Answer:
The correct answer is option B. coal
Explanation:
Coal is made of remains of organic material including trees and other vegetation which got trapped beneath the earth’s surface or at the bottom of the swamps. After burial below the ground the organic material was acted upon by the high temperature and pressure in the absence of air to form peat. Peat after further processing for a longer period of time converted into coal
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:
1340.2MW
Explanation:
Hi!
To solve this problem follow the steps below!
1 finds the maximum maximum power, using the hydraulic power equation which is the product of the flow rate by height by the specific weight of fluid
W=αhQ
α=specific weight for water =9.81KN/m^3
h=height=220m
Q=flow=690m^3/s
W=(690)(220)(9.81)=1489158Kw=1489.16MW
2. Taking into account that the generator has a 90% efficiency, Find the real power by multiplying the ideal power by the efficiency of the electric generator
Wr=(0.9)(1489.16MW)=1340.2MW
the maximum possible electric power output is 1340.2MW
Answer:
σ = 1.09 mm
Explanation:
Step 1: Identify the given parameters
rod diameter = 20 mm
stiffness constant (k) = 55 MN/m = 55X10⁶N/m
applied force (f) = 60 KN = 60 X 10³N
young modulus (E) = 200 Gpa = 200 X 10⁹pa
Step 2: calculate length of the rod, L
K = \frac{A*E}{L}K=
L
A∗E
L = \frac{A*E}{K}L=
K
A∗E
A=\frac{\pi d^{2}}{4}A=
4
πd
2
d = 20-mm = 0.02 m
A=\frac{\pi (0.02)^{2}}{4}A=
4
π(0.02)
2
A = 0.0003 m²
L = \frac{A*E}{K}L=
K
A∗E
L = \frac{(0.0003142)*(200X10^9)}{55X10^6}L=
55X10
6
(0.0003142)∗(200X10
9
)
L = 1.14 m
Step 3: calculate the displacement of the rod, σ
\sigma = \frac{F*L}{A*E}σ=
A∗E
F∗L
\sigma = \frac{(60X10^3)*(1.14)}{(0.0003142)*(200X10^9)}σ=
(0.0003142)∗(200X10
9
)
(60X10
3
)∗(1.14)
σ = 0.00109 m
σ = 1.09 mm
Therefore, the displacement at the end of A is 1.09 mm
Answer:
5501 kg/m^3
Explanation:
The value of g at the Earth's surface is
where G is the gravitational constant
M is the Earth's mass
is the Earth's radius
Solving the formula for M, we find the value of the Earth's mass:
The Earth's volume is (approximating the Earth to a perfect sphere)
So, the average density of the Earth is
