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WITCHER [35]
1 year ago
11

using a lever to move a rock reduces the force a person must exert. what does it do to the work the person must do?

Physics
1 answer:
morpeh [17]1 year ago
3 0

Using a lever to move a rock reduces the force a person must exert . It also reduces the work (effort ) the person must do.

Using a lever decreases the input force required to lift up the load, for example if a force of 900N is required to lift the load without a lever . Using the lever the input force decreases to 500N and as we all know by the formulae of work done :

Work done ∝  force applied

By this equation we can conclude that if the force to be applied on the load reduces by the use of a lever therefore the work to be done on the load will also reduce .

You can get a additional information about the lever system below :

brainly.com/question/3522307

#SPJ4

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C. turtles with genes for long necks had a better chance of surviving to reach reproductive age.

Explanation:

The turtles that had long necks were more fit to the environmnet in which they were lovated and were able to grow larger and have more reproductive time because of their ability to feed on grass and small shrubs, this helped them always haev food available, and made them the dominant gene eventually.

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3 years ago
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Find the rms speed of the molecules of a sample of n2 (diatomic nitrogen) gas at a temperature of 31.5°c.
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The rms speed can be calculated using the following rule:
rms = sqrt ((3RT) / (M)) where:
R is the gas constant =  8.314 J/mol-K
T is the temperature = 31.5 + 273 = 304.5 degrees kelvin
M is the molar mass = 2*14 = 28 grams = 0.028 kg

Substitute with the givens to get the rms speed as follows:
rms speed = sqrt [(3*8.314*304.5) / (0.028)] = 520.811 m/sec
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3 years ago
A train travels 100 km east and then 75 km west. What is the magnitude of<br> its displacement?
tamaranim1 [39]

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6 0
3 years ago
If a proton and an electron are released when they are 7.00×10−10 m apart (typical atomic distances), find the initial accelerat
Ugo [173]

Answer:

The acceleration of the proton is 2.823 x 10¹⁷ m/s²

The acceleration of the electron is 5.175 x 10²⁰ m/s²

Explanation:

Given;

distance between the electron and proton, r = 7 x 10⁻¹⁰ m

mass of proton, m_p = 1.67 x 10⁻²⁷ kg

mass of electron, m_e = 9.11 x 10⁻³¹ kg

The attractive force between the two charges is given by Coulomb's law;

F = \frac{k(q_p)(q_e)}{r^2}

where;

k is Coulomb's constant = 9 x 10⁹ Nm²/c²

F = \frac{k(q_p)(q_e)}{r^2} \\\\F = \frac{(9*10^9)(1.602*10^{-19})(1.602*10^{-19})}{(7*10^{-10})^2} \\\\F = 4.714 *10^{-10} \ N

Acceleration of proton is given by;

F = ma

F = m_pa_p\\\\a_p = \frac{F}{m_p}\\\\a_p = \frac{4.714*10^{-10}}{1.67*10^{-27}}\\\\a_p = 2.823 *10^{17} \ m/s^2

Acceleration of the electron is given by;

F = m_ea_e\\\\a_e = \frac{F}{m_e}\\\\a_e = \frac{4.714*10^{-10}}{9.11*10^{-31}}\\\\a_e = 5.175 *10^{20} \ m/s^2

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I believe the answer is free electrons
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