The carrying capacity of a biological species in an environment is the maximum population size of the species that the environment can sustain indefinitely, given the food, habitat, water, and other necessities available in the environment.
The velocity of the electron after moving a distance of 1cm in the electric field is 5.95×10⁶m.
<h3>What is Electric field?</h3>
Electric field is the physical field that surrounds a charge.
<h3>How to find final velocity of the electron when it moves some distance in a certain electric field?</h3>
- From Newton's second law, the acceleration the electron will be
a=F/m=qE/m
- where q= charge of electron
E= electric field
m= mass of electron
=(−1.60×10^−19C)(3×10³N/C)/(9.11×10^-31kg)
=10¹⁵×0.526m/s²
- The kinematics equation v²=v0²+2a(Δx)
- where v=final velocity of the electron
v0=initial velocity of the electron =5×10⁶m/s
a=acceleration of the electron =10¹⁵×0.526m/s²
Δx=distance moved by the electron in east direction =1cm=10^-2m
- Now v^2=(5×10⁶)²+2×10¹⁵×0.526×10^-2
=25×10¹²+10.52×10¹²
=35.52×10¹²
- Now velocity of electron=5.95×10⁶m/s.
Thus , we can conclude that the velocity of the electron after moving a distance of 1cm in the electric field is 5.95×10⁶m.
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Ans: Beat Frequency = 1.97HzExplanation:
The fundamental frequency on a vibrating string is

<span> -- (A)</span>
<span>here, T=Tension in the string=56.7N,
L=Length of the string=0.66m,
m= mass = 8.3x10^-4kg/m * 0.66m = 5.48x10^-4kg </span>
Plug in the values in Equation (A)
<span>so </span>

<span> = 197.97Hz </span>
<span>the beat frequency is the difference between these two frequencies, therefore:
Beat frequency = 197.97 - 196.0 = 1.97Hz
-i</span>
Answer: 71.7 KJ
Explanation:
The rotational kinetic energy of a rotating body can be written as follows:
Krot = ½ I ω2
Now, any point on the rim of the flywheel, is acted by a centripetal force, according to Newton’s 2nd Law, as follows:
Fc = m. ac
It can be showed that the centripetal acceleration, is related with the angular velocity and the radius, as follows:
ac = ω2 r
We know that this acceleration has a limit value, so , we can take this limit to obtain a maximum value for the angular velocity also.
As the flywheel is a solid disk, the rotational inertia I is just ½ m r2.
Replacing in the expression for the Krot, we have:
Krot= ½ (1/2 mr2.ac/r) = ¼ mr ac = ¼ 67.0 Kg. 1.22 m . 3,510 m/s2 = 71. 7 KJ
I am quite sure the first one is Friction, but I am not sure about the second one. Is it wind?