Answer:
Angular acceleration = 5 rad /s ^2
Kinetic energy = 0.391 J
Work done = 0.391 J
P =6.25 W
Explanation:
The torque is given as moment of inertia × angular acceleration
angular acceleration = torque/ moment of inertia
= 10/2= 5 rad/ s^2
The kinetic energy is = 1/2 Iw^2
w = angular acceleration/time
=5/8= 0.625 rad /s
1/2 × 2× 0.625^2
=0.391 J
The work done is equal to the kinetic energy of the motor at this time
W= 0.391 J
The average power is = torque × angular speed
= 10× 0.625
P = 6.25 W
Answer:
v = 3.7 m/s
Explanation:
As the swing starts from rest, if we choose the lowest point of the trajectory to be the zero reference level for gravitational potential energy, and if we neglect air resistance, we can apply energy conservation as follows:
m. g. h = 1/2 m v²
The only unknown (let alone the speed) in the equation , is the height from which the swing is released.
At this point, the ropes make a 30⁰ angle with the vertical, so we can obtain the vertical length at this point as L cos 30⁰, appying simply cos definition.
As the height we are looking for is the difference respect from the vertical length L, we can simply write as follows:
h = L - Lcos 30⁰ = 5m -5m. 0.866 = 4.3 m
Replacing in the energy conservation equation, and solving for v, we get:
v = √2.g.(L-Lcos30⁰) = √2.9.8 m/s². 4.3 m =3.7 m/s
Answer:
Distance of the object from eye is approx 4.52 m
Explanation:
As we know that the object subtend a small angle on both the eyes which is given as
![\theta = 0.995 degree](https://tex.z-dn.net/?f=%5Ctheta%20%3D%200.995%20degree)
now we know that the distance between two eyes is given as
d = 7.50 cm
so we have
![angle = \frac{arc}{Radius}](https://tex.z-dn.net/?f=angle%20%3D%20%5Cfrac%7Barc%7D%7BRadius%7D)
so here the radius is same as the distance from eye while arc is the distance between two eyes
so we have
![0.95 \frac{\pi}{180} = \frac{7.50}{R}](https://tex.z-dn.net/?f=0.95%20%5Cfrac%7B%5Cpi%7D%7B180%7D%20%3D%20%5Cfrac%7B7.50%7D%7BR%7D)
![R = 452 cm](https://tex.z-dn.net/?f=R%20%3D%20452%20cm)
Possible beat frequencies with tuning forks of frequencies 255, 258, and 260 Hz are 2, 3 and 5 Hz respectively.
The beat frequency refers to the rate at which the volume is heard to be oscillating from high to low volume. For example, if two complete cycles of high and low volumes are heard every second, the beat frequency is 2 Hz. The beat frequency is always equal to the difference in frequency of the two notes that interfere to produce the beats. So if two sound waves with frequencies of 256 Hz and 254 Hz are played simultaneously, a beat frequency of 2 Hz will be detected. A common physics demonstration involves producing beats using two tuning forks with very similar frequencies. If a tine on one of two identical tuning forks is wrapped with a rubber band, then that tuning forks frequency will be lowered. If both tuning forks are vibrated together, then they produce sounds with slightly different frequencies. These sounds will interfere to produce detectable beats. The human ear is capable of detecting beats with frequencies of 7 Hz and below.
A piano tuner frequently utilizes the phenomenon of beats to tune a piano string. She will pluck the string and tap a tuning fork at the same time. If the two sound sources - the piano string and the tuning fork - produce detectable beats then their frequencies are not identical. She will then adjust the tension of the piano string and repeat the process until the beats can no longer be heard. As the piano string becomes more in tune with the tuning fork, the beat frequency will be reduced and approach 0 Hz. When beats are no longer heard, the piano string is tuned to the tuning fork; that is, they play the same frequency. The process allows a piano tuner to match the strings' frequency to the frequency of a standardized set of tuning forks.
Learn more about beat frequency here : brainly.com/question/14157895
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