Answer:
solution given:
acceleration (a)=?
initial velocity (u)=3m/s
final velocity (v)=6m/s
distance (s)=90m
we have
v²=u²+2as
substituting value
6²=3²+2*a*90
36=9+180a
36-9=180a
a=25/180
<u>a=0.1388m/s²</u>
Answer:OB=58.3m
Explanation:
So here cow wanders 30m in north and turns 22 degrees in right side and moves 40m more, as shown in figure given.
now take the starting point as a origin such that cow moves in x-y co-ordinate axis.
As shown in figure length OA is the length when cow moves in north or y direction. Later she takes 22 degrees turn to right and moves 40m more.
So the final displacement is the length of cow from the origin that is length OB.
now co-ordinates of B are [40cos22°,40sin22°+30] i.e [37.084,44.984]
now displacement of cow= length of OB
= ![\sqrt{[37.084]^{2}+[44.984]^{2} }](https://tex.z-dn.net/?f=%5Csqrt%7B%5B37.084%5D%5E%7B2%7D%2B%5B44.984%5D%5E%7B2%7D%20%20%7D)
=
OB =
<span>We can use Coulomb's law to find the force F acting on the proton that is released.
F = k x Q1 x Q2 / r^2
k = 9 x 10^9
Q1 is the charge on one proton which is 1.6 x 10^{-19} C
Q2 is the same charge on the other proton
r is the distance between the protons
F = (9x10^9) x (1.6 x 10^{-19} C) x (1.6 x 10^{-19} C) / (10^{-3})^2
F = 2.304 x 10^{-22} N
We can use the force to find the acceleration.
F = ma
a = F / m
a = (2.304 x 10^{-22} N) / (1.67 x 10^{-27} kg)
a = 1.38 x 10^5 m/s^2
The initial acceleration of the proton is 1.38 x 10^5 m/s^2</span>
Equations of motion (EoM) use EoM <span>v2=u2+2ax</span> to establish velocities at positions shown in blue in drawing from EoM v=u+at for final 1 second of flight time, we can say v=u+g(1) <span><span>2gH−−−−√</span>=<span><span>2g1625H</span>−−−−−−√</span>+g</span><span> then, solve for H [in terms of g]
</span>
I believe your answer would be D- Physical properties of matter effect the chemical properties of matter.
Hope this helps. Tell me if I'm right.
