Answer:
9. A 1500kg car traveling +6m/s with a 2000kg truck at rest. The vehicles collide, but do not stick together. The car has a velocity -3m/s after the collision. What is the velocity of the truck? a. What type of collision occurred above?
Answer:
Explanation:
90 rpm = 90 / 60 rps
= 1.5 rps
= 1.5 x 2π rad /s
angular velocity of flywheel
ω = 3π rad /s
Let I be the moment of inertia of flywheel
kinetic energy = (1/2) I ω²
(1/2) I ω² = 10⁷ J
I = 2 x 10⁷ / ω²
=2 x 10⁷ / (3π)²
= 2.2538 x 10⁵ kg m²
Let radius of wheel be R
I = 1/2 M R² , M is mass of flywheel
= 1/2 πR² x t x d x R² , t is thickness , d is density of wheel .
1/2 πR⁴ x t x d = 2.2538 x 10⁵
R⁴ = 2 x 2.2538 x 10⁵ / πt d
= 4.5076 x 10⁵ / 3.14 x .1 x 7800
= 184
R= 3.683 m .
diameter = 7.366 m .
b ) centripetal accn required
= ω² R
= 9π² x 3.683
= 326.816 m /s²
Answer:
B
Explanation:
this is because the neutrons do not have a charge, the things that have charge in an atom are electrons and protons.
and in the nucleus of an atom, there are protons and neutrons so you can see that A is not the answer
if you see the periodic table, you will know that the number of electrons and protons are equal, so the charges cancel each other out, hence the charge of an atom will be neutral
let me give you a tip which I got from my teacher, never write there is no charge in the atom, this suggests that there is no protons or electrons.
instead, write, the it is neutral
hope it helps if not please report it so that someone else gets to try it out
Answer:
a) KE = 888.26J
b) N = 294.5 turns
Explanation:
For the kinetic energy:

The inertia is:

So, the kinetic energy will be:

Now, friction force is:
Ff = μ*N = 0.80*5N = 4N
The energy balance would be:
Kf - Ko = Wf where Kf=0; Ko = 888.26J; and Wf is the work done by friction force.
Wf = -Ff*d = -Ff*N*2*π*R where N is the amount of turns it gives.
Replacing these values into the energy balance:
0-888.26=-4*N*2*π*0.12
-888.26=-0.96*π*N
N=294.5 turns
Split the operation in two parts. Part A) constant acceleration 58.8m/s^2, Part B) free fall.
Part A)
Height reached, y = a*[t^2] / 2 = 58.8 m/s^2 * [7.00 s]^2 / 2 = 1440.6 m
Now you need the final speed to use it as initial speed of the next part.
Vf = Vo + at = 0 + 58.8m/s^2 * 7.00 s = 411.6 m/s
Part B) Free fall
Maximum height, y max ==> Vf = 0
Vf = Vo - gt ==> t = [Vo - Vf]/g = 411.6 m/s / 9.8 m/s^2 = 42 s
ymax = yo + Vo*t - g[t^2] / 2
ymax = 1440.6 m + 411.6m/s * 42 s - 9.8m/s^2 * [42s]^2 /2
ymax = 1440.6 m + 17287.2m - 8643.6m = 10084.2 m
Answer: ymax = 10084.2m