Can someone give me an objective and subjective statement example please
1 answer:
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Answer:
7.11x10^-3
Explanation:
We are to get the volume rate of flows
1/2pv1² + pl = 1/2pv2²
Such that A1V1 = A2V2
V1 = A2V2/A1
From the attachment I uploaded, we have a formula named equation 1 from which I have plugged in these values
P2 = 33000
P2 = 24000
P = 1000
r2 = 2.25
r1 = 4
When we put these values into the equation,
V2 = 4.47
A2V2 = pi(0.0225)²x4.47
= 7.7x19^-3m³/s
Answer:
a) v² = v₀² + 2 g h, b) t = v₀/g (1+ √ (1 + 2gh/ v₀²))
Explanation:
a) This is an exercise that we can solve using conservation of energy.
Starting point. High point
Em₀ = K + U = ½ m v₀² + m gh
Final point. Soil
= K = ½ m v²
energy is conserved because there is no friction
Emo = Em_{f}
½ m v₀² + m g h = ½ m v²
v² = v₀² + 2 g h
b) the time it takes to reach the ground can be calculated with kinematics
let's create a reference frame with positive upward direction
v = vo - g t
when it reaches the ground it has a velocity v, the initial velocity is downwards v₀ = -v₀
v = -v₀ - gt
t = - (v + v₀) / g
we substitute the velocity values calculated in the previous part
t = - (√(v₀² + 2 g h) + vo) / g
we will simplify the equation a bit
t = - v₀/g (1+ √ (1 + 2gh/ v₀²))
c) is now thrown vertically upward with the same initial velocity vo.
To find the final velocity we use the conservation of energy where the velocity is squared, so it does not matter if it is positive or negative, therefore in this section the value should be the same as in part a
v = √ (v₀² + 2gh)
d) for this part if there is change since the speed is not squared
v₀ = v₀
v = v₀ - gt
t = (v₀ - v) / g
t = (v₀ - √(v₀² + 2 g h)) / g
t = v₀/g (1 - √(1 + 2gh / v₀²))
Answer:
x = 2.044 m
Explanation:
given data
initial vertical component of velocity = Vy = 2sin18
initial horizontal component of velocity = Vx = 2cos18
distance from the ground yo = 5m
ground distance y = 0
from equation of motion
solving for t
t = 1.075 sec
for horizontal motion
x = 2cos18*1.075
x = 2.044 m
