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Ira Lisetskai [31]
3 years ago
9

A 1.75 kg stone falls from the top of a cliff and strikes the ground at a speed of 58.8 m/s. What is the height of the cliff??

Physics
1 answer:
Reil [10]3 years ago
6 0

Answer:

172.9m

Explanation:

h = 1/2 gt^

First calculat for t using

v = gt

t = v/g = 58.8/10

= 5.88secs

now h = 1/2 x 10 x 5.88^2

h =1/2 x 10 x 34.57

= 345.74/2

= 172.9m

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Rubidium forms the positive ion Rb+. Does this ion have more or fewer electrons than the neutral atom?
ohaa [14]

Answer:

The ion has fewer electrons

Explanation:

As Rubidium forms a positive ion Rb⁺ the number of electrons becomes fewer for the atom.

  • A positive charge on an ion indicates that the number of protons for the atom is greater than the number of electrons.
  • This leaves a net positive charge on the atom.
  • For a negatively charged ion, the number of electrons is more than the number of protons because they have gained additional electrons.
  • In a neutral atom, the number of protons and electrons are the same.
7 0
3 years ago
Dividing up scarce resources refers to resource _____. conservation costs distribution/9919485/7cdfc77f?utm_source=registration
stiks02 [169]
The answer is c. Distrbution
8 0
3 years ago
12 seconds after starting from rest a frewly falling cantaloupe has a speed of
adoni [48]

Answer:

<em>The cantaloupe has a speed of 117.6 m/s</em>

Explanation:

<u>Free Fall Motion</u>

It occurs when an object falls under the sole influence of gravity. Any object that is being acted upon solely by the force of gravity is said to be in a state of free fall. Free-falling objects do not face air resistance.

If an object is dropped from rest in a free-falling motion, it falls with a constant acceleration called the acceleration of gravity, which value is g = 9.8 m/s^2.

The final velocity of a free-falling object after a time t is given by:

vf=g.t

The cantaloupe has been dropped from rest. We are required to find the speed after t=12 seconds.

Calculate the final speed:

vf=9.8 * 12 = 117.6 m/s

The cantaloupe has a speed of 117.6 m/s

3 0
3 years ago
If a person weighs 1000 N on the surface of the Earth, how much will they weigh on the surface of the Moon? What “g” value shoul
Dafna1 [17]

Answer:

Hence the weight of the person on the moon is 162.4, and the value of g used is 1.624 m/s²

Explanation:

from the question,

W = mg........................ Equation 1

Where W = weight of the man on Earth, m = mass of the man, g = acceleartion due to gravity of the man

make m the subject of the equation

m = W/g.............. Equation 2

Given: W = 1000 N,

Constant: g = 10 m/s²

Therefore,

m = 1000/10

m = 100 kg

Weight on the moon

W' = mg'

W' = 100(1.624)

W' = 162.4 N.

Hence the weight of tthe person on the moon is 162.4, and the value of g used is 1.624 m/s²

5 0
3 years ago
The gravitational force law, deduced by Newton in the 1660's, is remarkably similar to Coulomb's law. Recall that the universal
viktelen [127]

Answer:

a.F=7.83\times 10^{-51} N

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F=\frac{GM_1M_2}{R^2}

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Mass of an electron,M_1=9.11\times 10^{-31} kg

Mass of proton,M_2=1836\times 9.11\times 10^{-31} kg

Distance between electron and proton,R=3.602nm=3.602\times 10^{-9} m

1nm=10^{-9} m

a.Substitute the values then  we get

F=\frac{6.67\times 10^{-11}\times 9.11\times 10^{-31}\times 1836\times 9.11\times 10^{-31}}{(3.602\times 10^{-9})^2}

F=7.83\times 10^{-51} N

b.We know that like charges repel to each other and unlike charges attract to each other.

Proton and electron are unlike charges therefore, the force between proton and electron is attractive.

4 0
3 years ago
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