Here we go.
My abbreviations; KE = Kinetic Energy; GPE = Gravitational Potential Energy.
So first off, we know the fish has KE right when the bird releases it. Why? Because it has horizontal velocity after released! So let’s calculate it:
KE = 1/2(m)(V)^2
KE = 1/2(2)(18)^2
KE = 324 J
Nice!
We also know that the fish has GPE at its maximum height before release:
GPE = mgh
GPE = (2)(9.81)(5.40)
GPE = 105.95 J
Now, based on the *queue dramatic voice* LAW OF CONSERVATION OF ENERGY, we know all of the initial energy of the fish will be equal to the amount of final energy. And since the only form of energy when it hits the water is KE, we can write:
KEi + GPEi = KEf
(Remember - we found the initial energies before!)
(324) + (105.95) = KEf
KEf = 429.95J
And that’s you’re final answer! Notice how this value is MORE than the initial KE from before (324 J) - this is because all of the initial GPE from before was transformed into more KE as the fish fell (h decreased) and sped up (V increased).
If this helped please like it and comment!
As the container starts to heat up, so will the neon gas. Heat is nothing but energy, and when you add energy to a gas, it will start vibrating much faster and hit the edges of the container at a higher rate and a faster velocity. Therefore, it's possible to deduce that the container will most likely rupture and/or "explode".
Answer:
the mass of water is 0.3 Kg
Explanation:
since the container is well-insulated, the heat released by the copper is absorbed by the water , therefore:
Q water + Q copper = Q surroundings =0 (insulated)
Q water = - Q copper
since Q = m * c * ( T eq - Ti ) , where m = mass, c = specific heat, T eq = equilibrium temperature and Ti = initial temperature
and denoting w as water and co as copper :
m w * c w * (T eq - Tiw) = - m co * c co * (T eq - Ti co) = m co * c co * (T co - Ti eq)
m w = m co * c co * (T co - Ti eq) / [ c w * (T eq - Tiw) ]
We take the specific heat of water as c= 1 cal/g °C = 4.186 J/g °C . Also the specific heat of copper can be found in tables → at 25°C c co = 0.385 J/g°C
if we assume that both specific heats do not change during the process (or the change is insignificant)
m w = m co * c co * (T eq - Ti co) / [ c w * (T eq - Tiw) ]
m w= 1.80 kg * 0.385 J/g°C ( 150°C - 70°C) /( 4.186 J/g°C ( 70°C- 27°C))
m w= 0.3 kg
Answer: 8.1 x 10^24
Explanation:
I(t) = (0.6 A) e^(-t/6 hr)
I'll leave out units for neatness: I(t) = 0.6e^(-t/6)
If t is in seconds then since 1hr = 3600s: I(t) = 0.6e^(-t/(6 x 3600) ).
For neatness let k = 1/(6x3600) = 4.63x10^-5, then:
I(t) = 0.6e^(-kt)
Providing t is in seconds, total charge Q in coulombs is
Q= ∫ I(t).dt evaluated from t=0 to t=∞.
Q = ∫(0.6e^(-kt)
= (0.6/-k)e^(-kt) evaluated from t=0 to t=∞.
= -(0.6/k)[e^-∞ - e^-0]
= -0.6/k[0 - 1]
= 0.6/k
= 0.6/(4.63x10^-5)
= 12958 C
Since the magnitude of the charge on an electron = 1.6x10⁻¹⁹ C, the number of electrons is 12958/(1.6x10^-19) = 8.1x10^24 to two significant figures.
