A 256 Hz tuning fork is resonating in a closed tube on a warm day when the speed of sound is 346 m/s. What is the length of the
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Answer : The average speed of the sprinter is, 34.95 Km/hr
Solution :
Average velocity : It is defined as the distance traveled by the time taken.
Formula used for average velocity :
where,
= average velocity
d = distance traveled = 200 m
t = time taken = 20.6 s
Now put all the given values in the above formula, we get the average velocity of the sprinter.
conversion :
(1 Km = 1000m)
(1 hr = 3600 s)
Therefore, the average speed of the sprinter is, 34.95 Km/hr
This question is incomplete, the complete question is;
In a softball windmill pitch, the pitcher rotates her arm through just over half a circle, bringing the ball from a point above her shoulder and slightly forward to a release point below her shoulder and slightly forward. (Figure 1) shows smoothed data for the angular velocity of the upper arm of a college softball pitcher doing a windmill pitch; at time t = 0 her arm is vertical and already in motion. For the first 0.15 s there is a steady increase in speed, leading to a final push with a greater acceleration during the final 0.05 s before the release. In each part of the problem, determine the corresponding quantity during the first 0.15 s of the pitch.
Angular Velocity at time 0s = 12 rad/s
Angular Velocity at time 0.15s = 24 rad/s
a) What is the angular acceleration?
b) If the ball is 0.60 m from her shoulder, what is the tangential acceleration of the ball? This is the key quantity here--it's a measure of how much the ball is speeding up. Express your answer in m/s2 and in units of g
Answer:
a) the angular acceleration is 80 rad/s²
b) the tangential acceleration of the ball is;
- a = 48 m/s²
- a = 4.9 g
Explanation:
Given the data in the question;
from the graph below;
Angular Velocity at time 0s = 12 rad/s
Angular Velocity at time 0.15s = 24 rad/s
a) What is the angular acceleration;
Angular acceleration ∝ = ( -
) / dt
we substitute
Angular acceleration ∝ = ( 24 - 12 ) / 0.15
Angular acceleration ∝ = 12 / 0.15
Angular acceleration ∝ = 80 rad/s²
Therefore, the angular acceleration is 80 rad/s²
b)
If the ball is 0.60 m from her shoulder, i.e s = 0.6 m
the tangential acceleration of the ball will be;
a = ∝ × s
we substitute
a = 80 × 0.6
a = 48 m/s²
a = ( 48 / 9.8 )g
a = 4.9 g
Therefore, the tangential acceleration of the ball is;
- a = 48 m/s²
- a = 4.9 g
Answer:
Light of a shorter wavelength should be used.
Explanation:
This is studied in the phenomenon called photoelectric effect, in which light is able to release electrons from a metal, said electrons are called photoelectrons .
The experiments that have been carried out show that <u>increasing or decreasing the intensity of the light will not cause the photoelectrons to be emitted</u>, what will cause the photoelectrons to be emitted is to increase the frequency of the incident light.
And a higher frequency corresponds to a shorter wavelength according to the equation:
(where is frequency,
the speed of light, and
the wavelength)
So the answer is that the wavelength of the light must be shortened to cause the emission of electrones.