The distance an object falls from rest through gravity is
D = (1/2) (g) (t²)
Distance = (1/2 acceleration of gravity) x (square of the falling time)
We want to see how the time will be affected
if ' D ' doesn't change but ' g ' does.
So I'm going to start by rearranging the equation
to solve for ' t '.
D = (1/2) (g) (t²)
Multiply each side by 2 : 2 D = g t²
Divide each side by ' g ' : 2 D/g = t²
Square root each side: t = √ (2D/g)
Looking at the equation now, we can see what happens
to ' t ' when only ' g ' changes:
-- ' g ' is in the denominator; so bigger 'g' ==> shorter 't'
and smaller 'g' ==> longer 't' .
-- They don't change by the same factor, because 1/g is inside
the square root. So 't' changes the same amount as √1/g does.
Gravity on the surface of the moon is roughly 1/6 the value
of gravity on the surface of the Earth.
So we expect ' t ' to increase by √6 = 2.45 times.
It would take the same bottle (2.45 x 4.95) = 12.12 seconds
to roll off the same window sill and fall 120 meters down to the
surface of the Moon.
Normal force, friction force, gravitational force
To solve this problem it is necessary to apply an energy balance equation in each of the states to assess what their respective relationship is.
By definition the energy balance is simply given by the change between the two states:
Our states are given by
In this way the energy balance for the states would be given by,
Therefore the states of energy would be
Lowest : 0.9eV
Middle :7.5eV
Highest: 8.4eV
Answer: 4.17m
Explanation:
The observer at C will hear a sound on no sound upon whether the interference is constructive or destructive.
If the listeners hears sounds it is caled constructive interference but if he hears no sound its called destructive interference.
So
d2 - d1 = (n *lamba)/ 2
Where n=1,3,5
lamda=v/f =349/62.8
lamda=5.56m
d2= d1 + nlamda/2
d2= 1 + 5.56/2
d2= 3.78m
X'= 1 cos 60= 0.5m
Y= 1 sin60= 0.866m
X"^2 + Y^2 =d2^2
X" =√(y^2 - d2^2)
X"=√(3.78^2 - 0.886^2)
X"= 3.67m
So therefore the closest that speaker A can be to speaker B so the listener does not hear any sound is X' + X"= 0.5 + 3.67
4.17m
The weight of a column of air with cross-sectional area 4. 5 m^2 extending from earth's surface to the top of the atmosphere is, 4.56*10^5N.
To find the answer, we have to know about the pressure.
<h3>How to find the weight of a column of air?</h3>
- As we know that the expression of pressure as,
where; F is the force, here it is equal to the weight of the air column, and A is the area of cross section.
- It is given that, the air column is extending from earth's surface to the top of the atmosphere, thus, the pressure will be atmospheric pressure,
- From this, the value of weight will be,
Thus, we can conclude that, the weight of a column of air with cross-sectional area 4. 5 m^2 extending from earth's surface to the top of the atmosphere is, 4.56*10^5N.
Learn more about the pressure here:
brainly.com/question/12830237
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