<span>When picking up a load, the correct fork spacing must be spaced in an evenly manner in which the centre stringer of the pallet and the balance of the load should be spaced evenly in which makes the picking up the load to be correct and well-balanced.</span>
Answer:
- 3.46 m
Explanation:
initial position, xi = 3.37 m
displacement, Δx = - 6.83 m
Let the final position is xf.
So, displacement = final position - initial position
Δx = xf - xi
- 6.83 = xf - 3.37
xf = 3.37- 6.83
xf = - 3.46 m
Thus, the final position of the squirrel is - 3.46 m.
To determine how far the car goes, we have to know the distance covered by the 75 revolutions of the tires of the car. To do this, we have to calculate for the perimeter of the tire, then multiply it to the number of revolutions.
Distance =
where pi*d is the circumference of the tire
Distance = 75 * pi * 0.90 m
Distance = 212.06 m
The car covers a distance of
212.06 m as it slows down.
Answer:
The minimum thickness of the material is 114.5 nm
Explanation:
The remaining part of the question is as follows
What should the minimum thickness of the material be for it to be nonreflecting for light of wavelength 600 nm (in a vacuum)? Please give details of your reasoning.
Both reflections occur against a surface of higher index of refraction, so there is a phase shift at each reflection. The phase shifts cancel each other out, so if we want the emerging rays to be out of phase, we must have
Given
Index of refraction = 1.31
Index of refraction = 1.43
As we know
Here t is the thickness
represents the phase shift
n is the refraction index
Substituting the given values we get
nm
As per the question the length of the string is given as 2 metre.
A ball is attached to the string.
The ball is taken above the head and the ball starts swinging.
Swinging motion of a body may be oscillatory or circular. Here the motion is circular.
Here the length of the string will be considered as the radius of the circular path. we are asked to calculate the distance traveled by the ball in each circular rotation.
As the motion is circular, the path covered by the ball in each rotation will be equal to the circumference of the circle.
The circumference of the circle is given as
where C is the circumference.
hence C=2×3.14×2 metre
=.12.56 metre.
Hence distance covered by the ball in each circular rotation is 12.56 metre.
Here the displacement is zero. one should not be confused between distance and displacement.