Acceleration = (0.2 x g) = 1.96m/sec^2.
<span>Accelerating force on 1kg. = (ma) = 1.96N. </span>
<span>1kg. has a weight (normal force) of 9.8N. </span>
<span>Coefficient µ = 1.96/9.8 = 0.2 minimum. </span>
<span>Coefficient is a ratio, so holds true for any value of mass to find accelerating force acting. </span>
<span>e.g. 75kg = (75 x g) = 735N. </span>
<span>Accelerating force = (735 x 0.2) = 147N</span>
I believe this would be an example of positive acceleration as the initial velocity of the rocket is less than the final velocity, indicating that the rocket is accelerating and thus is positive.
Answer:
(a) a = (2i + 4.5j) m/s^2
(b) r = ro + vot + (1/2)at^2
Explanation:
(a) The acceleration of the particle is given by:
vo: initial velocity = (3.00i -2.00j) m/s
v: final velocity = (9.00i + 7.00j) m/s
t = 3s
by replacing the values of the vectors and time you obtain:
![\vec{a}=\frac{1}{3s}[(9.00-3.00)\hat{i}+(7.00-(-2.00))\hat{j}]\\\\\vec{a}=(2\hat{i}+4.5\hat{j})m/s^2](https://tex.z-dn.net/?f=%5Cvec%7Ba%7D%3D%5Cfrac%7B1%7D%7B3s%7D%5B%289.00-3.00%29%5Chat%7Bi%7D%2B%287.00-%28-2.00%29%29%5Chat%7Bj%7D%5D%5C%5C%5C%5C%5Cvec%7Ba%7D%3D%282%5Chat%7Bi%7D%2B4.5%5Chat%7Bj%7D%29m%2Fs%5E2)
(b) The position vector is given by:
where vo = (3.00i -2.00j) m/s and a = (2.00i + 4.50j)m/s^2
Answer
The second rock will land 2.4s after the first rock
Explanation:
Given that
Height of the building s=50m
We assume that the first rock is acting with gravity so that a=9.81m/s
And initial velocity u=0
Applying the equation of motion
S=ut+1/2at²
50=0*t+1/2(9.81)t²
50=4.905t²
t²=50/4.905
t²=10.19
t=√10.19
t=3.19sec
For the second rock initial velocity u=8m/s and v=0 and a=9.81
Applying the equation of motion
v=u+at
0=8+9.81t
t=-8/9.81
t=0.81sec
Hence the second rock will land 2.4s after the first rock
I.e
3.19-0.81
