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DiKsa [7]
3 years ago
7

A force that keeps an object at a fixed angle

Physics
1 answer:
Helga [31]3 years ago
3 0
I'm pretty sure it's called, "gravity". or gravy :) 
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Consider a uniform hoop of radius r and mass m rolling without slipping. which is larger, its translational kinetic energy or it
OlgaM077 [116]
<span>translational kinetic energy is larger than its rotational kinetic energy</span>
3 0
3 years ago
An electron is released from rest at a distance of 6.00 cm from a proton. If the proton is held in place, how fast will the elec
lana66690 [7]

Answer:

91.87 m/s

Explanation:

<u>Given:</u>

  • x = initial distance of the electron from the proton = 6 cm = 0.06 m
  • y = initial distance of the electron from the proton = 3 cm = 0.03 m
  • u = initial velocity of the electron = 0 m/s

<u>Assume:</u>

  • m = mass of an electron = 9.1\times 10^{-31}\ kg
  • v = final velocity of the electron
  • e = magnitude of charge on an electron = 1.6\times 10^{-19}\ C
  • p = magnitude of charge on a proton = 1.6\times 10^{-19}\ C

We know that only only electric field due to proton causes to move from a distance of 6 cm from proton to 3 cm distance from it. This means the electric force force does work on the electron to move it from one initial position to the final position which is equal to the change in potential energy of the electron due to proton.

Now, according to the work-energy theorem, the total work done by the electric force on the electron due to proton is equal to the kinetic energy change in it.

\therefore \textrm{Kinetic energy change}= \textrm{Change in potential energy}\\\Rightarrow \dfrac{1}{2}m(v^2-u^2)= \dfrac{kpe}{y}-\dfrac{kpe}{x}\\\Rightarrow \dfrac{1}{2}m(v^2-(0)^2)= \dfrac{kpe}{0.03}-\dfrac{kpe}{0.06}\\\Rightarrow \dfrac{1}{2}mv^2= \dfrac{100kpe}{3}-\dfrac{100kpe}{6}\\\Rightarrow \dfrac{1}{2}mv^2= \dfrac{100kpe}{6}\\

\Rightarrow v^2= \dfrac{100kpe\times 2}{6m}\\\Rightarrow v^2= \dfrac{100kpe}{3m}\\\Rightarrow v^2= \dfrac{100\times 9\times 10^9\times 1.6\times 10^{-19}\times 1.6\times 10^{-19}}{3\times 9.1\times 10^{-31}}\\\Rightarrow v^2=8.44\times 10^3\\\Rightarrow v=91.87\ m/s\\

Hence, when the electron is at a distance of c cm from the proton, it moves with a velocity of 91.87 m/s.

8 0
3 years ago
why does diving 30m below sea level affect our bodies more than being in a building 30m above sea level
Genrish500 [490]
Imagine you are in a swimming pool 30m deep. Assuming you know that water is denser than air, you would know that the 30m of water above you will carry more weight, and press down on your body. Say you were in a swimming pool 60m deep, you would be sandwiched between 30m of water pressing down on you, and the upthrust created by the 30m of water below you.

In a building 30m up, the pressure will be regulated, as you are in a building. The floor will be strong enough to support the weight of the body, and the body will not recoil into itself.
5 0
3 years ago
When Dr. Montero was observing the endoplasmic reticulum of a cell using an electron microscope, she noticed that it was covered
Vesnalui [34]
The answer is Convoluted endoplasmic reticulum
7 0
3 years ago
An acorn falls from an oak tree. You note that it
Vaselesa [24]

Answer:9.8 m/s²

Explanation:

It was going at 9.8m/s² as this is the acceleration of an object due to gravity

when an object falls it accelerates at a consant and uniform speed which is 9.8m/s²

6 0
3 years ago
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