Answer:
The farther star will appear 4 times fainter than the star that is near to the observer.
Explanation:
Since it is given that the luminosity of the 2 stars is same thus they radiate the same energy per unit time
Consider a spherical wave front of energy 'E' that leaves both the stars (Both radiate 'E' as they have same luminosity)
This Energy is spread over the whole surface area of sphere Thus when the wave front is at a distance 'r' the energy per unit surface area is given by
For the star that is twice away from the earth the distance is '2r' thus we will receive an energy given by
Hence we sense it as 4 times fainter than the nearer star.
A sound wave leaves the loudspeaker. As it travels, it experiences a temporary increase in wavelength and then returns to its original wavelength. The sound wave traveled through a helium balloon (helium is less dense than air could explain this change in wavelength
The pattern of disruption brought on by energy moving away from the sound source is known as a sound wave. Longitudinal waves are what makeup sound. This indicates that the direction of energy wave propagation and particle vibrational propagation are parallel. The atoms oscillate when they are put into vibration.
A high-pressure and a low-pressure zone are created in the medium as a result of this constant back and forth action. Compressions and rarefactions, respectively, are terms used to describe these high- and low-pressure zones. The sound waves go from one medium to another as a result of these regions being transmitted to the surrounding media.
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The energy that was lost due to air resistance while she was bouncing is determined as 3,360 J.
<h3>Conservation of energy</h3>
The amount of energy lost due to air resistance while she was bouncing is determined from the principle of conservation of energy.
ΔE = P.E - Ux
ΔE = mgh - ¹/₂kx²
ΔE = (50)(9.8)(16) - ¹/₂(35)(16)²
ΔE = 3,360 J
Thus, the energy that was lost due to air resistance while she was bouncing is determined as 3,360 J.
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Answer:
a) I = 13.38 kg m / s, b) F = 1,373 10³ N
Explanation:
The impulse is given by the relation
I = ∫ F dt = Δp
I = p_f -p₀
I = m (v_f - v₀)
take the ball's exit direction as positive, whereby the ball velocities
v₀ = -90mph, the final velocity v_f = + 54 m / s
Let's reduce the units to
I = 0.142 [54- (-40.23) ]
the SI system
v₀ = - 90 mph (1609.34 m / 1 mile) (1h / 3600 s = -40.23 m / s
m = 142 g (1kg / 1000) = 0.142 kg
we calculate
I = 0.142 [54- (-40) ]
I = 13.38 kg m / s
b) let's use the definition of momentum
I = ∫ F .dt
I = F ∫ dt
F = I / t
F = 13.38 / 0.008
F = 1,373 10³ N
