<u>Answer:</u>
Work input = Work output * Work against friction is your answer so C
<u>Explanation:</u>
I hope this helps you :)
Answer:
The new period will be √6 *T
Explanation:
period ,T=2π√(L/g) ................equation 1
where T is the period on earth
gravitational acceleration on the moon is g/6
T1 = 2π√[L/(g/6)]
T1=2π√(6L/g) ...............equation 2
divide equation 2 by 1
T1/T =2π√(6L/g)÷2π√(L/g)
T1/T =√(6L/L)
T1/T =√6
T1 = √6 *T
Answer:
Area = 20 m²
Explanation:
Given the following data;
Force = 2500 N
Pressure = 125 Pa
To find the area on which it rest;
Mathematically, pressure is given by the formula;
Making area the subject of formula, we have;
Substituting into the formula, we have;
Area = 20 m²
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Square root of (4^2 + 4^2) = 4*squareRoot(2)
Answer:
x2 = 64 revolutions.
it rotate through 64 revolutions in the next 5.00 s
Explanation:
Given;
wheel rotates from rest with constant angular acceleration.
Initial angular speed v = 0
Time t = 2.50
Distance x = 8 rev
Applying equation of motion;
x = vt +0.5at^2 ........1
Since v = 0
x = 0.5at^2
making a the subject of formula;
a = x/0.5t^2 = 2x/t^2
a = angular acceleration
t = time taken
x = angular distance
Substituting the values;
a = 2(8)/2.5^2
a = 2.56 rev/s^2
velocity at t = 2.50
v1 = a×t = 2.56×2.50 = 6.4 rev/s
Through the next 5 second;
t2 = 5 seconds
a2 = 2.56 rev/s^2
v2 = 6.4 rev/s
From equation 1;
x = vt +0.5at^2
Substituting the values;
x2 = 6.4(5) + 0.5×2.56×5^2
x2 = 64 revolutions.
it rotate through 64 revolutions in the next 5.00 s
