.8), and h is the max height of the ball. What we do not have here is that height. We need to first solve for it using one-dimensional equations. What we have to know above all else, is that the final velocity of an object at its max height is always 0. That allows us to use the equation
where vf is the final velocity and v0 is the initial velocity. We will find out how long it takes for the object to reach that max height first and then use that time to find out what that max height is. Baby steps here...
0 = 21.5 + (-9.8)t and
-21.5 = -9.8t so
t = 2.19 seconds (Keep in mind that if I used the rules correctly for sig fig's, the answer you SHOULD get is not one shown, so I had to adjust the sig fig's and break the rules. But you know what they say about rules...)
Now we will use that time to find out the max height of the object in the equation
Δx = and filling in:
Δx = which simplifies down a bit to
Δx = 47.1 - 23.5 so
Δx = 23.6 meters.
Now we can plug that in to the PE equation to find the PE of the object:
PE = (.19)(9.8)(23.6) so
PE = 43.9 J
Answer:
The speed v of the particle at t=5.00 seconds = 43 m/s
Explanation:
Given :
mass m = 5.00 kg
force f(t) = 6.00t2−4.00t+3.00 N
time t between t=0.00 seconds and t=5.00 seconds
From mathematical expression of Newton's second law;
Force = mass (m) x acceleration (a)
F = ma
...... (1)
acceleration (a) ......(2)
substituting (2) into (1)
Hence, F
Integrating both sides
The force is acting on the particle between t=0.00 seconds and t=5.00 seconds;
......(3)
Substituting the mass (m) =5.00 kg of the particle, equation of the varying force f(t)=6.00t2−4.00t+3.00 and calculating speed at t = 5.00seconds into (3):
v = 43 meters per second
The speed v of the particle at t=5.00 seconds = 43 m/s