Input work = 9.63×10³ J.
Output work = 3.0×10³ J
By definition,
Efficiency = (Output work)/(Input work)
= (3.0×10³)/(9.63×10³)
= 0.31 = 31%
Answer: 31%
Omitting the 1 will not change the value of the number, but will change the units at the end of the problem
Answer:
1.67m
5m/s
Explanation:
Given parameters:
Wavelength of the wave = 3m
Speed of the wave = 5m/s
Unknown:
Frequency of the wave = ?
Speed of the water waves = ?
Solution:
The distance between crest and the adjacent trough of water waves is known as the wavelength of a wave.
To find the frequency ;
V = f∧
V is the speed of the wave
f is the frequency
∧ is the wavelength
Insert the parameters and find the frequency;
f = V/ ∧ = 5 / 3 = 1.67Hz
The rate at which the wave passed a given point is the speed of the wave and it is 5m/s
Answer:
gravitational force and outward pressure
Explanation:
The stars are very massive stellar objects, so they have a strong gravitational force, which drives the star to contract itself, but also in stars there are nuclear reactions such as fusion of hydrogen and other elements, that releases energy and creates a pressure from the center to the star exterior, an outward pressure that goes against the gravitational force. So when a star is stable these two forces exist in equilibrium or in balance, in which the star does not collapse by gravity or disintegrate by its outward pressure.
Answer:
Addition of shells increases the distance of outer electrons from the nucleus.
Explanation:
Shielding effect is known as the attraction between the nucleus and an electron of any atom. In other words, it is the reduction in effective nuclear charge on an electron cloud.
Addition of electron shells results in the shielding of electron from nucleus. As the number of electron shells increases then farther will be the electrons placed from the nucleus and hence it will become easier to expel the electrons from outer shells with only little amount of ionization energy.
So, the amount of ionization energy require will be indirectly proportional to the shielding effect because more the shielding of electrons from the nucleus less will be the ionization energy require to expel the electrons.
