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.
Answer: By Contributor. Measuring the volume of an irregularly shaped object using geometry is often difficult and complicated. The easiest way to do this is by using the water displacement method. Often taught in chemistry or other science classes, this method is known for its simplicity and accuracy.
Explanation:
The pH of the buffer solution is 7.30 for 0.172 m in Hypochlorous acid and 0.131 m in Sodium hypochlorite.
<h3>What is a buffer solution?</h3>
A weak acid and its conjugate base, or a weak base and its conjugate acid, are mixed together to form a solution called a buffer solution, which is based on water as the solvent. They do not change in pH when diluted or when modest amounts of acid or alkali are added to them. A relatively little amount of a strong acid or strong base has little effect on the pH of buffer solutions. As a result, they are employed to maintain a steady pH.
According to the question:
Ka = 3.8×10⁻⁸
pKa = - log (Ka)
= - log(3.8×10⁻⁸)
= 7.42
pH = pKa + log {[conjugate base]/[acid]}
= 7.42+ log {0.131/0.172}
= 7.302
To know more about buffer solutions, visit:
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Downward force acting on the ball is 19.6N
Net force acting on the ball is 1960V N
<u>Explanation:</u>
<u />
Given:
Mass of the ball, m = 2kg
Density of ball, σ = 800 kg/m³
Density of water, ρ = 1000 kg/m³
Downward force acting by the ball in the vessel = mg
where, g = 9.8m/s²
F = 2 X 9.8
F = 19.6N
Net force acting on the ball:
Fnet = (ρ - σ) Vg
where,
V is the volume of water
Fnet = (1000 - 800) V X 9.8
Fnet = 1960V N
If the volume is known, then substitute the value of V to find the net force.
Thus, Downward force acting on the ball is 19.6N
Net force acting on the ball is 1960V N
