Answer:
h = 13.06 m
Explanation:
Given:
- Specific gravity of gasoline S.G = 0.739
- Density of water p_w = 997 kg/m^3
- The atmosphere pressure P_o = 101.325 KPa
- The change in height of the liquid is h m
Find:
How high would the level be in a gasoline barometer at normal atmospheric pressure?
Solution:
- When we consider a barometer setup. We dip the open mouth of an inverted test tube into a pool of fluid. Due to the pressure acting on the free surface of the pool, the fluid starts to rise into the test-tube to a height h.
- The relation with the pressure acting on the free surface and the height to which the fluid travels depends on the density of the fluid and gravitational acceleration as follows:
P = S.G*p_w*g*h
Where, h = P / S.G*p_w*g
- Input the values given:
h = 101.325 KPa / 0.739*9.81*997
h = 13.06 m
- Hence, the gasoline will rise up to the height of 13.06 m under normal atmospheric conditions at sea level.
The government should put it support in a combination of sources, as no
source in the present can fully provide all energy requirements.
Answer:
the speed after 3 seconds is 10 m/s
Explanation:
The computation of the speed is shown below:
As we know that
V = U + at
Here,
U = 34 m/s
a = - 8 m/s²
t = 3 Sec
V = velocity after 3 sec
V = 34 + (-8)3
= 34 - 24
V = 10 m/s
Hence, the speed after 3 seconds is 10 m/s
Answer:
Current = dQ/dt
or I = dQ/dt
Where I represents current.
Which is the rate of flow of charge.
Q=4 + 2t + t²
dQ/dt = 2 + 2t --- This is the relation that gives the instantaneous current.
At time t=2sec
dQ/dt = I = 2 + 2t
= 2 + 2(2)
=2 + 4
= 6A.
Answer:
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