A spherical Christmas tree ornament is 8.00 cm in diameter. What is the magnification of an object placed 12.0 cm away from the
1 answer:
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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:
Spring's displacement, x = -0.04 meters.
Explanation:
Let the spring's displacement be x.
Given the following data;
Mass of each shrew, m = 2.0 g to kilograms = 2/1000 = 0.002 kg
Number of shrews, n = 49
Spring constant, k = 24 N/m
We know that acceleration due to gravity, g is equal to 9.8 m/s².
To find the spring's displacement;
At equilibrium position:
Fnet = Felastic + Fg = 0
But, Felastic = -kx
Total mass, Mt = nm
Fg = -Mt = -nmg
-kx -nmg = 0
Rearranging, we have;
kx = -nmg
Making x the subject of formula, we have;
Substituting into the formula, we have;
x = -0.04 m
Therefore, the spring's displacement is -0.04 meters.