<u>Answer:</u>
Option A is the correct answer.
<u>Explanation:</u>
Let the east point towards positive X-axis and north point towards positive Y-axis.
First walking 1.2 km north, displacement = 1.2 j km
Secondly 1.6 km east, displacement = 1.6 i km
Total displacement = (1.6 i + 1.2 j) km
Magnitude = 
Angle of resultant with positive X - axis = 
= 36.87⁰ east of north.
Answer:
When we double the angular velocity the maximum acceleration 
will changes by a factor of 4.
Explanation:
Given the angular frequency 
of the simple harmonic oscillator is doubled.
We need to find the change in the maximum acceleration of the oscillator.
Now, according to the problem, the angular frequency got doubled.
Let us plug 
. Then the maximum acceleration will be 
We can see, when we double the angular velocity the maximum acceleration will changes by a factor of 4.
It's a bit of a trick question, had the same one on my homework. You're given an electric field strength (1*10^5 N/C for mine), a drag force (7.25*10^-11 N) and the critical info is that it's moving with constant velocity(the particle is in equilibrium/not accelerating).
<span>All you need is F=(K*Q1*Q2)/r^2 </span>
<span>Just set F=the drag force and the electric field strength is (K*Q2)/r^2, plugging those values in gives you </span>
<span>(7.25*10^-11 N) = (1*10^5 N/C)*Q1 ---> Q1 = 7.25*10^-16 C </span>
Answer:
conservation of momentum, general law of physics according to which the quantity called momentum that characterizes motion never changes in an isolated collection of objects; that is, the to
Displacement s = (u+v)*t/2 (t refers to delta time)
= (0.45 + 2.7)*6/2
= 3.15*3
= 9.45 m
