Determine how much farther a person can jump on the moon as compared to the earth if take off speed and angle are the same

1 answer:

8 0

From Newton's Three Laws of Motion, derived formulas are already conveniently presented for a rectilinear motion at constant acceleration. One of its equations is

y = (Vf² - Vi²)/2a, where
y is the vertical height travelled by the object
Vf is the final velocity
Vi is the initial velocity
a is the acceleration

Now, when a man jumps, the only force acting on him is gravity pulling him down. When he reaches his maximum height, eventually his velocity will reach zero. So, Vf = 0. Suppose all parameters with subscript 1 refers to man jumping on Earth and those with subscript 2 refers to the man jumping on moon. Since initial velocity and angle is said to be the same, when we find the ratio of x₂/x₁, the terms (Vf²-Vi²) cancels out leaving us with

x₂/x₁ = a₂/a₁

It is common knowledge that gravity on Earth is 9.81 m/s². According to literature, the gravity on the moon is 1.62 m/s². Thus,

x₂/x₁ = a₁/a₂ = 9.81/1.62 = 6
x₂ = 6x₁

Therefore, the man jumping on the moon can reach 6 times higher than in Earth.

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Calculate the change in the energy of an electron that moves from the n = 3 level to the n = 2 level. What type of light is emit

marissa [1.9K]

Answer:

Red light

Explanation:

The energy emitted during an electron transition in an atom of hydrogen is given by

$E=E_0 (\frac{1}{n_2^2}-\frac{1}{n_1^2})$

where

$E_0 = 13.6 eV$ is the energy of the lowest level

n1 and n2 are the numbers corresponding to the two levels

Here we have

n1 = 3

n2 = 2

So the energy of the emitted photon is

$E=(13.6) (\frac{1}{2^2}-\frac{1}{3^2})=1.9 eV$

Converting into Joules,

$E=(1.9 eV)(1.6\cdot 10^{-19} J/eV)=3.0\cdot 10^{-19} J$

And now we can find the wavelength of the emitted photon by using the equation

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

where h is the Planck constant and c is the speed of light. Solving for $\lambda$ ,

$\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34})(3\cdot 10^8)}{3.0\cdot 10^{-19}}=6.63\cdot 10^{-7} m = 663 nm$

And this wavelength corresponds to red light.

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2 years ago

Determine the field strength, E, experienced by a test charge, q, if a charge of 7.0 × 10-5 coulombs is placed on q and a force

kkurt [141]

Formula for feild strength= F/q
q=7.0^10-5 coulombs
F=5.2 N
E=5.2 / 7.0^10-5
E=

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3 years ago

Read 2 more answers

Both the lens and the cornea of the eye have a primary function of

3241004551 [841]

Answer: B. bending light

Explanation:

The phenomenom of vision in human eye is thanks to refraction (when light changes its direction as it passes through one medium to another), and this is what the cornea and the lens do.

When the ray of light encounters the eye, the first thing it finds is the cornea, which bends this ray and begins to form an image, then light passes through the pupil, which is in charge of regulating the amount of light that enters in the eye.

After light travels through pupil it passes through the lens, where the rays of light change the direction again in order to focus the formed image on the retina.

At this point it is important to note the formed image is downward, then the retina transforms light into electrical impulses that are sent to the brain through the optic nerve and finally the brain interprets these messages, and forms a right upward image.

In the image attached these parts can be seen.