Data !
hope this helped <3
It was <em>(1/2) (Net force on the cart) m/s²) </em>.
(a) 3.56 m/s
(b) 11 - 3.72a
(c) t = 5.9 s
(d) -11 m/s
For most of these problems, you're being asked the velocity of the rock as a function of t, while you've been given the position as a function of t. So first calculate the first derivative of the position function using the power rule.
y = 11t - 1.86t^2
y' = 11 - 3.72t
Now that you have the first derivative, it will give you the velocity as a function of t.
(a) Velocity after 2 seconds.
y' = 11 - 3.72t
y' = 11 - 3.72*2 = 11 - 7.44 = 3.56
So the velocity is 3.56 m/s
(b) Velocity after a seconds.
y' = 11 - 3.72t
y' = 11 - 3.72a
So the answer is 11 - 3.72a
(c) Use the quadratic formula to find the zeros for the position function y = 11t-1.86t^2. Roots are t = 0 and t = 5.913978495. The t = 0 is for the moment the rock was thrown, so the answer is t = 5.9 seconds.
(d) Plug in the value of t calculated for (c) into the velocity function, so:
y' = 11 - 3.72a
y' = 11 - 3.72*5.913978495
y' = 11 - 22
y' = -11
So the velocity is -11 m/s which makes sense since the total energy of the rock will remain constant, so it's coming down at the same speed as it was going up.
Answer:
B) I1 = 1680 kg.m^2 I2 = 1120 kg.m^2
C) V = 0.84m/s T = 29.92s
D) ω2 = 0.315 rad/s
Explanation:
The moment of inertia when they are standing on the edge:
where M is the mass of the merry-go-round.
I1 = 1680 kg.m^2
The moment of inertia when they are standing half way to the center:
I2 = 1120 kg.m^2
The tangencial velocity is given by:
V = ω1*R = 0.84m/s
Period of rotation:
T = 2π / ω1 = 29.92s
Assuming that there is no friction and their parents are not pushing anymore, we can use conservation of the angular momentum to calculate the new angular velocity:
I1*ω1 = I2*ω2 Solving for ω2:
ω2 = I1*ω1 / I2 = 0.315 rad/s
Answer:
The SI unit of force is the newton, symbol N. The base units relevant to force are: The metre, unit of length — symbol m. The kilogram, unit of mass — symbol kg.
