Answer:
Amount of concrete need to make slab = 1,500 feet³
Explanation:
Given:
Length of slab = 50 feet
Width of slab = 30 feet
Height of slab = 1 feet
Find:
Amount of concrete need to make slab
Computation;
Amount of concrete need to make slab = Volume of cuboid
Volume of cuboid = (l)(b)(h)
Amount of concrete need to make slab = (50)(30)(1)
Amount of concrete need to make slab = 1,500 feet³
Answer:
The costs to run the dryer for one year are $ 9.03.
Explanation:
Given that the clothes dryer in my home has a power rating of 2250 Watts, and to dry one typical load of clothes the dryer will run for approximately 45 minutes, and in Ontario, the cost of electricity is $ 0.11 / kWh, to calculate the costs to run the dryer for one year the following calculation must be performed:
1 watt = 0.001 kilowatt
2250/45 = 50 watts per minute
45 x 365 = 16,425 / 60 = 273.75 hours of consumption
50 x 60 = 300 watt = 0.3 kw / h
0.3 x 273.75 = 82.125
82.125 x 0.11 = 9.03
Therefore, the costs to run the dryer for one year are $ 9.03.
I think the answer is B. 10D
Answer:
a) m=336.18N
b) Vn=16.67m/kmol
Vm=0.1459m^3/kg
Explanation:
To calculate the mass of the octane(m):
Number of mole of octane (n) =0.3kmol(given)
Molarmass of octane (M) =114.23kg/kmol
m=n*M
m=(0.3kmol)*(114.23kg/kmol)
m=34.269kg
To calculate for the weight of octane(W):
W=g*m
W=(9.81m/s^2)*(34.269kg)
W=336.18N
b) For specific volumes of Vn and Vm:
Given volume of octane (V) =5m^3
Vm=V/m
Vm=5m^3/34.269kg
Vm=0.1459m^3/kg
And Vn will be :
Vn=V/m=5m^3/0.3kmol
Vn=16.67m/Kmol
Therefore, the answers are:
a) m=336.18N
b) Vn=16.67m/kmol
Vm=0.1459m^3/kg
Answer:
# Initialize a dictionary with the keys
contestants = {"Darci Lynne":0, "Angelica Hale":0, "Angelina Green":0};
# Repeatedly prompt the user for a contestant name to vote for
while True:
# Prompting user for contestant name
cName = input("Enter contestant name to vote: ");
# Checking for Done
if cName.lower() == "done":
break;
# Checking in dictionary
if cName in contestants.keys():
# Updating vote value
contestants[cName] += 1
# New entry
else:
contestants[cName] = 1
# Printing header
print("\n%-20s %-15s\n" %("Contestant Name", "Votes Casted"))
# Printing results
for contestant in contestants:
print("%-23s %-15d" %(contestant, contestants[contestant]))
