Let A→=(150iˆ+270jˆ) mm , B→=(300iˆ−450jˆ) mm , and C→=(−100iˆ−250jˆ) mm . Find scalars r and s, if possible, such that R→=rA→+s
1 answer:
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Answer:
# kelvin_to_celsius function is defined
# it has value_kelvin as argument
def kelvin_to_celsius(value_kelvin):
# value_celsius is initialized to 0.0
value_celsius = 0.0
# value_celsius is calculated by
# subtracting 273.15 from value_kelvin
value_celsius = value_kelvin - 273.15
# value_celsius is returned
return value_celsius
# celsius_to_kelvin function is defined
# it has value_celsius as argument
def celsius_to_kelvin(value_celsius):
# value_kelvin is initialized to 0.0
value_kelvin = 0.0
# value_kelvin is calculated by
# adding 273.15 to value_celsius
value_kelvin = value_celsius + 273.15
# value_kelvin is returned
return value_kelvin
value_c = 0.0
value_k = 0.0
value_c = 10.0
# value_c = 10.0 is used to test the function celsius_to_kelvin
# the result is displayed
print(value_c, 'C is', celsius_to_kelvin(value_c), 'K')
value_k = 283.15
# value_k = 283.15 is used to test the function kelvin_to_celsius
# the result is displayed
print(value_k, 'is', kelvin_to_celsius(value_k), 'C')
Explanation:
Image of celsius_to_kelvin function used as guideline is attached
Image of program output is attached.
Answer:
no of unit is 17941
Explanation:
given data
fixed cost = $338,000
variable cost = $143 per unit
fixed cost = $1,244,000
variable cost = $92.50 per unit
solution
we consider here no of unit is = n
so here total cost of labor will be sum of fix and variable cost i.e
total cost of labor = $33800 + $143 n ..........1
and
total cost of capital intensive = $1,244,000 + $92.5 n ..........2
so here in both we prefer cost of capital if cost of capital intensive less than cost of labor
$1,244,000 + $92.5 n < $33800 + $143 n
solve we get
n >
n > 17941
and
cost of producing less than selling cost so here
$1,244,000 + $92.5 n < 197 n
solve it we get
n >
n > 11904
so in both we get greatest no is 17941
so no of unit is 17941
Answer:
Tso = 28.15°C
Explanation:
given data
t2 = 21 mm
ki = 0.026 W/m K
t1 = 9 mm
kp = 180 W/m K
length of the roof is L = 13 m
net solar radiation into the roof = 107 W/m²
temperature of the inner surface Ts,i = -4°C
air temperature is T[infinity] = 29°C
convective heat transfer coefficient h = 47 W/m² K
solution
As when energy on the outer surface at roof of a refrigerated truck that is balance as
Q = .....................1
Q = .....................2
now we compare both equation 1 and 2 and put here value
solve it and we get
Tso = 28.153113
so Tso = 28.15°C