Answer:
When a pilot pushes the top of the right pedal, it activates the brakes on the right main wheel/wheels, and when the pilot pushes the top of the left rudder pedal, it activates the brake on the left main wheel/wheels. The brakes work in a rather simple way: they convert the kinetic energy of motion into heat energy.
Explanation:
When a pilot pushes the top of the right pedal, it activates the brakes on the right main wheel/wheels, and when the pilot pushes the top of the left rudder pedal, it activates the brake on the left main wheel/wheels. The brakes work in a rather simple way: they convert the kinetic energy of motion into heat energy.
Answer:
option B
Explanation:
given,
heating tap water from 16° C to 50° C
at the average rate of 0.2 kg/min
the COP of this heat pump is 2.8
power output = ?
the required power input is 0.169 kW or 0.17 kW
hence, the correct answer is option B
Answer:
Average heat flux=3729.82 W/
Explanation:
Answer:
hello your question has a missing part below is the missing part
Consider the string length equal to
answer : 2cos(2t) sin(2x) - 10cos(10t)sin(10x)
Explanation:
Given string length =
distorted function f(x) = 2sin(2x) - 10sin(10x)
Determine the wave formed in the string
attached below is a detailed solution of the problem
True
False
Answer: true
Explanation:
Answer:
The maximum load (in N) that may be applied to a specimen with this cross sectional area is
Explanation:
We know that the stress at which plastic deformation begins is 267 MPa.
We are going to assume that the stress is homogeneously distributed. As a definition, we know that stress (P) is force (F) over area(A), as follows:
(equation 1)
We need to find the maximum load (or force) that may be applied before plastic deformation. And the problem says that the maximum stress before plastic deformation is 267 MPa. So, we need to find the value of load when P= 267 MPa.
Before apply the equation, we need to convert the units of area in . So,
And then, from the equation 1,
Answer:
Explanation:
The annual worth of land application = $ 110,000 ( A/P , 10% , 5 years ) + $ 95,000 - $ 15,000 (A/F , 10% , 5 years )
The annual worth of land application = $ 110,000 X 0.263797 + $ 95,000 - $ 15,000 X 0.163797
The annual worth of land application = $ 29,017.54 + $ 95,000 - $ 2,456.95
The annual worth of land application = $ 121,560.59
The annual worth of land Incineration = $ 800,000 ( A/P , 10% , 6 years ) + $ 60,000 - $ 250,000 X (A/F , 10% , 6 years )
The annual worth of land Incineration = $ 800,000 X 0.229607 + $ 60,000 - $ 250,000 X 0.129607
The annual worth of land Incineration = $ 211,283.85
The annual worth of contract = $ 190,000
The annual worth of contract = $ 190,000
The land application has the least cost , hence it is preferred .