Answer:
Answered
Explanation:
W = ∫ F dx = ∫ P dV = ∫ PA dx = PA* H = P∆V
∆V = W/P
= 1 kJ / 1200 kPa = 0.0 833 m3
Both cases the height is H = ∆V/A
IN case 1 ( sectional area = 0.01 m^2)
in case 2 (sectional area = 0.03 m^2)
Answer:
5.14m/s
Explanation:
First we need to get the Maximum height reached by the rocket;
H = u²sin² theta/2g
H = 25²sin² 40/2(9.8)
H = 625(0.5878)/19.6
H = 18.74m
Get the velocity after 1.5secs
Using the equation of motion
H = ut + 1/2gt²
18.74 = u(1.5)+1/2(9.8)(1.5)²
18.74 = 1.5u + 4.9(2.25)
18.74 = 1.5u + 11.025
1.5u = 18.74 - 11.025
1.5u = 7.715
u = 7.715/1.5
u = 5.14m/s
Hence the velocity after 1.5secs is 5.14m/s
Answer:
<em>Rifting is defined as <u>the splitting apart of a single tectonic plate into two or more tectonic plates separated by divergent plate boundaries</u>. The rifting of a continental tectonic plate creates normal fault valleys, small tilted block mountains, and volcanism. The process is illustrated in Fig. 5.4.</em>
Answer:
U = 25 J
Explanation:
The energy in a set of charges is given by
U =
in this case we have three charges of equal magnitude
q = q₁ = q₂ = q₃
with the configuration of an equilateral triangle all distances are worth
d = a
U = k ( )
we substitute
15 = k q² (3 / a)
k q² /a = 5
For the second configuration a load is moved to the measured point of the other two
d₁₃ = a
The distance to charge 2 that is at the midpoint of the other two is
d₁₂ = d₂₃ = a / 2
U = k (\frac{q_1q_2}{ r_1_2 } + \frac{q_1q_3}{r_1_3} + \frac{q_2q_3}{r_2_3})
U = k q² ( )
U = (kq² /a) 5
substituting
U = 5 5
U = 25 J