General internal energy expression:
Uf = (Ui-W) + Q
Where Uf = Final internal energy, Ui = Initial internal energy, W = Work done, Q = Heat added
But,
Uf -Ui = ΔU = 36 J, W = 17 J. Then,
ΔU = -W +Q => Q = ΔU+W = 36+17 = 53 J
Therefore, heat added to the system is 53 Joules
The distance from the centre of the rule at which a 2N weight must be suspend from A is 29.3 cm.
<h3>Distance from the center of the meter rule</h3>
The distance from the centre of the rule at which a 2N weight must be suspend from A is calculated as follows;
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20 A (30 - x)↓ x ↓ 20 cm B 30 cm
2N 0.9N
Let the center of the meter rule = 50 cm
take moment about the center;
2(30 - x) + 0.9(x)(30 - x) = 0.9(20)
(30 - x)(2 + 0.9x) = 18
60 + 27x - 2x - 0.9x² = 18
60 + 25x - 0.9x² = 18
0.9x² - 25x - 42 = 0
x = 29.3 cm
Thus, the distance from the centre of the rule at which a 2N weight must be suspend from A is 29.3 cm.
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Answer:
0.255
Explanation:
The following data were obtained from the question:
Force (F) = 57 N
Mass (m) = 22.8 Kg
Coefficient of static friction (µ) =...?
Next, we shall determine the normal reaction (R). This is illustrated below:
Mass (m) = 22.8 Kg
Acceleration due to gravity (g) = 9.8 m/s²
Normal reaction (R) =?
R = mg
R = 22.8 x 9.8
R = 223.44 N
Finally, we can obtain the coefficient of static friction (µ) as follow:
Force (F) = 57 N
Normal reaction (R) = 223.44 N
Coefficient of static friction (µ) =...?
F = µR
57 = µ x 223.44
Divide both side by 223.44
µ = 57/223.44
µ = 0.255
Therefore, the coefficient of static friction (µ) is 0.255.
Answer:
The net force is 91780.8 N.
Explanation:
mass, m = 8950 kg
Radius, R = 9.33 miles = 15015.2 m
Time, T = 0.123 h = 442.8 s
There are two forces acting on the plane.
Horizontal force is the centripetal force and the vertical force is the weight.
The net force is
