The answer is B because the water molecules
IF the toss was straight upward, then the kinetic energy it got
from the toss is the gravitational potential energy it has at the top,
where it stops rising and starts falling.
Potential energy = (mass) x (gravity) x (height)
= (0.15 kg) x (9.8 m/s²) x (20 m)
= 29.4 kg-m²/s² = 29.4 joules .
Answer:
Period of the signal.
Explanation:
So, this question is all about a concept in physics or astronomy which is called or known as Radiation Astronomy and Galactic Nuclei that are active. This concept talks most about Quasars; a powerful radiating object which derives its power from black holes.
When You take a look at Quasars, we get the to know that the more you think you can see, the more they move away from us.
Thus, when "You are observing the radiation from a distant active galaxy and you notice that the amplitude of the signal varies in strength regularly over a certain period. The maximum possible size for the source of this radiation can now be calculated from the "PERIOD OF THE SIGNAL.
NB: not the amplitude but the period.
Answer:
Explanation:
Gravitational force between two objects having mass m₁ and m₂ at a distance R
F = G m₁ m₂ / R²
Force between baby and father F₁ = 6.67x10⁻¹¹ x 4.1 x 120 / .18²
= 1.01 x 10⁻⁶ N
b )
Force between baby and Jupiter
F₂ = 6.67x10⁻¹¹ x 1.9x 10²⁷ x 4.1 / ( 6.29 x 10¹¹ )²
= 1.31 x 10⁻⁶ N
c )
Ratio = 1.01 / 1.31
= .77
Answer:
Cannot be determined from the given information
Explanation:
Given the following data;
Velocity = 24 m/s
Period = 3 seconds
To find the amplitude of the wave;
Mathematically, the amplitude of a wave is given by the formula;
x = Asin(ωt + ϕ)
Where;
x is displacement of the wave measured in meters.
A is the amplitude.
ω is the angular frequency measured in rad/s.
t is the time period measured in seconds.
ϕ is the phase angle.
Hence, the information provided in this exercise isn't sufficient to find the amplitude of the waveform.
However, the given parameters can be used to calculate the frequency and wavelength of the wave.
