Answer:
Explanation:
Let T be the tension
For linear motion of hoop downwards
mg -T = ma , m is mass of the hoop . a is linear acceleration of CG of hoop .
For rotational motion of hoop
Torque by tension
T x R , R is radius of hoop.
Angular acceleration be α,
Linear acceleration a = α R
So TR = I α
= I a / R
a = TR² / I
Putting this value in earlier relation
mg -T = m TR² / I
mg = T ( 1 + m R² / I )
T = mg / ( 1 + m R² / I )
mg / ( 1 + R² / k² )
Tension is less than mg or weight because denominator of the expression is more than 1.
Part A:
Acceleration can be calculated by dividing the difference of the initial and final velocities by the given time. That is,
a = (Vf - Vi) / t
where a is acceleration,
Vf is final velocity,
Vi is initial velocity, and
t is time
Substituting,
a = (9 m/s - 0 m/s) / 3 s = 3 m/s²
<em>ANSWER: 3 m/s²</em>
Part B:
From Newton's second law of motion, the net force is equal to the product of the mass and acceleration,
F = m x a
where F is force,
m is mass, and
a is acceleration
Substituting,
F = (80 kg) x (3 m/s²) = 240 kg m/s² = 240 N
<em>ANSWER: 240 N </em>
Part C:
The distance that the sprinter travel is calculated through the equation,
d = V₀t + 0.5at²
Substituting,
d = (0 m/s)(3 s) + 0.5(3 m/s²)(3 s)²
d = 13.5 m
<em>ANSWER: d = 13.5 m</em>
<em>Convert 1nanosecond in to its SI init</em>
<em>In SI units, nano is 1000th part of micro which in turn is 1000th part of mini which in turn is 1000th part of main unit. Now, when you affix nano to any unit, here in case, second, it means that you are referring to 1000th part of 1000th part of 1000th part of second or in short, 1000000000th(10^9) part of a second.</em>
<em>In SI units, nano is 1000th part of micro which in turn is 1000th part of mini which in turn is 1000th part of main unit. Now, when you affix nano to any unit, here in case, second, it means that you are referring to 1000th part of 1000th part of 1000th part of second or in short, 1000000000th(10^9) part of a second.So to convert nanosecond into second, just multiply the nanosecond with 0.000000001 or (10^-9)</em>
The question for this problem would be the minimum headphone delay, in ms, that will cancel this noise.
The 200 Hz. period = (1/200) = 0.005 sec. It will need to be delayed by 1/2, so 0.005/2, that is = 0.0025 sec. So converting sec to ms, will give us the delay of:Delay = 2.5 ms.
