<h2>
The seagull's approximate height above the ground at the time the clam was dropped is 4 m</h2>
Explanation:
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 0 m/s
Acceleration, a = 9.81 m/s²
Time, t = 3 s
Substituting
s = ut + 0.5 at²
s = 0 x 3 + 0.5 x 9.81 x 3²
s = 44.145 m
The seagull's approximate height above the ground at the time the clam was dropped is 4 m
Radio waves in a vacuum travel at the speed of light because they are a type of electromagnetic radiation like a light has been measured as traveling at 3×10^8 m/s in a vacuum.
Charged particles that are accelerating, like time-varying electric currents, are what produce radio waves. Radio and television signals are transmitted using radio waves, and microwaves used in radar and microwave ovens are also radio waves. Radio waves are emitted by a lot of celestial bodies, including pulsars. High RF exposure levels have the potential to heat biological tissue and raise body temperature. The body's inability to handle or remove the extra heat that could be generated by high RF exposure in humans could result in tissue damage.
To learn more about radio waves please visit -
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1) all matter is made of atoms , atoms are indivisible and indestructible 2) compounds are formed by a combination of two or more different kinds of atoms 3) a chemical reaction is an arrangement of atoms
More force needs to be applied
Answer:
54%
Explanation:
So, we have that the "magnitude of its displacement from equilibrium is greater than (0.66)A—''. Thus, the first step to take in answering this question is to write out the equation showing the displacement in simple harmonic motion which is = A cos w×t.
Therefore, we will have two instances t the displacement that is to say at a point 2π/w - a2 and the second point at a = a2.
Let us say that 2π/w = A, then, we have that a = A cos ^-1 (0.66)/2π. Also, we have that a2 = A/2 - A cos^- (0.66) / 2π.
The next thing to do is to calculate or determine the total length of of the required time. Thus, the total length is given as:
2a1 + ( A - 2a2) = 2A{ cos^-1 (0.66)}/ π.
Therefore, the total percentage of the period does the mass lie in these regions = 100 × {2a1 + ( A - 2a2) }/A = 2 { cos^-1 (0.66)}/ π × 100 = 54%.
Thus, the total percentage of the period does the mass lie in these regions = 54%.
