Condyloid joints are, logically enough, created by joints at condyle bone markings.

A bird is flying with a speed of 18.0 m/s over water when it accidentally drops a 2.00 kg fish. If the altitude of the bird is 5

Alina [70]

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!

4 0

3 years ago

Which has the least potential energy

makvit [3.9K]

Answer:

The plasma state would have the least potential energy and the greatest kinetic energy.

Explanation:

i just took the test and i go it right

4 0

3 years ago

Read 2 more answers

Steam enters a well-insulated nozzle at 200 lbf/in.2 , 500F, with a velocity of 200 ft/s and exits at 60 lbf/in.2 with a velocit

Ede4ka [16]

Answer:

$386.2^{\circ}F$

Explanation:

We are given that

$P_1=200lbf/in^2$

$P_2=60lbf/in^2$

$v_1=200ft/s$

$v_2=1700ft/s$

$T_1=500^{\circ}F$

$Q=0$

$C_p=1BTU/lb^{\circ}F$

We have to find the exit temperature.

By steady energy flow equation

$h_1+v^2_1+Q=h_2+v^2_2$

$C_pT_1+\frac{P^2_1}{25037}+Q=C_pT_2+\frac{P^2_2}{25037}$

$1BTU/lb=25037ft^2/s^2$

Substitute the values

$1\times 500+\frac{(200)^2}{25037}+0=1\times T_2+\frac{(1700)^2}{25037}$

$500+1.598=T_2+115.4$

$T_2=500+1.598-115.4$

$T_2=386.2^{\circ}F$

7 0

4 years ago

Radium-226 emits alpha (α), beta (β) and gamma (γ) radiation.

Furkat [3]

Answer:

Explanation:

alpha

Alpha Radiation (α): A large, unstable nucleus decays to produce a smaller, more stable nucleus and an alpha particle (identical to a helium nucleus, ⁴₂He or ⁴₂α).

It has a very high ionizing energy and low penetrating power. It can be stopped by paper skin

Beta Radiation (β): A neutron in an unstable nucleus decays, forming a proton and emitting a beta (β) particle (identical to an electron, ⁰₋₁e or ⁰₋₁b) and resulting in a more stable nucleus.

It has high ionizing energy and penetrating power. It can be stopped by aluminium sheet

Gamma Radiation (γ): An unstable nucleus releases energy in the form of a high energy photon (no mass)to become more stable; this often accompanies other forms of radioactivity.

It has very high penetrating power and very low ionizing energy. It can be stopped by lead block.

3 0

4 years ago

When the voltage

ValentinkaMS [17]

The current will increase at the same rate the voltage does.

6 0

3 years ago