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2 years ago

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PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE H

ELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP
I NEED TO GIVE THIS IN TODAY!!!

1 answer:

Masteriza [31]2 years ago

5 0

Answer:

a) greater

smaller

Explanation:

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Prof. Kopp is working on an experiment located in an abandoned mine near Duluth, MN. The experiment is located approximately 713

MA_775_DIABLO [31]

Answer:

The average speed of the elevator going down in the abandoned mine is 17.722mph.

Explanation:

If the elevator takes 90 seconds to descend a height of 713m, the average speed of the elevator is:

$v_{av}=x_T/t_T=713m/90s=7.922m/s$

And if 1m/s is 2.23694mph, the average speed is:

$v_{av}=7.922m/s=17.722mph$ .

8 0

3 years ago

Photons of wavelength 65.0 pm are Compton-scattered from a free electron which picks up a kinetic energy of 0.84 keV from the co

ElenaW [278]

Answer:

λ = 65.6 pm

Explanation:

Given that

λo = 65 pm

The initial energy of the electron

$E_o=\dfrac{hC}{\lambda_0 }$

Now by putting the values

$E_o=\dfrac{hC}{\lambda_0 }$

$E_o=\dfrac{6.67\times 10^{-34}\times 3\times 10^8}{65\times 10^{-12}}$

$E_o=3.05\times 10^{-15}\ J$

$E_o=\dfrac{3.05\times 10^{-15}}{1.6\times 10^{-19}\times 10^3}\ KeV$

Eo=19.06 KeV

Given that kinetic energy KE= 0.84 KeV

Therefore the final energy

E= Eo - KE

E = 19.06 - 0.84 KeV

E= 18.22 KeV

The wavelength λ can be find as

$E=\dfrac{hC}{\lambda}$

$\lambda=\dfrac{hC}{E}$

$\lambda=\dfrac{6.67\times 10^{-34}\times 3\times 10^8}{19.06 \times 10^3\times 1.6\times 10^{-19}}$

λ = 6.56 x 10⁻¹¹ m

λ = 65.6 pm

3 0

3 years ago

Identify the atom with the following ground-state electron configuration for its valence shell 4s23d104p6

Annette [7]

Since the greatest number in the given electron configuration of the atom is 4, it means that the element is found in the fourth energy level. The number of electrons found in the valence shell is 2+10+6 = 18. From this answer and using the periodic table, we can see that the atom is that of Krypton (Kr).

4 0

3 years ago

Read 2 more answers

A 57 kg boy and a 41 kg girl use an elastic rope while engaged in a tug-of-war on a frictionless icy surface. If the acceleratio

Sergeeva-Olga [200]

Answer:

Acceleration of the boy a₁:

$a_{1} = 1.87 \frac{m}{s^{2} }$

Explanation:

Conceptual analysis

We apply Newton's second law to the boy and the girl:

F = m*a (Formula 1)

F : Force in Newtons (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Nomenclature

m₁ : boy mass

m₂ : girl mass

a₁ : boy acceleration

a₂ : girl acceleration

F₁ : boy acceleration

F₂ : girl acceleration

Known data

m₁ = 57 kg

m₂ = 41 kg

a₂ = 2.6 m/s²

Problem development

We apply to Newton's third law of action and reaction, then:

F₁ = F₂ , We apply the formula (1):

m₁*a₁ = m₂*a₂

$a_{1} = \frac{m_{2}* a_{2} }{m_{1} }$

$a_{1} = \frac{41* 2.6 }{ 57 }$

$a_{1} = 1.87 \frac{m}{s^{2} }$

5 0

3 years ago

A technician wraps wire around a tube of length 38 cm having a diameter of 8.3 cm. When the windings are evenly spread over the

seropon [69]

Answer:

The inductance of this solenoid is 5.51 mH.

Explanation:

Given that,

length of the tube, l = 38 cm = 0.38 m

Diameter of the tube, d = 8.3 cm

Radius of the tube, r = 4.15 cm

The number of turns, N = 555

We need to find the inductance of this solenoid. It is given by the formula as :

$L=\dfrac{\mu N^2A}{l}\\\\L=\dfrac{4\pi \times 10^{-7}\times 555^2\times \pi \times (4.15\times 10^{-2})^2}{0.38}\\\\L=0.00551\ H\\\\L=5.51\ mH$

So, the inductance of this solenoid is 5.51 mH. Hence, this is the required solution.

7 0

3 years ago