Answer:
Explanation:
Given
acceleration of object
(b)For maximum positive displacement velocity must be zero at that instant
i.e.
substitute the value of t
Answer:
Let us consider the case of a bus turning around a corner with a constant velocity, as the bus approaches the corner, the velocity at say point A is Va, and is tangential to the curve with direction pointing away from the curve. Also, the velocity at another point say point B is Vb and is also tangential to the curve with direction pointing away from the curve.<em> </em><em>Although the velocity at point A and the velocity at point B have the same magnitude, their directions are different (velocity is a vector quantity), and hence we have a change in velocity. By definition, an acceleration occurs when we have a change in velocity, so the bus experiences an acceleration at the corner whose direction is away from the center of the corner</em>.
The acceleration is not aligned with the direction of travel because<em> the change in velocity is at a tangent (directed away) to the direction of travel of the bus.</em>
Answer: 11 m/s
vinitial=2 m/s
time=3 s
acceleration = 3 m/s^2
vfinal = ?
The key here is that it is a constant acceleration, so we can use the constant acceleration equations. The easiest one to use would be:
vfinal=vinitial + a*t
We need vfinal, so algebraically we are ready to put in numbers into the equation:
vfinal=vinitial + a*t = 2 m/s + (3 m/s^2)*(3 s ) = 11 m/s is the final velocity
Answer:
Option B
Explanation:
The phase difference is found by subtracting the 2.3m for the receiver from the other speaker which is 2.9m hence
Phase difference= 2.9-2.3= 0.6
Newton's second law ...Force = momentum change/time.momentum change = Forcextme.also, F=ma -> a=F/m - the more familiar form of Newton's second law
using one of the kinematic equations for m ... V=u+at; u=0; a=F/m -> V=(F/m)xt.-> t=mV/F using one of the kinematic equations for 2m ... V=u+at; u=0; a=F/2m -> V=(F/2m)xt. -> t=2mV/F (twice as long, maybe ?)
I think I've made a mistake somewhere below, but I think that the principle is right ...using one of the kinematic equations for m ... s=ut + (1/2)at^2); s=d;u=0;a=F/m; t=1; -> d=(1/2)(F/m)=F/2musing one of the kinematic equations for 2m ... s=ut + (1/2)at^2); s=d;u=0;a=F/2m; t=1; -> d=(1/2)(F/2m)=F/4m (half as far ????? WHAT ???)
