Answer:
Explanation:
The total fluid mass can be obtained by multiplying the mass flow rate by the time flow rate.
Mass flow rate is given as
m = ρAv
Where
m is mass flow rate
ρ is density
A is area and it is given as πr²
v is velocity
Then,
M = mt
Where M is mass and t is time
Them,
M = ρAv × t
M = ρ× πr² × v × t
Given that, .
Radius of pipe is
r = 0.089m
velocity of pipe is
v = 3.3m/s
Time taken is
t = 1 hour = 3600 seconds
Density of water is
ρ = 1000kg/m³
M = ρ× πr² × v × t
M = 1000 × π × 0.089² × 3.3 × 3600
M = 295,628.52 kg
M = 2.96 × 10^5 kg
Explanation:
Distance travelled = Area under the line
= ut + ½ (v-u)t
Acceleration (a) = (v-u)/t and so (v-u) = at
Therefore,
Distance travelled (s) = ut + ½ (v-u)t = ut + ½ (at)t = ut + ½ at²
Thus,proved.
Answer:
car1: a=3.1m/s^2 , car2: a=6.1m/s^2
(a) 1/2*3.1*t^2= 1/2*6.1*(t-0.9)^2
1.55t^2= 3.05(t^2-1.8t+0.81)= 3.05t^2-5.49t+2.4705
1.5t^2-5.49t+2.4705= 0
t= 3.13457 = 3.14[s] after.
(b) d= 1/2*3.1*3.13457^2= 15.23[m] approx.
(c) car1: v=at = 3.1*3.13457= 9.717m/s
car2: v=at = 6.1*(3.13457-0.9)= 13.631m/s
13.631-9.717= 3.914 = 3.91[m/s] faster than car1.
Answer:
u/2 √(1 + 3 cos² θ)
Explanation:
The object is thrown at an angle θ, so the velocity has two components, vertical and horizontal.
Initially, the vertical component is u sin θ and the horizontal component is u cos θ.
At the maximum height, the vertical component is 0 and the horizontal component is u cos θ.
The mean vertical velocity is:
(u sin θ + 0) / 2 = u/2 sin θ
The mean horizontal velocity is:
(u cos θ + u cos θ) / 2 = u cos θ
The net mean velocity can be found with Pythagorean theorem:
v² = (u/2 sin θ)² + (u cos θ)²
v² = u²/4 sin² θ + u² cos² θ
v² = u²/4 (1 − cos² θ) + u² cos² θ
v² = u²/4 (1 − cos² θ) + u²/4 (4 cos² θ)
v² = u²/4 (1 − cos² θ + 4 cos² θ)
v² = u²/4 (1 + 3 cos² θ)
v = u/2 √(1 + 3 cos² θ)
