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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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Which of the following statement(s) about chemical bonding

spin [16.1K]

Answer:B

Explanation:

6 0

2 years ago

When 136g of glycine are dissolved in of a certain mystery liquid , the freezing point of the solution is lower than the freezin

loris [4]

The given question is incomplete. The complete question is:

When 136 g of glycine are dissolved in 950 g of a certain mystery liquid X, the freezing point of the solution is 8.2C lower than the freezing point of pure X. On the other hand, when 136 g of sodium chloride are dissolved in the same mass of X, the freezing point of the solution is 20.0C lower than the freezing point of pure X. Calculate the van't Hoff factor for sodium chloride in X.

Answer: The vant hoff factor for sodium chloride in X is 1.9

Explanation:

Depression in freezing point is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=T_f^0-T_f=8.2^0C$ = Depression in freezing point

$K_f$ = freezing point constant

i = vant hoff factor = 1 ( for non electrolyte)

m= molality = $\frac{136g\times 1000}{950g\times 75.07g/mol}=1.9$

$8.2^0C=1\times K_f\times 1.9$

$K_f=4.32^0C/m$

Now Depression in freezing point for sodium chloride is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=20.0^0C$ = Depression in freezing point

$K_f$ = freezing point constant

m= molality = $\frac{136g\times 1000}{950g\times 58.5g/mol}=2.45$

$20.0^0C=i\times 4.32^0C\times 2.45$

$i=1.9$

Thus vant hoff factor for sodium chloride in X is 1.9

3 0

2 years ago

A 25.00 g solid sample of Ca(OH)2 was added into 1250 mL of 0.400 M HCl aqueous solution. The temperature of the solution was de

daser333 [38]

Answer:

86.735 kJ

Explanation:

Simply multiply the change in temperature by the Ccal;

(36.6 - 20.0)×5.225 = 86.735

5 0

2 years ago

What is the concentration of a solution of 10. Moles of copper nitrate in 5.0 liters of solution

Alja [10]

Answer:

10.5

Explanation:

well we added together so thats the result

3 0

2 years ago

How many significant figures does .150, 20.20, and 25.00 have?

viva [34]

.150 = 3
20.20 = 4
25.00 = 4
If a 0 does not have a number or a period after it, it is not significant.
If the 0 is behind the decimal point, it is always significant.

6 0

3 years ago