Since this is a distance/time graph, the speed at any time is the slope
of the part of the graph that's directly over that time on the x-axis.
At time t1 = 2.0 s
That's in the middle of the first segment of the graph,
that extends from zero to 3 seconds.
Its slope is 7/3 . v1 = 7/3 m/s .
At time t2 = 4.0 s
That's in the middle of the horizontal part of the graph
that runs from 3 to 6 seconds.
Its slope is zero.
v2 = zero .
At time t3 = 13 s.
That's in the middle of the part of the graph that's sloping down,
between 11 and 16 seconds.
Its slope is -3/5 . v3 = -0.6 m/s .
We solve this using special
relativity. Special relativity actually places the relativistic mass to be the
rest mass factored by a constant "gamma". The gamma is equal to 1/sqrt
(1 - (v/c)^2). <span>
We want a ratio of 3000000 to 1, or 3 million to 1.
</span>
<span>Therefore:
3E6 = 1/sqrt (1 - (v/c)^2)
1 - (v/c)^2 = (0.000000333)^2
0.99999999999999 = (v/c)^2
0.99999999999999 = v/c
<span>v= 99.999999999999% of the speed of light ~ speed of light
<span>v = 3 x 10^8 m/s</span></span></span>
Answer:
solved
Explanation:
a) F_net = (F2 - F3)i - F1 j
b) |Fnet| = sqrt( (F2 - F3)^2 + F1^2)
= sqrt( (9- 5)^2 + 1^2)
= 4.123 N
c) θ = tan^-1( (Fnet_y/Fnet_x)
= tan^-1( -1/(9-5) )
= -14.036°
Answer:
Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.
Explanation:
The corrosion rate is the rate of material remove.The formula for calculating CPR or corrosion penetration rate is
K= constant depends on the system of units used.
W= weight =485 g
D= density =7.9 g/cm³
A = exposed specimen area =100 in² =6.452 cm²
K=534 to give CPR in mpy
K=87.6 to give CPR in mm/yr
mpy
=37.4mpy
mm/yr
=0.952 mm/yr
Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.
Answer:
Explanation:
According to the statement of the problems, the following identity exists:
After some algebraic handling, the ratio is obtained:
