I'm pretty sure its d. because depends what tipe of dog it is there are some that see different colors so I'm pretty sure its d.
(a) The velocity of the object on the x-axis is 6 m/s, while on the y-axis is 2 m/s, so the magnitude of its velocity is the resultant of the velocities on the two axes:

And so, the kinetic energy of the object is

(b) The new velocity is 8.00 m/s on the x-axis and 4.00 m/s on the y-axis, so the magnitude of the new velocity is

And so the new kinetic energy is

So, the work done on the object is the variation of kinetic energy of the object:
A bird in the hand is worth two in the bush
or
actions speak louder than words
or
her bark is worse than her bite
i hope this helps these are examples of aphorism
The new period will be 2.486 days.
<h3>What is the period?</h3>
The period is found as the ratio of the angular displacement and the angular velocity. Its unit is the second and is denoted by t. The value of time needed to complete the rotation is the total period.
Given data;
Mass of a star,m= 1.210×10³¹ kg
The time period for one rotation of the star, T = 20.30 days
D' = 0.350 D
R' = 0.350 R
From the law of conservation of angular momentum;

Hence, the new period will be 2.486 days.
To learn more about the period, refer to the link;
brainly.com/question/569003
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