Answer:
F - fr = ma
, N - W = 0
Explanation:
In this exercise we are asked to identify the forces that act on the jack, for this we will use Newton's second law
On the x axis
We have two forces: the friction and the force of the girl who pulls the cat
F - fr = ma
On the y axis
There are two forces: normal and weight
N - W = 0
A diagram of these forces can be seen in the attachment
I’m going to assume initial velocity is 0.
Use Newton’s second law:
F = m•a
F/m = a
14.0/32.5kg= 28/65 m/s^2
Use constant SUVAT acceleration formulae:
S- displacement - what we need to find out
U - initial velocity - 0
V
A - 28/65 m/s^2
T - 10 seconds
S = ut + 1/2at^2
Since u = 0
S = 1/2at^2
1/2• 28/65 • 10^2 = 21.5metres~
Answer is 21.5 metres
~Hoodini, here to help.
For a white dwarf star<span> to become explosively active, </span>the gap<span> between the dwarf </span>and therefore the<span> companion </span>should<span> be </span>sufficiently little<span> that the white dwarf's </span>force field will<span> pull matter </span>aloof from<span> the surface of the companion. </span>because of<span> the rotation of the </span>positional representation system<span>, the matter flowing through the mass-transfer stream from the companion star forms a </span>flat <span>disk, </span>known as associate degree<span> accretion disk, </span>that<span> orbits </span>round the white dwarf star<span>.</span>
Convection currents in the mantle
