A spool of thread, free to unwind, is on a horizontal surface (with friction) and pulled directly upward (+y) at all times. The
spool rolls without slipping on the floor. The spool has a mass M, moment of inertia I, and a radius of R. Assume the string is wrapped around the spool at a radius r (where r < R). spool rolling on the floor
Which equation below is a correct expression of Newton’s Second Law ( ΣF=ma ) in the x (horizontal) -direction for this situation. Assume all the forces below are magnitudes and that to the right is considered the positive direction.
a. F_T - Ff= Max
b. F_T+Ff= Max
c. Ff= Max
Which equation below is a correct expression of Newton’s Second Law ( ΣF=ma ) in the x (horizontal) -direction for this situation. Assume all the forces below are magnitudes and that to the right is considered the positive direction.
a. F_T - Ff= Max
b. F_T+Ff= Max
c. Ff= Max
1 answer:
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Answer:
0.5 eV
Explanation:
= Initial potential energy =
= Final potential energy
= Initial width
= Final width =
Energy of an electron in a one-dimensional trap is given by
From the equation we get
So,
The ground state energy will be 0.5 eV
Answer:
the tension in the part of the cord attached to the textbook is 7.4989 N
Explanation:
Given the data in the question;
As illustrated in the image below;
first we determine the value of the acceleration,
along vertical direction; we use the second equation of motion;
y = ut + a
t²
we substitute;
0 m/s for u, 1.29 m for y, 0.850 s for t,
1.29 = 0×0.850 + ×a
×(0.850)²
1.29 = 0.36125a
a = 1.29 / 0.36125
a = 3.5709 m/s²
Now when the text book is moving with acceleration , the dynamic equation will be;
T₁ = m₁a
where m₁ is the mass of the text book ( 2.10 kg )
a is the vertical acceleration ( 3.5709 m/s² )
so we substitute
T₁ = 2.10 × 3.5709
T₁ = 7.4989 N
Therefore, the tension in the part of the cord attached to the textbook is 7.4989 N
