Answer:
F=480.491 N
Explanation:
Given that
mass ,m = 22 kg
Angular speed ω = 40 rev/min
ω =4.18 rad/s
The radius r= 1.25 m
We know that centripetal force is given as
F=m ω² r
Now by putting the values in the above equation we get
F=480.491 N
Therefore the centripetal force on the child will be 480.491 N.
Work done is 0.442J
<u>Explanation:</u>
Given:
Spring constant, k = 33 N/m
Distance, x₁ = 0.15m
Additional distance, x₂ = 0.072 m
Total distance = 0.15 + 0.072 m
= 0.222 m
Work done, W = ?
We can calculate work done by the formula
On substituting the value we get:
![W = \frac{1}{2}k [(x_2)^2 - (x_1)^2]\\ \\W = \frac{1}{2} X 33[(0.222)^2 - (0.15)^2]\\ \\W = \frac{1}{2}X 33 [ 0.0493 - 0.0225]\\ \\W = 0.442 J](https://tex.z-dn.net/?f=W%20%3D%20%5Cfrac%7B1%7D%7B2%7Dk%20%5B%28x_2%29%5E2%20%20-%20%28x_1%29%5E2%5D%5C%5C%20%5C%5CW%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20X%2033%5B%280.222%29%5E2%20-%20%280.15%29%5E2%5D%5C%5C%20%5C%5CW%20%3D%20%5Cfrac%7B1%7D%7B2%7DX%2033%20%5B%200.0493%20-%200.0225%5D%5C%5C%20%5C%5CW%20%3D%200.442%20J)
Therefore, work done is 0.442J
Answer:
25kg
Explanation:
The formula for acceleration due to gravity is:
Here, G represents a constant and M represents the mass of the object. We don't need to know the actual values to solve this, then: We can just multiply by the amount they are multiplied relative to earth.
Multiplying this by the weight of the person on Earth of 40kg, you get 0.625*40=25kg.
Hope this helps!
Let's start with the total amount of energy available for the whole scenario:
Some kind of machine gave the coaster a bunch of potential energy by
dragging it up to the top of a 45m hill,and that's the energy is has to work with.
Potential energy = (M) (G) (H) = (800) (9.8) (45) = 352,800 joules
It was then given an extra kick ... enough to give it some kinetic energy, and
start it rolling at 4 m/s.
Kinetic energy = (1/2) (M) (V)² = (1/2) (800) (4)² = 6,400 joules
So the coaster starts out with (352,000 + 6,400) =<em> </em><u><em>359,200 joules</em></u><em> </em>of energy.
There's no friction, so it'll have <u>that same energy</u> at every point of the story.
=================================
Skip the loop for a moment, because the first question concerns the hill after
the loop. We'll come back to it.
The coaster is traveling 10 m/sat the top of the next hill. Its kinetic energy is
(1/2) (M) (V)² = (400) (10)² = 40,000 joules.
Its potential energy at the top of the hill is (359,200 - 40,000) = 319,200.
PE = (M) (G) (H)
319,200 = (800) (9.8) (H)
H = (319,200) / (800 x 9.8) = <em>40.71 meters</em>
=================================
Now back to the loop:
You said that the loop is 22m high at the top. The PE up there is
PE = (M) (G) (H) = (800) (9.8) (22) = 172,480 joules
So the rest is now kinetic. KE = (359,200 - 172,480) = 186,720 joules.
KE = (1/2) (M) (V)² = 186,720
(400) (V)² = 186,720
V² = 186,720 / 400 = 466.8
V = √466.8 = <em>21.61 m/s</em>
===============================
Now it looks like there should be another question ... that's why they
bothered to tell you that the end is 4m off the ground. They must
want you to find the coaster's speed when it gets to the end.
At 4m off the ground, PE = (M) (G) (H) = (800) (9.8) (4) = 31,360 joules.
The rest will be kinetic. KE = (359,200 - 31,360) = 327,840 joules
KE = (1/2) (M) (V)² = 327,840
400 V² = 327,840
V² = 327,840 / 400 = 819.6
V = √819.6 = <em>28.63 m/s</em> at the end
=======================================
If the official answers in class are a little bit different from these,
it'll be because they used some different number for Gravity.
I used '9.8' for gravity, but very often, they use '10' .
If the official answers in class are way way different from these,
then I made one or more big mistakes somewhere. Sorry.
