Answer:
The temperature of the gas is 1197.02 K
Explanation:
From ideal gas law;
PV = nRT
Where;
P is the pressure of the gas
V is the volume of the gas
R is ideal gas constant = 8.314 L.kPa/mol.K
T is the temperature of the gas
n is the number of moles of gas
Volume of the gas in the cylindrical container = πr²h
Given;
r = 6/2 = 3 cm = 0.03 m
h = 11 cm = 0.11 m
V = π × (0.03)² × 0.11 = 3.11 × 10⁻⁴ m³ = 0.311 L
number of moles of oxygen gas = Reacting mass / molar mass


Therefore, the temperature of the gas is 1197.02 K
Answer:
(a) coefficient of friction = 0.451
This was calculated by the application of energy conservation principle (the total sum of energy in a closed system is conserved)
(b) No, it comes to a stop 5.35m short of point B. This is so because the spring on expanding only does a work of 43 J on the block which is not enough to meet up the workdone of 398 J against friction.
Explanation:
The detailed step by step solution to this problems can be found in the attachment below. The solution for part (a) was divided into two: the motion of the body from point A to point B and from point B to point C. The total energy in the system is gotten from the initial gravitational potential energy. This energy becomes transformed into the work done against friction and the work done in compression the spring. A work of 398J was done in overcoming friction over a distance of 6.00m. The energy used in doing so is lost as friction is not a conservative force. This leaves only 43J of energy which compresses the spring. On expansion the spring does a work of 43J back on the block is only enough to push it over a distance of 0.65m stopping short of 5.35m from point B.
Thank you for reading and I hope this is helpful to you.
Answer:
4.9 cm
Explanation:
From Hook's Law,
F = ke......................... Equation 1
Where F= force, e = extension, k = spring constant.
Note: the Force acting on the the spring is the weight of the mass.
W = mg.
F = mg.................... Equation 2
Where m = mass, g = acceleration due to gravity
Substitute equation 2 into equation 1
mg = ke
make e the subject of the equation
e = mg/k............... Equation 3.
Given: m = 2 kg, g = 9.8 m/s², k = 400 N/m
e = (2×9.8)/400
e = 19.6/400
e = 0.049 m
e = 4.9 cm
Answer:
<h2>9.375Nm</h2>
Explanation:
The formula for calculating torque τ = Frsin∅ where;
F = applied force (in newton)
r = radius (in metres)
∅ = angle that the force made with the bar.
Given F= 25N, r = 0.75m and ∅ = 30°
torque on the bar τ = 25*0.75*sin30°
τ = 25*0.75*0.5
τ = 9.375Nm
The torque on the bar is 9.375Nm
Answer:2.53*10^-10F
Explanation:
C=£o£r*A/d
Where £ is the permitivity of a constant
£o= 8.85*10^-12f/m
£r=6.3
A=150mm^2=0.015m^2
d=3.3mm= 0.0033m
C=8.85*10^-12*6.3*0.015/0.0033
C=8.85*6.3*10^-12*0.015/0.0033
C=55.755*0.015^-12/0.003
C=8.36/3.3*10^-13+3
C=2.53*10^-10F