A. a bicycle chain is an example of a compound machine
Answer: car B has travelled 4times as far as Car A
d=vi*t+1/2at^2
No initial velocity so equation becomes;
d=1/2at^2 and the acceleration is the same between both only time is different;
Car A d=1/2a(1)^2
Car B d=1/2a(2)^2
Car A d= 1^2=1
Car B d= 2^2=4
Car B d=4*Car A
So car B has travelled 4 times as far as car A
F = ma so u can plug in the given numbers and solve:
F = (2)(3)
Answer:
<h2>4.1 N</h2>
Explanation:
The force acting on an object given it's mass and acceleration can be found by using the formula
force = mass × acceleration
From the question we have
force = 0.205 × 20
We have the final answer as
<h3>4.1 N</h3>
Hope this helps you
In order to answer these questions, we need to know the charges on
the electron and proton, and then we need to know the electron's mass.
I'm beginning to get the creepy feeling that, in return for the generous
5 points, you also want me to go and look these up so I can use them
in calculations ... go and collect my own straw to make the bricks with,
as it were.
Ok, Rameses:
Elementary charge . . . . . 1.6 x 10⁻¹⁹ coulomb
negative on the electron
plussitive on the proton
Electron rest-mass . . . . . 9.11 x 10⁻³¹ kg
a). The force between two charges is
F = (9 x 10⁹) Q₁ Q₂ / R²
= (9 x 10⁹ m/farad) (-1.6 x 10⁻¹⁹C) (1.6 x 10⁻¹⁹C) / (5.35 x 10⁻¹¹m)²
= ( -2.304 x 10⁻²⁸) / (5.35 x 10⁻¹¹)²
= 8.05 x 10⁻⁸ Newton .
b). Centripetal acceleration =
v² / r .
A = (2.03 x 10⁶)² / (5.35 x 10⁻¹¹)
= 7.7 x 10²² m/s² .
That's an enormous acceleration ... about 7.85 x 10²¹ G's !
More than enough to cause the poor electron to lose its lunch.
It would be so easy to check this work of mine ...
First I calculated the force, then I calculated the centripetal acceleration.
I didn't use either answer to find the other one, and I didn't use " F = MA "
either.
I could just take the ' F ' that I found, and the 'A' that I found, and the
electron mass that I looked up, and mash the numbers together to see
whether F = M A .
I'm going to leave that step for you. Good luck !
