Answer:
Explanation 118 = (1/2) * 0.15 * v² 118 = 0.075 * v² v² = 1573.33 m/s ... since KE = m/2*V^2 , then : V = √2KE/m = √20*118/1.5 = 39.67 m//sec ( 142.8 km/h ; 88.75 mph).:
The correct answer is A. 32.5
Mechanical advantage is the ratio of force that is input into a machine to the force output.
Mechanical advantage of a wheel and axle is calculated by dividing the radius of the wheel by that of the axle.
MA=R/r where R is the radius of the wheel and r is the radius of the axle.
Substituting for the values in the question gives:
MA=26cm/0.8cm
=32.5
A. Move 2 m east and then 12 m east; displacement is 14 m east and the distance is 14 m
B. Move 10 m east and then 12 m west, the displacement is 2 m west and the distance is 22 m.
C. Move 8 m west and then 16 m east; the displacement is 8 m east and the distance is 24 m
D. Move 12 m west and then 8 m east; the displacement is 4 m and the distance is 20 m
Answer:
the period of the motion will increase by√2.
Explanation:
Given that the motion you are talking about is circular motion of a mass attached to the end of a string. I speculated that from the word usage in the question(e.g radius instead of length, tension, period). Given this is so we will have to recall the formula for the centripetral force Fc acting on the object which will be equal in magnitude to the tension in the string and will be given by,
if we want the above defined tension to remain constant when we double the mass and keep the radius of the string constant, the the w(angular frequency) must change which is related to the period by the below equation which will also change,
to find out by how much the period will change we see that from the first equation that if we double the mass making it 2m then the <em>w</em>² will have to decrease by 2 that is it will become <em>w</em>²/2, at the same time keeping r constant since it says that in the question. We now absorb the 2 inside the square and we get,
we can clearly see that the new period has become,
where T is the old period. So the new period is √2 times the old period given by the equation above.