How much of Earth is made of mantle?
1 answer:
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Answer:
m
Explanation:
At 10am, the minute hand and hour hand are ' 2 hours apart', since the minute hand is at 12pm and hour hand is at 10am.
Angle between the two hands = 2/12 * 360
= 60°
Arc Length =
=
This question is incomplete, the complete question is;
Model a hydrogen atom as a three-dimensional potential well with U₀ = 0 in the region 0 < x < L, 0 < y < L and 0 < z < L, and infinite otherwise, with L = 1.0 × 10⁻¹⁰ m.
Which of the following is NOT one of the lowest three energy levels of an electron in this model?
a. 283 eV
b. 339 eV
c. 113 eV
d. 226 eV
Answer:
the lowest three energy are; 113 eV, 225 eV, and 339 eV.
Hence Option a) 283 eV is not among the three lowest energy
Explanation:
Given the data in the question;
Three dimension cube or particle in a cubic box
the energy value is given by;
=
× π²h"² / 2ml²
where h" = h/2π and h is Planck's constant ( 6.626 × 10⁻³⁴ m² kg / s )
m is mass of electron ( 9.1 × 10⁻³¹ kg )
l is length of side of box ( 1.0 × 10⁻¹⁰ m )
for ground level ( )
so
× π²h"² / 2ml²
since h" = h/2π
× π²h² / (2π)²2ml²
so we substitute
= ( 1² + 1² + 1² ) × [ π²( 6.626 × 10⁻³⁴ )² ] / [ (2π)² × 2 × 9.1 × 10⁻³¹ kg × ( 1.0 × 10⁻¹⁰)² ]
= 3 × [ (4.333188779 × 10⁻⁶⁶) / ( 7.185072 × 10⁻⁴⁹ ) ]
= 3 × [ 6.03082165 × 10⁻¹⁸ ]
Now, we know that electric charge = 1.602 x 10⁻¹⁹
so
= 3 × [ (6.03082165 × 10⁻¹⁸) / (1.602 x 10⁻¹⁹) ]
= 3 × [ 37.645578 ]
= 112.9 ≈ 113 eV
=
× π²h² / (2π)²2ml²
we substitute
= ( 1² + 1² + 2² ) × [ 37.645578 ]
= 6 × [ 37.645578 ]
= 225.87 ≈ 226 eV
=
× π²h² / (2π)²2ml²
we substitute
= ( 2² + 2² + 1² ) × [ 37.645578 ]
= 9 × [ 37.645578 ]
= 338.8 ≈ 339 eV
Therefore, the lowest three energy are; 113 eV, 225 eV, and 339 eV.
Hence Option a) 283 eV is not among the three lowest energy
Answer:
15.67 m/s
Explanation:
The ball has a projectile motion, with a horizontal uniform motion with constant speed and a vertical accelerated motion with constant acceleration g=9.8 m/s^2 downward.
Let's consider the vertical motion only first: the vertical distance covered by the ball, which is S=50 m, is given by
where t is the time of the fall. Substituting S=50 m and re-arranging the equation, we can find t:
Now we now that the ball must cover a distance of 50 meters horizontally during this time, in order to fall inside the carriage; therefore, the velocity of the carriage should be: