The position of an object at any time t is given by s(t)=3t4−40t3+126t2−9. i. Determine the velocity of the object at any time t
1 answer:
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Answer:
The tunnel probability for 0.5 nm and 1.00 nm are and
respectively.
Explanation:
Given that,
Energy E = 2 eV
Barrier V₀= 5.0 eV
Width = 1.00 nm
We need to calculate the value of
Using formula of
Put the value into the formula
(a). We need to calculate the tunnel probability for width 0.5 nm
Using formula of tunnel barrier
Put the value into the formula
(b). We need to calculate the tunnel probability for width 1.00 nm
Hence, The tunnel probability for 0.5 nm and 1.00 nm are and
respectively.
Answer:
The minimum velocity is V=5.98[m/s]
Explanation:
In this problem the athlete jumps by printing a force to raise his mass at an initial speed, which we must determine. These two initial amounts of energy that we have, the kinetic energy (velocity with which it jumps) and the work done (mass of the athlete by the jumped distance) must be equal to the kinetic energy when its velocity is equal to 0.65[m/s]. By raising this equation, the initial speed can be cleared.
We have to keep in mind that the weight is acting down and the movement is one up so the work done by the force is negative
Therefore we have:
Reorganizing each of the members of the equation and clearing the final velocity.