Answer:
Maximum acceleration in the simple harmonic motion will be
Explanation:
We have given amplitude of simple harmonic motion is A = 0.43 m
Time period of the oscillation is T = 3.9 sec
We have to find the maximum acceleration
For this we have to find the angular frequency
Angular frequency will be equal to 
Maximum acceleration is given by 
So maximum acceleration in the simple harmonic motion will be

★ A grey hound pursues a hare and takes 5 leaps for every 6 leaps of the hare, but 3 leaps of the hound are equal to 5 leaps of the hare.

★ The speed of the hound and the hare

★ The speed of the hound and the hare = 25:18

As it's given that a grey hound pursues a hare and takes 5 leaps for every 6 leaps of the hare, but 3 leaps of the hound are equal to 5 leaps of the hare.
So firstly let us assume a metres as the distance covered by the hare in one leap.
Ok now let's talk about 5 leaps,.! As it's cleared that the hare cover the distance of 5a metres.
But 3 leaps of the hound are equal to 5 leaps of the hare.
Henceforth, (5/3)a meters is the distance that is covered by the hound.
Now according to the question,
Hound pursues a hare and takes 5 leaps for every 6 leaps of the hare..! (Same interval)
Now the distance travelled by the hound in it's 5 leaps..!
Now the distance travelled by the hare in it's 6 leaps..!
Now let us compare the speed of the hound and the hare. Let us calculate them in the form of ratio..!
5 What is the angular displacement at the end of the 25-mm-diameter shaft and the linear displacement of point A of Figure P5.5
<h3>What is
displacement ?</h3>
A displacement is a vector in geometry and mechanics that has a length equal to the shortest distance between a point P's initial and final positions. It calculates the length and angle of the net motion, or total motion, in a straight line from the starting point to the destination of the point trajectory. The translation that links the starting point and the ending point can be used to spot a displacement.
The final location xf of a point relative to its beginning position xi, or a relative position (derived from the motion), is another way to express a displacement. The difference between the end and beginning positions can be used to define the equivalent displacement vector
To learn more about displacement from the given link:
brainly.com/question/321442
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Answer:
D
Explanation:i think but dont get mad if im wrong
The answer is D. They have a higher frequency than X-rays.
Gamma-rays falls in the electromagnetic spectrum above, Gamma-rays have frequencies greater than 10¹⁸ cycles per second, or Hertz (Hz)