Answer:
The amount of work that must be done to compress the gas 11 times less than its initial pressure is 909.091 J
Explanation:
The given variables are
Work done = 550 J
Volume change = V₂ - V₁ = -0.5V₁
Thus the product of pressure and volume change = work done by gas, thus
P × -0.5V₁ = 500 J
Hence -PV₁ = 1000 J
also P₁/V₁ = P₂/V₂ but V₂ = 0.5V₁ Therefore P₁/V₁ = P₂/0.5V₁ or P₁ = 2P₂
Also to compress the gas by a factor of 11 we have
P (V₂ - V₁) = P×(V₁/11 -V₁) = P(11V₁ - V₁)/11 = P×-10V₁/11 = -PV₁×10/11 = 1000 J ×10/11 = 909.091 J of work
Answer:
The value of the linear coefficient of thermal expansion is : α=1.01 *10⁻⁵ (ºC)⁻¹
Explanation:
Li = 0.2m
ΔL = 0.2 mm = 0.0002m
T1 = 21ºC
T2 = 120ºC
ΔT =99ºC
α =ΔL/(Li*ΔT)
α =0.0002m /(0.2m * 99ºC)
α = 1.01 *10⁻⁵ (ºC)⁻¹
Answer:
x = 11.23 m
Explanation:
For this interesting exercise, we must use angular kinematics, linear kinematics and the relationship between angular and linear quantities.
Let's reduce to SI system units
θ = 155 rev (2pi rad / rev) = 310π rad
α = 2.00rev / s2 (2pi rad / 1 rev) = 4π rad / s²
Let's look for the angular velocity at the time the piece is released, with starting from rest the initial angular velocity is zero (wo = 0)
w² = w₀² + 2 α θ
w =√ 2 α θ
w = √(2 4pi 310pi)
w = 156.45 rad / s
The relationship between angular and linear velocity
v = w r
v = 156.45 0.175
v = 27.38 m / s
In this part we have the linear speed and the height that it travels to reach the floor, so with the projectile launch equations we can find the time it takes to arrive
y = 
t - ½ g t²
As it leaves the highest point its speed is horizontal
y = 0 - ½ g t²
t = √ (-2y / g)
t = √ (-2 (-0.820) /9.8)
t = 0.41 s
With this time we calculate the horizontal distance, because the constant horizontal speed
x = vox t
x = 27.38 0.41
x = 11.23 m
Answer:
C
Explanation:
To solve this question, we will need to develop an expression that relates the diameter 'd', at temperature T equals the original diameter d₀ (at 0 degrees) plus the change in diameter from the temperature increase ( ΔT = T):
d = d₀ + d₀αT
for the sphere, we were given
D₀ = 4.000 cm
α = 1.1 x 10⁻⁵/degrees celsius
we have D = 4 + (4x(1.1 x 10⁻⁵)T = 4 + (4.4x10⁻⁵)T EQN 1
Similarly for the Aluminium ring we have
we were given
d₀ = 3.994 cm
α = 2.4 x 10⁻⁵/degrees celsius
we have d = 3.994 + (3.994x(2.4 x 10⁻⁵)T = 3.994 + (9.58x10⁻⁵)T EQN 2
Since @ the temperature T at which the sphere fall through the ring, d=D
Eqn 1 = Eqn 2
4 + (4.4x10⁻⁵)T =3.994 + (9.58x10⁻⁵)T, collect like terms
0.006=5.18x10⁻⁵T
T=115.7K
