Answer: Tl = - 13.3°C
the lowest outdoor temperature is - 13.3°C
Explanation:
Given that;
Temperature of Th = 21°C = 21 + 273 = 294 K
the rate at which heat lost is Qh = 5400 kJ/h°C
the power input to heat pump Wnet = 6 kw
The COP of a reversible heat pump depends on the temperature limits in the cycle only, and is determined by;
COPhp = Th/(Th - Tl)
COPhp = Qh/Wnet
Qh/Wnet = Th/(Th -Tl)
the amount of heat loss is expressed as
Qh = 5400/3600(294 - Tl)
the temperature of sink
( 5400/3600(294 - Tl)) / 6 = 294 / ( 294 - Tl)
now solving the equation
Tl = 259.7 - 273
Tl = - 13.3°C
so the lowest outdoor temperature is - 13.3°C
Answer:
Are you smart You seem smart
Explanation:
im ur dad
Answer and Explanation:
The main objective that the torsion test serves is the determination of the material behavior or the behavior of the test sample when subjected to torsional stresses or forces due to the application of moments that results in shear stress along the axis.
Plasticity is the property of elastic material and tension or shear stresses leads to plasticity in a material where these links are the weakest, that gives torsion test a major advantage over the tension test.
Torsion tests are performed on materials to deduct properties like the shear modulus of elasticity, the torsional strength, and the MOR, i.e., Modulus of Rupture.
This test can be used to obtain larger strain values of strain without any complexity as that in tension test.
This test provides a curve of shear-stress-shear strain which is more significant in determining the plasticity as compared to the curve of stress-strain in tension test.
Maximum torque for a given value of maximum stress will be 2 times higher in torsion as that of tension.
In torsion, for plastic flow, the threshold value of shear stress is achieved before the threshold value of normal stress for fracture whereas in tension the critical value of normal stress is achieved sooner than the critical shear for plastic flow.
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,
commutative property of addition
identity property of multiplication
associative property of addition
commutative property of multiplication
The property of realnumbers is shown below is associative property of addition. The correct option is C.
According to the associativeproperty of addition, you can arrange the addends in several ways without changing the result.
According to the commutative property of addition, you can rearrange the addends without altering the result.
When more than two numbers are added together or multiplied, the outcome is always the same, regardless of how the numbers are arranged.
This is known as the associativeproperty. As an illustration, 2 (7 6) = (2 7) 6. 2 + (7 + 6) = (2 + 7) + 6.
Thus, the correct option is C.
For more details regarding associative property, visit:
#SPJ1
Answer:
C
Explanation:
Answer:
At the middle of the DC load
Explanation:
For the Q point of an amplifier to have the Largest linear output. the Q plant has to be biased at the middle of the DC load line, this is because when the input voltage is low the transistor will be in the cutoff region while when the input voltage is very high the transistor will be in the saturation, hence when the Q point is biased at the middle it is will be higher linearly in relation to the active region
☐ E-W
☐ NW-SE
☐ NE-SW
Answer:
☐ NE-SW
Explanation:
Based on the description, the rock direction is North East - South West (NE-SW). Rocks generally can expand or compress depending on the type and magnitude of stress applied on the rocks. However, if the applied stress is sufficiently high, cracks and fractures will be created on the rock and it can ultimately lead to the formation of particles.