The position at time t is
x(t) = 0.5t³ - 3t² + 3t + 2
When the velocity is zero, the derivative of x with respect to t is zero. That is,
x' = 1.5t² - 6t + 3 = 0
or
t² - 4t + 2 = 0
Solve with the quadratic formula.
t = (1/2) [ 4 +/- √(16 - 8)] = 3.4142 or 0.5858 s
When t =0.5858 s, the position is
x = 0.5(0.5858³) - 3(0.5858²) + 3(0.5858) + 2 = 2.828 m
When t=3.4142 s, the position is
x = 0.5(3.4142³) - 3(3.4142²) + 3(3.4142) + 2 = -2.828 m
Reject the negative answer.
Answer:
The velocity is zero when t = 0.586 s, and the distance is 2.83 m
When the acceleration is zero, the second derivative of x with respect to t is zero. That is,
3t - 6 = 0
t = 2
The distance traveled is
x = 0.5(2³) - 3(2²) + 3(2) + 2 = 0
Answer:
When the acceleration is zero, t = 2 s, and the distance traveled is zero.
The answer to your question is b) the natural law
Answer:
Explanation:
The theory of propagation of error in case of addition and subtraction states that maximum errors are added in absolute terms in both the operation of addition and subtraction . So in this case the subtracted value will be
10.5775 - 10.3005
= .2770 g
errors will be added ie in subtracted value we can find error to the tune of
.0002 + .0002 = .0004 g
So the subtracted value will be written as
.2770 ± .0004
Refraction. refraction is always accompanied by a wavelength and speed change
Answer:
Pressure = force/area
= 650N/0.12m^2
= 5416.7 to 1 d.p
I got 650N by taking the gravitational field on earth to be 10N/kg.
Weight = mass x gravitational field
= 65 x 10N/kg
= 650
Answer is 5416.7 pascals (Pa)
Or
5308.3 Pa
if gravitational field was taken to be 9.8N/kg
Hope this helps!
