The maximum height that is reached by an object thrown or jumped at a certain initial velocity can be calculated through the equation,
2ad = Vf² - Vi²
Vf is zero (0) in this equation because this is the point at the velocity at the maximum height of the object.
Substituting the known values,
2(a)(-16) = 0 - (3.6)²
The value of a from the equation is 0.405 m/s².
<em>Answer: 0.405 m/s²</em>
Answer:
D). energy resulting from the attraction between two masses.
Explanation:
Answer:
c- Time
Explanation:
The time will tell depending on the person who is exercising, the rate of time, the workout time. & limits on weights....
As per kinematics equation we are given that
now we are given that
a = 2.55 m/s^2
now we need to find x
from above equation we have
so it will cover a distance of 93.2 m
Displacement from the center line for minimum intensity is 1.35 mm , width of the slit is 0.75 so Wavelength of the light is 506.25.
<h3>How to find Wavelength of the light?</h3>
When a wave is bent by an obstruction whose dimensions are similar to the wavelength, diffraction is observed. We can disregard the effects of extremes because the Fraunhofer diffraction is the most straightforward scenario and the obstacle is a long, narrow slit.
This is a straightforward situation in which we can apply the
Fraunhofer single slit diffraction equation:
y = mλD/a
Where:
y = Displacement from the center line for minimum intensity = 1.35 mm
λ = wavelength of the light.
D = distance
a = width of the slit = 0.75
m = order number = 1
Solving for λ
λ = y + a/ mD
Changing the information that the issue has provided:
λ = 1.35 * 10^-3 + 0.75 * 10^-3 / 1*2
=5.0625 *10^-7 = 506.25
so
Wavelength of the light 506.25.
To learn more about Wavelength of the light refer to:
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