Answer:
The angular velocity is
Explanation:
From the question we are told that
The mass of the child is ![m_c = 46.2 \ kg](https://tex.z-dn.net/?f=m_c%20%20%3D%20%2046.2%20%5C%20kg)
The radius of the merry go round is ![r = 1.9 \ m](https://tex.z-dn.net/?f=r%20%3D%20%201.9%20%5C%20m)
The moment of inertia of the merry go round is ![I_m = 130.09 \ kg \cdot m^2](https://tex.z-dn.net/?f=I_m%20%3D%20%20130.09%20%5C%20%20kg%20%5Ccdot%20%20m%5E2)
The angular velocity of the merry-go round is ![w = 2.4 \ rad/s](https://tex.z-dn.net/?f=w%20%3D%20%202.4%20%5C%20rad%2Fs)
The position of the child from the center of the merry-go-round is ![x = 0.779 \ m](https://tex.z-dn.net/?f=x%20%3D%200.779%20%5C%20m)
According to the law of angular momentum conservation
The initial angular momentum = final angular momentum
So
![L_i = L_f](https://tex.z-dn.net/?f=L_i%20%20%3D%20%20L_f)
=> ![I_i w_i = I_fw_f](https://tex.z-dn.net/?f=I_i%20w_i%20%20%3D%20%20I_fw_f)
Now
is the initial moment of inertia of the system which is mathematically represented as
![I_i = I_m + I_{b_1}](https://tex.z-dn.net/?f=I_i%20%20%3D%20I_m%20%2B%20I_%7Bb_1%7D)
Where
is the initial moment of inertia of the boy which is mathematically evaluated as
![I_{b_i} = m_c * r](https://tex.z-dn.net/?f=I_%7Bb_i%7D%20%3D%20%20m_c%20%2A%20r)
substituting values
![I_{b_i} = 46.2 * 1.9^2](https://tex.z-dn.net/?f=I_%7Bb_i%7D%20%3D%20%2046.2%20%2A%20%201.9%5E2)
![I_{b_i} = 166.8 \ kg \cdot m^2](https://tex.z-dn.net/?f=I_%7Bb_i%7D%20%3D%20%20166.8%20%5C%20kg%20%5Ccdot%20m%5E2)
Thus
Thus
![I_i * w_i =L_i= 296.9 * 2.4](https://tex.z-dn.net/?f=I_i%20%2A%20w_i%20%20%3DL_i%3D%20%20296.9%20%2A%202.4)
![L_i = 712.5 \ kg \cdot m^2/s](https://tex.z-dn.net/?f=L_i%20%20%3D%20712.5%20%5C%20kg%20%5Ccdot%20m%5E2%2Fs)
Now
![I_f = I_m + I_{b_f }](https://tex.z-dn.net/?f=I_f%20%3D%20%20I_m%20%20%2B%20I_%7Bb_f%20%7D)
Where
is the final moment of inertia of the boy which is mathematically evaluated as
![I_{b_f} = m_c * x](https://tex.z-dn.net/?f=I_%7Bb_f%7D%20%3D%20%20m_c%20%2A%20x)
substituting values
![I_{b_f} = 46.2 * 0.779^2](https://tex.z-dn.net/?f=I_%7Bb_f%7D%20%3D%20%2046.2%20%2A%200.779%5E2)
![I_{b_f} = 28.03 kg \cdot m^2](https://tex.z-dn.net/?f=I_%7Bb_f%7D%20%3D%20%2028.03%20%20kg%20%5Ccdot%20m%5E2)
Thus
![I_f = 130.09 + 28.03](https://tex.z-dn.net/?f=I_f%20%3D%20%20130.09%20%2B%2028.03)
![I_f = 158.12 \ kg \ m^2](https://tex.z-dn.net/?f=I_f%20%3D%20%20158.12%20%5C%20kg%20%5C%20m%5E2)
Thus
![L_f = 158.12 * w_f](https://tex.z-dn.net/?f=L_f%20%20%3D%20%20158.12%20%2A%20w_f)
Hence
![712.5 = 158.12 * w_f](https://tex.z-dn.net/?f=712.5%20%20%3D%20%20158.12%20%2A%20w_f)
![w_f = 4.503 \ rad/s](https://tex.z-dn.net/?f=w_f%20%3D%204.503%20%5C%20%20rad%2Fs)
Answer:
The speed of the baseball is approximately 19.855 m/s
Explanation:
From the question, we have;
The frequency of the microwave beam emitted by the speed gun, f = 2.41 × 10¹⁰ Hz
The change in the frequency of the returning wave, Δf = +3190 Hz higher
The Doppler shift for the microwave frequency emitted by the speed gun which is then reflected back to the gun by the moving baseball is given by 2 shifts as follows;
![\dfrac{\Delta f}{f} = \dfrac{2 \cdot v_{baseball}}{c}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5CDelta%20f%7D%7Bf%7D%20%3D%20%5Cdfrac%7B2%20%5Ccdot%20v_%7Bbaseball%7D%7D%7Bc%7D)
![\therefore{\Delta f}{} = \dfrac{2 \cdot v_{baseball}}{c} \times f](https://tex.z-dn.net/?f=%5Ctherefore%7B%5CDelta%20f%7D%7B%7D%20%3D%20%5Cdfrac%7B2%20%5Ccdot%20v_%7Bbaseball%7D%7D%7Bc%7D%20%5Ctimes%20f)
Where;
Δf = The change in frequency observed, known as the beat frequency = 3190 Hz
= The speed of the baseball
c = The speed of light = 3.0 × 10⁸ m/s
f = The frequency of the microwave beam = 2.41 × 10¹⁰ Hz
By plugging in the values, we have;
![\therefore{\Delta f} = 3190 \ Hz = \dfrac{2 \cdot v_{baseball}}{3.0 \times 10^8 \ m/s} \times 2.41 \times 10^{10} \ Hz](https://tex.z-dn.net/?f=%5Ctherefore%7B%5CDelta%20f%7D%20%3D%203190%20%5C%20Hz%20%3D%20%20%5Cdfrac%7B2%20%5Ccdot%20v_%7Bbaseball%7D%7D%7B3.0%20%5Ctimes%2010%5E8%20%5C%20m%2Fs%7D%20%5Ctimes%202.41%20%5Ctimes%2010%5E%7B10%7D%20%5C%20Hz)
![v_{baseball} = \dfrac{3190 \ Hz \times 3.0 \times 10^8 \ m/s }{2.41 \times 10^{10} \ Hz \times 2} \approx 19.855 \ m/s](https://tex.z-dn.net/?f=v_%7Bbaseball%7D%20%3D%20%5Cdfrac%7B3190%20%5C%20Hz%20%5Ctimes%203.0%20%5Ctimes%2010%5E8%20%5C%20m%2Fs%20%7D%7B2.41%20%5Ctimes%2010%5E%7B10%7D%20%5C%20Hz%20%5Ctimes%202%7D%20%5Capprox%2019.855%20%5C%20m%2Fs)
The speed of the baseball,
≈ 19.855 m/s
Given the time and the horizontal velocity, we can simply
compute for the distance how far the ball travelled using the formula:
distance = velocity * time
<span>Since velocity is in units of m/s and time is seconds,
therefore we can directly get a unit in meters.</span>
Answer:
6250 J
Explanation:
Work is the distance times the component of the force along that distance.
The horizontal component of the force is:
25 N × cos(60°) = 12.5 N
So the work done is:
W = (500 m) (12.5 N)
W = 6250 J
Answer:
down here.
Explanation:
When the two samples of water (hot water and cold water) are put in contact, thermal energy is transferred from the hot water to the cold water. This occurs because heat always flows from a hotter object to a colder object. The heat is transferred by collision between the molecules: the molecules of the hot water have on average more kinetic energy than the molecules of cold water, so when they collide to each other, the molecules of hot water transfer energy to the molecules of cold water. As a result, the kinetic energy of the molecules of cold water increases, and therefore the temperature of the cold water increases, while the temperature of the hot water decreases. This process lasts until the molecules of the two samples have same average kinetic energy: when this occurs, the two samples have.