2 min = 120 sec
120/15 = 8
The black horse represents 8 seconds.
Answer:
The girl has greater tangential acceleration
Explanation:
The angular acceleration (
) of the merry go round is equal to the rate of the change of the angular velocity,
:

Since all the points of the merry go round complete 1 circle in the same time, the angular velocity of each point of the merry go round is the same, and so all the points also have the same angular acceleration.
The tangential acceleration instead is given by

where
is the angular acceleration
r is the distance from the centre of the merry go round
Since the girl is near the outer edge and the boy is closer to the centre, the value of r for the girl is larger than for the boy, so the girl has greater tangential acceleration.
Answer:
Explanation:
Let T be the tension .
Applying newton's second law on the downward movement of the bucket
mg - T = ma
On the drum , a torque of TR will be acting which will create an angular acceleration of α in it . If I be the moment of inertia of the drum
TR = Iα
TR = Ia/ R
T = Ia/ R²
Replacing this value of T in the other equation
mg - T = ma
mg - Ia/ R² = ma
mg = Ia/ R² +ma
a ( I/ R² +m)= mg
a = mg / ( I/ R² +m)
mg - T = ma
mg - ma = T
mg - m x mg / ( I/ R² +m) = T
mg - m²g / ( I/ R² +m ) = T
mg - mg / ( 1 + I / m R² ) = T
b ) T = Ia/ R²
I = TR² / a
c ) Moment of inertia of hollow cylinder
I = 1/2 M ( R² - R² / 4 )
= 3/4 x 1/2 MR²
= 3/8 MR²
I / R² = 3/8 M
a = mg / ( I/ R² +m)
a = mg / ( 3/8 M + m )
T = Ia/ R²
= 3/8 MR² x mg / ( 3/8 M + m ) x 1 /R²
= 
Answer:
The intensity of the sound in W/m² is 1 x 10⁻⁶ W/m².
Explanation:
Given;
intensity of the sound level, dB = 60 dB
The intensity of the sound in W/m² is calculated as;
![dB = 10 Log[\frac{I}{I_o} ]\\\\](https://tex.z-dn.net/?f=dB%20%3D%2010%20Log%5B%5Cfrac%7BI%7D%7BI_o%7D%20%5D%5C%5C%5C%5C)
where;
I₀ is threshold of hearing = 1 x 10⁻¹² W/m²
I is intensity of the sound in W/m²
Substitute the given values and for I;
![dB = 10 Log[\frac{I}{I_o} ]\\\\60 = 10 Log[\frac{I}{I_o} ]\\\\6 = Log[\frac{I}{I_o} ]\\\\10^6 = \frac{I}{I_o} \\\\I = 10^6 \ \times \ I_o\\\\I = 10^6 \ \times \ 1^{-12} \ W/m^2 \\\\I = 1\ \times \ 10^{-6} \ W/m^2](https://tex.z-dn.net/?f=dB%20%3D%2010%20Log%5B%5Cfrac%7BI%7D%7BI_o%7D%20%5D%5C%5C%5C%5C60%20%3D%2010%20Log%5B%5Cfrac%7BI%7D%7BI_o%7D%20%5D%5C%5C%5C%5C6%20%3D%20%20Log%5B%5Cfrac%7BI%7D%7BI_o%7D%20%5D%5C%5C%5C%5C10%5E6%20%3D%20%5Cfrac%7BI%7D%7BI_o%7D%20%5C%5C%5C%5CI%20%3D%2010%5E6%20%5C%20%5Ctimes%20%5C%20I_o%5C%5C%5C%5CI%20%3D%2010%5E6%20%5C%20%5Ctimes%20%5C%201%5E%7B-12%7D%20%5C%20W%2Fm%5E2%20%5C%5C%5C%5CI%20%3D%201%5C%20%5Ctimes%20%5C%2010%5E%7B-6%7D%20%5C%20W%2Fm%5E2)
Therefore, the intensity of the sound in W/m² is 1 x 10⁻⁶ W/m².