Answer:
The work done in stretching the spring is 8 lb.ft
Step-by-step explanation:
Given;
Applied force, F = 192 lb
extension of the spring, x = 3 ft
Determine the spring constant from the applied force and extension;
When the spring is stretched 6 inches beyond its natural length, the work done is calculated as follows;
x = 6 inches = 0.5 ft
Therefore, the work done in stretching the spring is 8 lb.ft
Answer: $4 per hour
Step-by-step explanation:
First, you need to isolate w. So you get w = 2/3 divided by 4.
I made the 2/3 a decimal, as it is easier to work with: 0.6666 divided by 4 is 0.167.
0.167 is 1/6.
w = 1/6, or 0.167.
Answer:
n=-4 x=-7
Step-by-step explanation:
Let's make,
mechanic #1's rate = x
mechanic #2's rate = y
* Their rate is dollars per hour ($/hr)
mechanic #1 worked for 20 hours (hr × $/hr = $)
20x = money earned by mech#1
and mechanic #2 worked for 5 hours
5y = money earned by mech#2
together they charged a total of $1150. So the amount of money earned by both mechanics.
20x + 5y = 1150
the sum of the two rates was $95 per hour.
x + y = 95
which means
x = 95 - y
plug (95 - y) in for "x" in the other equation to get everything in terms of one variable.
20(95 - y) + 5y = 1150
solve for y
1900 - 20y + 5y = 1150
1900 - 15y = 1150
-15y = 1150 - 1900
-15y = -750
y = -750/-15
y = 50 $/hr
Now use this to solve for x
x + y = 95
x + 50 = 95
x = 95 - 50
x = 45 $/hr
mech#1 charged 45$/hr
mech #2 charged 50$/hr
