Answer:
The frictional force acting on the block is 14.8 N.
Explanation:
Given that,
Weight of block = 37 N
Coefficients of static = 0.8
Kinetic friction = 0.4
Tension = 24 N
We need to calculate the maximum friction force
Using formula of friction force

Put the value into the formula


So, the tension must exceeds 29.6 N for the block to move
We need to calculate the frictional force acting on the block
Using formula of frictional force

Put the value in to the formula


Hence, The frictional force acting on the block is 14.8 N.
Answer:
Explanation:
Given that,
Initial angular velocity is 0
ωo=0rad/s
It has angular velocity of 11rev/sec
ωi=11rev/sec
1rev=2πrad
Then, wi=11rev/sec ×2πrad
wi=22πrad/sec
And after 30 revolution
θ=30revolution
θ=30×2πrad
θ=60πrad
Final angular velocity is
ωf=18rev/sec
ωf=18×2πrad/sec
ωf=36πrad/sec
a. Angular acceleration(α)
Then, angular acceleration is given as
wf²=wi²+2αθ
(36π)²=(22π)²+2α×60π
(36π)²-(22π)²=120πα
Then, 120πα = 8014.119
α=8014.119/120π
α=21.26 rad/s²
Let. convert to revolution /sec²
α=21.26/2π
α=3.38rev/sec
b. Time Taken to complete 30revolution
θ=60πrad
∆θ= ½(wf+wi)•t
60π=½(36π+22π)t
60π×2=58πt
Then, t=120π/58π
t=2.07seconds
c. Time to reach 11rev/sec
wf=wo+αt
22π=0+21.26t
22π=21.26t
Then, t=22π/21.26
t=3.251seconds
d. Number of revolution to get to 11rev/s
∆θ= ½(wf+wo)•t
∆θ= ½(0+11)•3.251
∆θ= ½(11)•3.251
∆θ= 17.88rev.
For the answer to the question above
Forecasting how a business might do in the future.
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Calculating Revenues.
Producing charts.
--Going past 5--
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Very (VERY) basic CRM for small businesses
I hope my answer helped you.
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