oscillating spring mass systems can be used to experimentally determine an unknown mass without using a mass balance. a student
1 answer:
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Answer:
Explanation:
Given that, .
Mass of ladder is 51kg
Then, it weight is
WL = mg = 51 × 9.81 = 500.31N
This weight will act at the midpoint of the ladder
Length of ladder is 15m
The ladder makes an angle 55°C with the horizontal
An object whose mass is 81kg is at 4m from the bottom of the ladder
Then, weight of object
Wo = mg = 81 × 9.81 = 794.61 N
Using newton second law
Check attachment
Ng is normal force on the ground
Ff is the horizontal frictional force
Nw Is the normal force on the wall
ΣFy = 0
Ng = Wo + WL
Ng = 794.61 + 500.31
Ng = 1294.92 N
Also
ΣFx = 0
Ff — Nw = 0
Then,
Ff = Nw
Now taking moment about point A.
Check attachment
using the principle of equilibrium
Sum of clockwise moment equals to sum of anti-clockwise moment
Also note that the Normal force on the wall is not perpendicular to the ladder, so we will resolve that and also the weights of ladder and weight of object
Clockwise = Anticlockwise
Wo•Cos60 × 4 + WL•Cos60 × 7.5 = Nw•Sin60 × 15
794.61Cos60 × 4 + 500.31Cos60 × 7.5 = Nw × Sin60 × 15
1589.22 + 1876.163 = 12.99•Nw
3465.383 = 12.99•Nw
Nw = 3465.383 / 12.99
Nw = 266.77 N
Since, Nw = Ff
Then, Ff = 266.77N
the horizontal force exerted by the ground on the ladder is 266.77 N
Answer:
Negative
Explanation:
First law of thermodynamic also known as the law of conservation of energy states that the total energy of an isolated system is constant; energy can be transformed from one form to another, but can be neither created nor destroyed.
The first law relates relates changes in internal energy to heat added to a system and the work done by a system by the conservation of energy.
The first law is mathematically given as ΔU = -
= Q - W
Where Q = Quantity of heat
W = Work done
From the first law The internal energy has the symbol U. Q is positive if heat is added to the system, and negative if heat is removed; W is positive if work is done by the system, and negative if work is done on the system.
Analyzing the pistol when it raises in isothermal and when it falls in isobaric state.The following can be said:
In the Isothermal compression of a gas there is work done on the system to decrease the volume and increase the pressure. For work to be done on the system it is a negative work done then.
In the Isobaric State An isobaric process occurs at constant pressure. Since the pressure is constant, the force exerted is constant and the work done is given as PΔV.If a gas is to expand at a constant pressure, heat should be transferred into the system at a certain rate.Isobaric is a fuction of heat which is Isothermal Provided the pressure is kept constant.
In Isobaric definition above it can be seen that " Heat should be transferred into the system ata certain rate. For heat to be transferred into the system work is deinitely been done on the system thereby favouring the negative work done.