The answer is two (4 -3)
Zero is not a SF
So, the angular frequency of the blades approximately <u>36.43π rad/s</u>.
<h3>Introduction</h3>
Hi ! Here I will discuss about the angular frequency or what is also often called the angular velocity because it has the same unit dimensions. <u>Angular frequency occurs, when an object vibrates (either moving harmoniously / oscillating or moving in a circle)</u>. Angular frequency can be roughly interpreted as the magnitude of the change in angle (in units of rad) per unit time. So, based on this understanding, the angular frequency can be calculated using the equation :
![\boxed{\sf{\bold{\omega = \frac{\theta}{t}}}}](https://tex.z-dn.net/?f=%20%5Cboxed%7B%5Csf%7B%5Cbold%7B%5Comega%20%3D%20%5Cfrac%7B%5Ctheta%7D%7Bt%7D%7D%7D%7D%20)
With the following condition :
= angular frequency (rad/s)
= change of angle value (rad)- t = interval of the time (s)
<h3>Problem Solving</h3>
We know that :
= change of angle value = 1,000 revolution = 1,000 × 2π rad = 2,000π rad/s >> Remember 1 rev = 2π rad/s.- t = interval of the time = 54.9 s.
What was asked :
= angular frequency = ... rad/s
Step by step :
![\sf{\omega = \frac{\theta}{t}}](https://tex.z-dn.net/?f=%20%5Csf%7B%5Comega%20%3D%20%5Cfrac%7B%5Ctheta%7D%7Bt%7D%7D%20)
![\sf{\omega = \frac{2,000 \pi}{54.9}}](https://tex.z-dn.net/?f=%20%5Csf%7B%5Comega%20%3D%20%5Cfrac%7B2%2C000%20%5Cpi%7D%7B54.9%7D%7D%20)
![\boxed{\sf{\omega \approx 36.43 \pi \: rad/s}}](https://tex.z-dn.net/?f=%20%5Cboxed%7B%5Csf%7B%5Comega%20%5Capprox%2036.43%20%5Cpi%20%5C%3A%20rad%2Fs%7D%7D%20)
<h3>Conclusion :</h3>
So, the angular frequency of the blades approximately 36.43π rad/s.
Answer:
1) t=1.743 sec
2)Vo=61.388 m/sec
3)the x component of its velocity just be- fore it strikes the ground is the same as the initial velocity of the ball that is=61.388 m/sec
4)Vf=17.08 m/s
Explanation:
1)From second equation of motion we get
h=Vit+(1/2)gt^2
here in case(a): Vi=0 m/s,h=14.9m,,put these values in above equation to find the time the ball is in motion
14.9=(0)*t+(1/2)(9.8)t^2
t^2=14.9/4.9
t^2=3.040 sec
t=1.743 sec
2) s=Vo*t
Putting values we get
107=Vo*1.743
Vo=61.388 m/sec
3)the x component of its velocity just be- fore it strikes the ground is the same as the initial velocity of the ball that is=61.388 m/sec
4)From third equation of motion we know that
Vf^2-Vi^2=2gh
here Vi=0 m/s,h=14.9 m
Vf^2=Vi^2+2gh=0+2(9.8)(14.9)
Vf^2=292.04
Vf=17.08 m/s
Answer: MR²
is the the moment of inertia of a hoop of radius R and mass M with respect to an axis perpendicular to the hoop and passing through its center
Explanation:
Since in the hoop , all mass elements are situated at the same distance from the centre , the following expression for the moment of inertia can be written as follows.
I = ∫ r² dm
= R²∫ dm
MR²
where M is total mass and R is radius of the hoop .
Answer:
JA
Explanation:
s of time, (b) the velocity and acceleration at t = 2.0 s, (c) the time at which the position is a maximum, (d) the time at which the velocity is zero, and (e) the maximum position. Assume all variable and constants are in SI units.