A uniform metal rod with of length 80cm and a mass of 3.2kg is supported horizontally by two vertical spring balances C and D. Balance C is 20cm from one end while D is 30cm from the other end would show the reading of 1.06 Kg and 2.13 kg respectively
<h3>What is gravity?</h3>
It can be defined as the force by which a body attracts another body towards its center as the result of the gravitational pull of one body and another, The gravity varies according to the mass and size of the body for example the force of gravity on the moon is the 1/6th times of the force of gravity on the earth.
As given in the problem, A uniform metal rod of the length of 80cm and mass of 3.2kg is supported horizontally by two vertical springs balance C and D. Balance C is 20cm from one end while D is 30cm from the other end
The weight of the rod acting downward is from the center of the rod at 40 cm
Let us suppose the reading on the spring balance C and D are P and Q respectively
By using the equilibrium for the vertical force
Fv=0
P + C = 3.2
By using the equilibrium for the moment around the left corner
20×P+ 50×Q= 40 ×3.2
By solving for both P and Q from the above two equations we would get
P =1.06 and Q = 2.13
Thus, the reading on the spring balance C and D would be 1.06 Kg and 2.13 kg respectively
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Answer:
Explanation:
number of turns, N = 149
radius, r = 2.15 cm
Area, A = πr² = 3.14 x 2.15 x 2.15 x 10^-4 = 1.45 x 10^-3 m^2
Change in magnetic field, ΔB = 95.5 - 50.5 = 45 mT = 45 x 10^-3 T
time, Δt = 0.165 second
induced emf
e = N x dФ/dt
where, dФ be the change in flux.
e = N x A x ΔB/Δt
e = 149 x 1.45 x 10^-3 x 45 x 10^-3 / 0.165
e = 0.058 V
Answer:
0.958891203 m/s²
Explanation:
N = Weight of crate = 900 N
= Coefficient of friction = 0.25
Force of friction acting on the force applied
Force used to pull the crate
The net force is
Acceleration is given by
The magnitude of the acceleration of the crate is 0.958891203 m/s²
Answer:
when volume and the number of particles are constant
Explanation:
Gay Lussac law states that when the volume of an ideal gas is kept constant, the pressure of the gas is directly proportional to the absolute temperature of the gas.
Mathematically, Gay Lussac's law is given by;
The ideal gas law is the equation PV = nRT
Where;
P is the pressure.
V is the volume.
n is the number of moles of substance.
R is the ideal gas constant.
T is the temperature.
Generally, raising the temperature of an ideal gas would increase its pressure when volume and the number of particles are constant.
This ultimately implies that, when volume and the number of particles are held constant, there would be a linear relationship between the temperature and pressure of a gas i.e temperature would be directly proportional to the pressure of the gas. Thus, an increase in the temperature of the gas would cause an increase in the pressure of the gas at constant volume and number of particles.
