Answer:
a) 2nd case rate of rotation gives the greater speed for the ball
b) 1534.98 m/s^2
c) 1515.04 m/s^2
Explanation:
(a) v = ωR
when R = 0.60, ω = 8.05×2π
v = 0.60×8.05×2π = 30.34 m/s
Now in 2nd case
when R = 0.90, ω = 6.53×2π
v = 0.90×6.53×2π = 36.92 m/s
6.35 rev/s gives greater speed for the ball.
(b) a = ω^2 R = (8.05×2π)^2 )(0.60) = 1534.98 m/s^2
(c) a = ω^2 R = (6.53×2π)^2 )(0.90) = 1515.05 m/s^2
To solve this problem it is necessary to apply the concepts related to frequency as a function of speed and wavelength as well as the kinematic equations of simple harmonic motion
From the definition we know that the frequency can be expressed as
Where,
Therefore the frequency would be given as
The frequency is directly proportional to the angular velocity therefore
Now the maximum speed from the simple harmonic movement is given by
Where
A = Amplitude
Then replacing,
Therefore the maximum speed of a point on the string is 3.59m/s
Answer:
the range or the ball is 48.81 m
Explanation:
given;
Nicole throws a ball at 25 m/s at an angle of 60 degrees abound the horizontal.
find:
What is the range of the ball?
solution:
let Ф = 25°
Vo = 25 m/s
<u>consider x-motion using time of fight: x = Vox * t</u>
where x = R = range
t =<u> 2 Voy </u>
g
R =<u> Vo² sin (2Ф)</u>
g
plugin values into the formula:
R = <u>(25)² sin (2*25) </u>
9.81
R = 48.81 m
therefore, the range or the ball is 48.81 m
sorry - late reply...just stumbled across tis...hope u can still use it :)
By the mirror equation: 1/di + 1/do = 1/f
<span>
</span>
<span>where di = distance to image = +12cm (+ for real image)</span>
and do = distance to object = +8cm
Substitute and solve for f, the focal length
<span><span>
1/12 + 1/8 = 1/f
</span><span>
1/f = (8 + 12) / 12 * 8 = 20/96
</span><span>
so f = 96/20 = 4.8 cm</span>
</span>
