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Levart [38]
3 years ago
12

The position of a particle moving along a coordinate line is s equals StartRoot 76 plus 6 t EndRoot​, with s in meters and t in

seconds. Find the rate of change of the​ particle's position at t equals 4 sec. The rate of change of the​ particle's position at t equals 4 sec is 104 4 m divided by sec.
Physics
1 answer:
Tanya [424]3 years ago
6 0

Answer:

= 3 /10

Explanation:

s = √(76 + 6t)

ds/dt = \frac{1}{2} (76 + 6t)^-^(^1^/^2^) . 6

at t = 4

ds / dt = \frac{1}{2} (76 + 6(4))^-^(^1^/^2^) . 6

  = \frac{1}{2} (100)^-^(^1^/^2^) . 6

  = 3 / 10

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makvit [3.9K]
The Earth revolves (orbits) around the Sun in one year. The Earth's rotation axis is tilted relative to the plane of its orbit around the Sun. This tilt of the Earth is responsible for the seasons as the Earth orbits the Sun. The Sun provides energy that sustains all life on Earth.
4 0
2 years ago
Model a hydrogen atom as a three-dimensional potential well with Uo = 0 in the region 0 &lt; x a. 283 eV <br> b. 339 eV <br> c.
denis23 [38]

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;

E_{nx,ny,nz = ( n_x^2 + n_y^2 + n_z^2 ) × 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 ( n_x = n_y = n_z = 1 )

so

( n_x^2 + n_y^2 + n_z^2 ) ×  π²h"² / 2ml²

since h" = h/2π

( n_x^2 + n_y^2 + n_z^2 ) × π²h² / (2π)²2ml²

so we substitute

E_{111 = ( 1² + 1² + 1² ) × [ π²( 6.626 × 10⁻³⁴ )² ] / [ (2π)² × 2 × 9.1 × 10⁻³¹ kg × ( 1.0 × 10⁻¹⁰)² ]

E_{111 = 3 × [ (4.333188779 × 10⁻⁶⁶) / ( 7.185072 × 10⁻⁴⁹ ) ]    

E_{111 = 3 × [ 6.03082165 × 10⁻¹⁸ ]

Now, we know that electric charge = 1.602 x 10⁻¹⁹

so

E_{111 = 3 × [ (6.03082165 × 10⁻¹⁸) / (1.602 x 10⁻¹⁹) ]

E_{111 = 3 × [ 37.645578 ]

E_{111 = 112.9 ≈ 113 eV

E_{211 = ( n_x^2 + n_y^2 + n_z^2 )  × π²h² / (2π)²2ml²

we substitute

E_{211 = ( 1² + 1² + 2² ) × [ 37.645578 ]

E_{211 = 6 × [ 37.645578 ]

E_{211 = 225.87 ≈ 226 eV

E_{221 = ( n_x^2 + n_y^2 + n_z^2 )  × π²h² / (2π)²2ml²

we substitute

E_{221 = ( 2² + 2² + 1² ) × [ 37.645578 ]

E_{211 = 9 × [ 37.645578 ]

E_{211 = 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

8 0
2 years ago
Based on their locations on the periodic table, which two elements would you
Nat2105 [25]

Answer:

i think it would be B. Aluminum, Al and D. Boron, B

Explanation:

since they're both in group 13 and they forms a 3+ ion

6 0
3 years ago
Read 2 more answers
Find the fundamental frequency and the next three frequencies that could cause standing-wave patterns on a string that is 30.0 m
maksim [4K]

Answer:

0.786 Hz, 1.572 Hz, 2.358 Hz, 3.144 Hz

Explanation:

The fundamental frequency of a standing wave on a string is given by

f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}

where

L is the length of the string

T is the tension in the string

\mu is the mass per unit length

For the string in the problem,

L = 30.0 m

\mu=9.00\cdot 10^{-3} kg/m

T = 20.0 N

Substituting into the equation, we find the fundamental frequency:

f=\frac{1}{2(30.0)}\sqrt{\frac{20.0}{(9.00\cdot 10^{-3}}}=0.786 Hz

The next frequencies (harmonics) are given by

f_n = nf

with n being an integer number and f being the fundamental frequency.

So we get:

f_2 = 2 (0.786 Hz)=1.572 Hz

f_3 = 3 (0.786 Hz)=2.358 Hz

f_4 = 4 (0.786 Hz)=3.144 Hz

6 0
3 years ago
6- A metal block measures 10 cm x 2 cm x 2 cm. what is its volume? how many blocks each
zalisa [80]
You need 5 blocks of the smaller object to contain the same amount of volume of the bigger object

3 0
3 years ago
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