Define the meaning of the phrase entry requirements to higher education insulations and state four reasons why knowing the entry
1 answer:
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Answer:
Explanation:
Given
mass of First Block
Temperature
mass of second block
Temperature
Heat capacity of aluminium c=899 J/kg-K
Final Temperature acquired by both blocks at steady state
Heat loss first block =Heat gain by second block
Answer:
a) The magnitude JJ of the current density for a copper cable is 5.91 × 10⁵A.m⁻²
b)The mass per unit length \lambdaλ for a copper cable is 0.757kg/m
c)The magnitude J of the current density for an aluminum cable is 3.5 × 10⁵A/m²
d)The mass per unit length \lambdaλ for an aluminum cable is 0.380kg/m
Explanation:
The expression for electric field of conductor is,
The general equation of voltage is,
V = iR
The expression for current density in term of electric field is,
Substitute (V/L) for E in the above equation of current density.
Substitute iR for V in equation (1)
Substitute 1.69 × 10⁸ Ω .m for p
50A for i
0.200Ω.km⁻¹ for (R/L) in eqn (2)
The magnitude JJ of the current density for a copper cable is 5.91 × 10⁵A.m⁻²
b) The expression for resistivity of the conductor is,
The expression for mass density of copper is,
m = dV
where, V is the density of the copper.
Substitute AL for V in equation of the mass density of copper.
m=d(AL)
m/L = dA
λ is use for (m/L)
substitute,
pL/R for A and λ is use for (m/L) in the eqn above
Substitute 0.200Ω.km⁻¹ for (R/L)
8960kgm⁻³ for d and 1.69 × 10⁸ Ω .m
c) Using the equation (2) current density for aluminum cable is,
p is the resistivity of the aluminum cable.
Substitute 2.82 × 10⁻⁸Ω.m for p ,
50A for i and 0.200Ω.km⁻¹ for (R/L)
The magnitude J of the current density for an aluminum cable is 3.5 × 10⁵A/m²
d) Using the equation (3) mass per unit length for aluminum cable is,
p is the resistivity and is the density of the aluminum cable.
Substitute 0.200Ω.km⁻¹ for (R/L), 2700 for d and 2.82 × 10⁻⁸Ω.m for p
The mass per unit length \lambdaλ for an aluminum cable is 0.380kg/m
Answer:
9) a = 25 [m/s^2], t = 4 [s]
10) a = 0.0875 [m/s^2], t = 34.3 [s]
11) t = 32 [s]
Explanation:
To solve this problem we must use kinematics equations. In this way we have:
9)
a)
where:
Vf = final velocity = 0
Vi = initial velocity = 100 [m/s]
a = acceleration [m/s^2]
x = distance = 200 [m]
Note: the final speed is zero, as the car stops completely when it stops. The negative sign of the equation means that the car loses speed or slows down as it stops.
0 = (100)^2 - (2*a*200)
a = 25 [m/s^2]
b)
Now using the following equation:
0 = 100 - (25*t)
t = 4 [s]
10)
a)
To solve this problem we must use kinematics equations. In this way we have:
Note: The positive sign of the equation means that the car increases his speed.
5^2 = 2^2 + 2*a*(125 - 5)
25 - 4 = 2*a* (120)
a = 0.0875 [m/s^2]
b)
Now using the following equation:
5 = 2 + 0.0875*t
3 = 0.0875*t
t = 34.3 [s]
11)
To solve this problem we must use kinematics equations. In this way we have:
10^2 = 2^2 + 2*a*(200 - 10)
100 - 4 = 2*a* (190)
a = 0.25 [m/s^2]
Now using the following equation:
10 = 2 + 0.25*t
8 = 0.25*t
t = 32 [s]