The pressure will 14. 0 g of co exert in a 3. 5 l container at 75°c is 4.1atm.
Therefore, option A is correct option.
Given,
Mass m = 14g
Volume= 3.5L
Temperature T= 75+273 = 348 K
Molar mass of CO = 28g/mol
Universal gas constant R= 0.082057L
Number of moles in 14 g of CO is
n= mass/ molar mass
= 14/28
= 0.5 mol
As we know that
PV= nRT
P × 3.5 = 0.5 × 0.082057 × 348
P × 3.5 = 14.277
P = 14.277/3.5
P = 4.0794 atm
P = 4.1 atm.
Thus we concluded that the pressure will 14. 0 g of co exert in a 3. 5 l container at 75°c is 4.1atm.
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Answer:
The beam used is a negatively charged electron beam with a velocity of
v = E / B
Explanation:
After reading this long statement we can extract the data to work on the problem.
* They indicate that when the beam passes through the plates it deviates towards the positive plate, so the beam must be negative electrons.
* Now indicates that the electric field and the magnetic field are contracted and that the beam passes without deviating, so the electric and magnetic forces must be balanced
q E = qv B
v = E / B
this configuration is called speed selector
They ask us what type of beam was used.
The beam used is a negatively charged electron beam with a velocity of v = E / B
Answer:
0.191 s
Explanation:
The distance from the center of the cube to the upper corner is r = d/√2.
When the cube is rotated an angle θ, the spring is stretched a distance of r sin θ. The new vertical distance from the center to the corner is r cos θ.
Sum of the torques:
∑τ = Iα
Fr cos θ = Iα
(k r sin θ) r cos θ = Iα
kr² sin θ cos θ = Iα
k (d²/2) sin θ cos θ = Iα
For a cube rotating about its center, I = ⅙ md².
k (d²/2) sin θ cos θ = ⅙ md² α
3k sin θ cos θ = mα
3/2 k sin(2θ) = mα
For small values of θ, sin θ ≈ θ.
3/2 k (2θ) = mα
α = (3k/m) θ
d²θ/dt² = (3k/m) θ
For this differential equation, the coefficient is the square of the angular frequency, ω².
ω² = 3k/m
ω = √(3k/m)
The period is:
T = 2π / ω
T = 2π √(m/(3k))
Given m = 2.50 kg and k = 900 N/m:
T = 2π √(2.50 kg / (3 × 900 N/m))
T = 0.191 s
The period is 0.191 seconds.
Answer:
Explanation:
As we know that there is no external torque on the system of two disc
then the angular momentum of the system will remains conserved
So we will have
now we have
also we have
now from above equation we have
now we have
