Answer:
D
Explanation:
correct me if I'm wrong on your answer and it's the right answer from my son
Answer:
The coefficient of friction in the hall is 0.038
Explanation:
Given;
mass of the Parker, m = 73.2 kg
applied force on the parker, F = 123 N
frictional force, Fs = 27.4 N
the coefficient of friction in the hall = ?
frictional force is given by;
Fs = μN
Where;
μ is the coefficient of friction
N is normal reaction = mg
Fs = μmg
μ = Fs / mg
μ = (27.4) / (73.2 x 9.8)
μ = 0.038
Therefore, the coefficient of friction in the hall is 0.038
Answer:
By 16.7% or 0.167 IPM
Explanation:
Substracting the final IPM (6.088) to the initial IPM (5.921) gives us the net difference, which is how much did it increase in IPM. Multiplying this number by 100 gives us the percentual increase in the feed rate.
<span>3.78 m
Ignoring resistance, the ball will travel upwards until it's velocity is 0 m/s. So we'll first calculate how many seconds that takes.
7.2 m/s / 9.81 m/s^2 = 0.77945 s
The distance traveled is given by the formula d = 1/2 AT^2, so substitute the known value for A and T, giving
d = 1/2 A T^2
d = 1/2 9.81 m/s^2 (0.77945 s)^2
d = 4.905 m/s^2 0.607542 s^2
d = 2.979995 m
So the volleyball will travel 2.979995 meters straight up from the point upon which it was launched. So we need to add the 0.80 meters initial height.
d = 2.979995 m + 0.8 m = 3.779995 m
Rounding to 2 decimal places gives us 3.78 m</span>
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
