Much energy as would Microraptor gui have to expend to fly with a speed of 10 m/s for 1.0 minutes is 486 J.
The first step is to find the energy that Microraptor must release to fly at 10 m/s for 1.0 minutes. The energy that Microraptor must expend to fly can be found using the relationship between Power and Energy.
P = E/t
Where:
P = power (W)
T = time (s)
Now, a minimum of 8.1 W is required to fly at 10 m/s. So, the energy expended in 1 minute (60 seconds) is
P = E/t
E = P x t
E = 8.1 x 60
E = 486 Joules
Thus, the energy that Microraptor must expend to fly at 10 m/s for 1.0 minutes is the 486 J.
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Answer:
Extraneous
Explanation:
Extraneous variables are any variables that you are not intentionally studying in your experiment or test
Answer:
Alaska: Hydrokinetic Energy Campbell CR9000X used for in-stream hydrokinetic device evaluation. Marine hydrokinetic energy power generation is an emerging sector in the renewable energy portfolio. Hydrokinetic devices convert the energy of waves, tidal currents, ocean currents or river currents into electrical power.
Answer:
The magnitude of momentum of the airplane is
.
Explanation:
Given that,
Mass of the airplane, m = 3400 kg
Speed of the airplane, v = 450 miles per hour
Since, 1 mile per hour = 0.44704 m/s
v = 201.16 m/s
We need to find the magnitude of momentum of the airplane. It is given by the product of mas and velocity such that,



or

So, the magnitude of momentum of the airplane is
. Hence, this is the required solution.
We can apply the law of conservation of energy here. The total energy of the proton must remain constant, so the sum of the variation of electric potential energy and of kinetic energy of the proton must be zero:

which means

The variation of electric potential energy is equal to the product between the charge of the proton (q=1eV) and the potential difference (

):

Therefore, the kinetic energy gained by the proton is

<span>And since the initial kinetic energy of the proton was zero (it started from rest), then this 1000 eV corresponds to the final kinetic energy of the proton.</span>