Answer: Skilled manpower is essential to carry out several development activities.
Explanation:
Answer:
a = 45 m/s/s
Explanation:
As we know that total mass of the rocket is
total mass of the fuel is given as
all the fuel is burnt in 15 s
so rate of the fuel burning is given as
now the thrust force on the rocket is given as
so we have
so we have
now acceleration is rate of change in velocity
so acceleration at t = 15 s
Answer:
p2 = 9.8×10^4 Pa
Explanation:
Total pressure is constant and PT = P = 1/2×ρ×v^2
So p1 + 1/2×ρ×(v1)^2 = p2 + 1/2×ρ×(v2)^2
from continuity we have ρ×A1×v1 = ρ×A2×v2
v2 = v1×A1/A2
and
r2 = 2×r1
then:
A2 = 4×A1
so,
v2 = (v1)/4
then:
p2 = p1 + 1/2×ρ×(v1)^2 - 1/2×ρ×(v2)^2 = p1 + 1/2×ρ×(v1)^2 - 1/2×ρ×(v1/4)^2
p2 = 3.0×10^4 Pa + 1/2×(1000 kg/m^3)×(12m/s)^2 - 1/2×(1000kg/m^3)×(12^2/16)
= 9.75×10^4 Pa
= 9.8×10^4 Pa
Therefore, the pressure in the wider section is 9.8×10^4 Pa
