Answer:
√(6ax)
Explanation:
Hi!
The question states that during a time t the motorcyle underwent a displacement x at constant acceleration a starting from rest, mathematically we can express it as:
x=(1/2)at^2
Then the we need to find the time t' for which the displacement is 3x
3x=(1/2)a(t')^2
Solving for t':
t'=√(6x/a)
Now, the velocity of the motorcycle as a function of time is:
v(t)=a*t
Evaluating at t=t'
v(t')=a*√(6x/a)=√(6*x*a)
Which is the final velocity
Have a nice day!
Answer:
F = - k (x-xo) a graph of the weight or applied force against the elongation obtaining a line already proves Hooke's law.
Explanation:
The student wants to prove hooke's law which has the form
F = - k (x-xo)
To do this we hang the spring in a vertical position and mark the equilibrium position on a tape measure, to simplify the calculations we can make this point zero by placing our reference system in this position.
Now for a series of known masses let's get them one by one and measure the spring elongation, building a table of weight vs elongation,
we must be careful when hanging the weights so as not to create oscillations in the spring
we look for the mass of each weight
W = mg
m = W / g
and we write them in a new column, we make a graph of the weight or applied force against the elongation and it should give a straight line; the slope of this line is sought, which is the spring constant.
The fact of obtaining a line already proves Hooke's law.
Answer:
He is warmed up now
Explanation:
His muscles are better and stretched now
