Answer:
a) 
b) 
c) 
d) 
Explanation:
Average translation kinetic energy (
) is given as
....................(1)
where,
k = Boltzmann's constant ; 1.38 × 10⁻²³ J/K
T = Temperature in kelvin
a) at T = 27.8° C
or
T = 27.8 + 273 = 300.8 K
substituting the value of temperature in the equation (1)
we have
b) at T = 143° C
or
T = 143 + 273 = 416 K
substituting the value of temperature in the equation (1)
we have
c ) The translational kinetic energy per mole of an ideal gas is given as:
here 
= Avagadro's number; ( 6.02×10²³ )
now at T = 27.8° C
d) now at T = 143° C
Answer:
Work done = 13605.44
Explanation:
Data provided in the question:
For elongation of 2.1 cm (0.021 m) work done by the spring is 3.0 J
The relation between Energy (U) and the elongation (s) is given as:
U = 
................(1)
where,
k is the spring constant
on substituting the valeus in the above equation, we get
3.0 = 
or
k = 13605.44 N/m
now
for the elongation x = 2.1 + 4.1 = 6.2 cm = 0.062 m
using the equation 1, we have
U = 
or
U = 26.149 J
Also,
Work done = change in energy
or
W = 26.149 - 3.0 = 23.149 J
Answer:
75 rad/s
Explanation:
The angular acceleration is the time rate of change of angular velocity. It is given by the formula:
α(t) = d/dt[ω(t)]
Hence: ω(t) = ∫a(t) dt
Also, angular velocity is the time rate of change of displacement. It is given by:
ω(t) = d/dt[θ(t)]
θ(t) = ∫w(t) dt
θ(t) = ∫∫α(t) dtdt
Given that: α (t) = (6.0 rad/s4)t² = 6t² rad/s⁴. Hence:
θ(t) = ∫∫α(t) dtdt
θ(t) = ∫∫6t² dtdt =∫[∫6t² dt]dt
θ(t) = ∫[2t³]dt = t⁴/2 rad
θ(t) = t⁴/2 rad
At θ(t) = 10 rev = (10 * 2π) rad = 20π rad, we can find t:
20π = t⁴/2
40π = t⁴
t = ⁴√40π
t = 3.348 s
ω(t) = ∫α(t) dt = ∫6t² dt = 2t³
ω(t) = 2t³
ω(3.348) = 2(3.348)³ = 75 rad/s
Answer:
gas
Explanation:
because It is fast and easy
This is something u are going to have to do
