The normal force is the supporting force that is exerted on an object that is in contact with another stable object.
Answer: Option C
<u>Explanation:
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Normal force is forward or upward pushing force acting on an object. Mostly the normal force acts as supporting force exerted on the object by the neighbouring stable object with which the object in question is in contact. So normal force falls under the category of contact forces.
Generally, normal force will be acting to support the weight of any object placed on another object. The best examples of normal forces are the weight of the book supported by table or by the pushing force of the wall on the person leaning on the wall.
Answer:
Explanation:
a)
Ff = μmgcosθ
Ff = 0.28(1600)(9.8)cos(-84)
Ff = 458.9217...
Ff = 460 N
b) ignoring the curves required at top and bottom which change the friction force significantly, especially at the bottom where centripetal acceleration will greatly increase normal forces and thus friction force.
W = Ffd
W = 458.9217(-49.4/sin(-84)
W = 22,795.6119...
W = 23 kJ
c) same assumptions as part b
The change in potential energy minus the work of friction will be kinetic energy.
KE = PE - W
½mv² = mgh - (μmgcosθ)d
v² = 2(gh - (μgcosθ)(h/sinθ))
v = √(2gh(1 - μcotθ))
v = √(2(9.8)(49.4)(1 - 0.28cot84))
v = 30.6552...
v = 31 m/s
Answer:
gravitational potential energy.
Explanation:
Gravitational potential energy (GPE) can be defined as an energy possessed by an object or body due to its position above the earth surface.
Mathematically, gravitational potential energy is given by the formula;
Where,
G.P.E represents gravitational potential energy measured in Joules.
m represents the mass of an object.
g represents acceleration due to gravity measured in meters per seconds square.
h represents the height measured in meters.
This ultimately implies that, anytime there is height, the object must have gravitational potential energy.
Hence, an object possesses gravitational potential energy due to its height (position) and the earth's gravitational force.
Answer:
Inductance as calculated is 13.12 mH
Solution:
As per the question:
Length of the coil, l = 12 cm = 0.12 m
Diameter, d = 1.7 cm = 0.017 m
No. of turns, N = 235
Now,
Area of cross-section of the wire, A =
We know that the inductance of the coil is given by the formula: