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3 years ago

Choose the best description of the attraction between you and your classmate.

Physics

1 answer:

Lerok [7]3 years ago

5 0

Answer:

The correct option is;

D. There is not enough information to answer this question

Explanation:

The universal gravitational constant = 6.67408 × 10⁻¹¹ 3³/(kg·s²)

For an in between distance of 1 m and equal masses of 60 kg, we have;

$F = G \times \dfrac{m_{1} \times m_{2}}{r^{2}} = 6.67408 \times 10^{-11} \times \dfrac{60 \times 60}{1^{2}} \approx 2.403 \times 10^{-7} N$

The gravitational attraction ≈ 2.403 × 10⁻⁷ N, which does not correspond with the answers, therefore, the best option is that there is not enough information to answer this question.

You might be interested in

Which factors affect the resistance of a material? Check all that apply.

shutvik [7]

Answer: Length, thickness, and temperature

Explanation:

I did it

6 0

3 years ago

A gun is fired with angle of elevation 30°. What is the muzzle speed if the maximum height of the shell is 544 m? (Round your an

Artemon [7]

Answer:

206.62313 m/s

Explanation:

u = Muzzle speed

g = Acceleration due to gravity = 9.8 m/s²

$\theta$ = Angle at which the bullet is fired = 30°

h = Maximum height = 544 m

Maximum height is given by

$h=\dfrac{u^2sin^2\theta}{2g}\\\Rightarrow u=\sqrt{\dfrac{2gh}{sin^2\theta}}\\\Rightarrow u=\sqrt{\dfrac{2\times 9.81\times 544}{sin^2(30)}}\\\Rightarrow u=206.62313\ m/s$

The muzzle speed is 206.62313 m/s

6 0

4 years ago

1. An electron in an atom absorbs a photon with an energy of 2.38 eV and jumps from the n = 2 to n = 4 energy level in the atom.

oksian1 [2.3K]

1. $1.0\cdot 10^{-6}m$

First of all, let's convert the energy of the absorbed photon into Joules:

$E=2.38 eV \cdot (1.6\cdot 10^{-19}J/eV)=1.98\cdot 10^{-19} J$

The energy of the photon can be rewritten as:

$E=\frac{hc}{\lambda}$

where

h is the Planck constant

c is the speed of light

$\lambda$ is the wavelength of the photon

Re-arranging the formula, we can solve to find the wavelength of the absorbed photon:

$\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{1.98\cdot 10^{-19} J}=1.0\cdot 10^{-6}m$

2. 1.24 eV

In this case, when the electron jumps from the n=4 level to the n=3 level, emits a photon with wavelength

$\lambda=1.66\cdot 10^{-6}m$

So the energy of the emitted photon is given by the formula used previously:

$E=\frac{hc}{\lambda}$

and using

$\lambda=1.66\cdot 10^{-6}m$

we find

$E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{1.0\cdot 10^{-6}m}=1.99\cdot 10^{-19}J$

converting into electronvolts,

$E=\frac{1.99\cdot 10^{-19} J}{1.6\cdot 10^{-19} J/eV}=1.24eV$

EDIT: an issue in Brainly does not allow me to add the last 2 parts of the solution - I have added them as an attachment to this post, check the figure in attachment.