Answer:
a-1 Graph is attached. The relation is linear.
a-2 The corresponding height for 68 kPa Pressure is 7.54 m
a-3 The corresponding weight for 68 kPa Pressure is 1394726kg
b The original height of the column is 5.98 m
Explanation:
Part a
a-1
The graph is attached with the solution. The relation is linear as indicated by the line.
a-2
By the equation
Here
- P is the pressure which is given as 68 kPa.
- ρ is the density of the oil whose SG is 0.92. It is calculated as
- g is the gravitational constant whose value is 9.8 m/s^2
- h is the height which is to be calculated
So the height of column is 7.54m
a-3
By the relation of volume and density
Here
- ρ is the density of the oil which is 920 kg/m^3
- V is the volume of cylinder with diameter 16m calculated as follows
Mass is given as
So the mass of oil leading to 68kPa is 1394726kg
Part b
Pressure variation is given as
Now corrected pressure is as
Finding the value of height for this corrected pressure as
The original height of column is 5.98m
Answer:
M = 0.730*m
V = 0.663*v
Explanation:
Data Given:
Conservation of Momentum:
Energy Balance:
Substitute Eq 2 into Eq 1
Using Eq 1
Answer:
the magnitude of the electric force on the projectile is 0.0335N
Explanation:
time of flight t = 2·V·sinθ/g
= (2 * 6.0m/s * sin35º) / 9.8m/s²
= 0.702 s
The body travels for this much time and cover horizontal displacement x from the point of lunch
So, use kinematic equation for horizontal motion
horizontal displacement
x = Vcosθ*t + ½at²
2.9 m = 6.0m/s * cos35º * 0.702s + ½a * (0.702s)²
a = -2.23 m/s²
This is the horizontal acceleration of the object.
Since the object is subject to only electric force in horizontal direction, this acceleration is due to electric force only
Therefore,the magnitude of the electric force on the projectile will be
F = m*|a|
= 0.015kg * 2.23m/s²
= 0.0335 N
Thus, the magnitude of the electric force on the projectile is 0.0335N
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