Explanation:
Given that,
Initial speed of the sports car, u = 80 km/h = 22.22 m/s
Final speed of the runner, v = 0
Distance covered by the sports car, d = 80 km = 80000 m
Let a is the acceleration of the sports car. It can be calculated using third equation of motion as :




Value of g, 


Hence, this is required solution.
Answer:
I think the answer is A. X: Mold Y: Cast
Explanation:
Hope that helps!!!
Answer:
hello your question has some missing values attached below is the complete question with the missing values
answer :
a) 0.083 secs
b) 0.33 secs
c) 3e^-4/3
Explanation:
Given that
g = 32 ft/s^2 , spring constant ( k ) = 2 Ib/ft
initial displacement = 1 ft above equilibrium
mass = weight / g = 4/32 = 1/8
damping force = instanteous velocity hence β = 1
a<u>)Calculate the time at which the mass passes through the equilibrium position.</u>
time mass passes through equilibrium = 1/12 seconds = 0.083
<u>b) Calculate the time at which the mass attains its extreme displacement </u>
time when mass attains extreme displacement = 1/3 seconds = 0.33 secs
<u>c) What is the position of the mass at this instant</u>
position = 3e^-4/3
attached below is the detailed solution to the given problem
Answer:
the equilibrium wage rate is 10 and the equilibrium quantity of labor is 1000 workers
Explanation:
The equilibrium wage rate and the equilibrium quantity of labor are found as the point where the equation of demand intercepts the equation of supply, so the equilibrium quantity of labor is:

15 - (1/200) L = 5 + (1/200) L
15 - 5 = (1/200) L + (1/200) L
10 = (2/200) L
(10*200)/2 = L
1000 = L
Then, the equilibrium wage rate is calculated using either the equation of demand for labor or the equation of supply of labor. If we use the equation of demand for labor, we get:
W = 15 - (1/200) L
W = 15 - (1/200) 1000
W = 10
Finally, the equilibrium wage rate is 10 and the equilibrium quantity of labor is 1000 workers