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2 years ago

How does air resistance affect the velocity of a falling object ?

1 answer:

4 0

Velocity the speed of something in a given direction.
Air resistance is like friction in air, it slows objects down.
When an object is moving faster, which would mean higher velocity, there is more air resistance trying to slow it down and keeping it from moving.

So in short, the more velocity there is, the greater the air resistance. And no matter what the velocity is, air resistance slows falling objects down.

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Find the area of a vegetable garden that is 4.83m long and 3.4m wide. Give your answer to the correct significant figure. (use t

sergejj [24]

Answer:

The area of vegetable garden is 16.422 m².

Explanation:

Given data:

Length = 4.83 m

Width = 3.4 m

Area = ?

Solution:

Formula:

Area = length × width

Area is measured in square unit.

By putting the values in formula:

Area = length × width

Area = 4.83 m × 3.4 m

Area = 16.422 m²

The area of vegetable garden is 16.422 m².

5 0

3 years ago

How to determine reaction order in reactant?

Artyom0805 [142]

In order to determine, Order of reaction, we have to add all the exponents written in the Chemical form, on the Reactant species.

Hope this helps!

8 0

3 years ago

Find the momentum of 5 kg bowling ball at 10 m/s.

leva [86]

30 kg m/s

momentum = mass x velocity = 10 x 3 m/s =30 kg m/s

5 0

3 years ago

What do you add to a base to weaken it

lesantik [10]

Answer:

you dont do anything because then you are supporting it

if u add something

Explanation:

6 0

3 years ago

When methane ( CH4 ) burns, it reacts with oxygen gas to produce carbon dioxide and water. The unbalanced equation for this reac

slega [8]

<h2>Answer:</h2> $1.33*10^{-2}grams$

<h2>Explanations</h2>

The complete balanced equation for the given reaction is expressed as;

$CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)$

Given the following parameters

Mass of CH4 = 5.90×10^−3 g = 0.0059grams

Determine the moles of methane

$\begin{gathered} moles\text{ of CH}_4=\frac{mass}{molar\text{ mass}} \\ moles\text{ of CH}_4=\frac{0.0059}{16.04} \\ moles\text{ of CH}_4=0.000368moles \end{gathered}$

According to stoichimetry, 1 mole of methane produces 2 moles of water, hence the moles of water required will be:

$\begin{gathered} moles\text{ of H}_2O=\frac{2}{1}\times0.000368 \\ moles\text{ of H}_2O=0.000736moles \end{gathered}$

Determine the mass of water produced

$\begin{gathered} Mass\text{ of H}_2O=moles\times molar\text{ mass} \\ Mass\text{ of H}_2O=0.000736\times18.02 \\ Mass\text{ of H}_2O=0.0133grams=1.33\times10^{-2}grams \end{gathered}$

Therefore the mass of water produced from the complete combustion of 5.90×10−3 g of methane is 1.33 * 10^-2grams

5 0

1 year ago