Ask question

Ask question

All categories

konstantin123 [22]

3 years ago

5

A runner stood at the 100-meter mark on a track. When the timer started, the runner sprinted north to the 200-meter mark. It too

k the runner 25 seconds to get to the 200-meter mark. What was the average velocity of the runner while he was sprinting?

2 answers:

Olegator [25]3 years ago

3 0

Answer:

8 m/s north

Explanation:

Eddi Din [679]3 years ago

3 0

The average velocity of the runner while he was sprinting is 4 m/s.

<u>Solution:</u>

Given the runner sprinted from 100m mark to 200m mark, which means the displacement of the runner is 200-100=100m.

The time taken to travel this 100 distance is 25 seconds.

$\text { Average velocity }=\frac{\text { displacement }}{\text { time }}$

$\Rightarrow \frac{100}{25}=4 \mathrm{m} / \mathrm{sec}$

Note: The direction of the average velocity is in the direction of the displacement

Therefore, average velocity = 4 m/s

Average velocity:

Average velocity of an object can be defined as the displacement divided by the time taken by the object.

You might be interested in

According to the iupac convention, alkyl substituents on a hydrocarbon chain should be listed in which order?

saveliy_v [14]

According to the iupac convention, alkyl substituents on a hydrocarbon chain should be listed in alphabetical without taking prefixes into name.

A hydrocarbon would be an organic molecule in organic chemistry that is made completely of hydrogen as well as carbon. A good example of group 14 hydrides includes hydrocarbons. In general, hydrocarbons lack color and are hydrophobic. Typically, their faint smells are compared to that of gasoline and lighter fluid.

The longest unbroken carbon chain should be found and named.
Name the groups that make up this chain but instead identify them.
Beginning at the end of the chain closest to a substituent group, fraction the chain sequentially.
Give each substituent group a corresponding number but also name to indicate its placement.

Therefore, according to the iupac convention, alkyl substituents on a hydrocarbon chain should be listed in alphabetical without taking prefixes into name.

To know more about hydrocarbon

brainly.com/question/12433190

#SPJ4

8 0

1 year ago

What is the percent by mass of potassium in KFe(CN)?

vladimir1956 [14]

Answer:

Explanation:

1. Add the atomic mass of all the elements.

39+55.8+12+14= 120.8

2.Divide atomic mass of potassium by total atomic mass

$39/120.8= .323$

2. Multiply by 100

$.323*100= %32.3$ %32.3

8 0

3 years ago

Monel metal is a corrosion-resistant copper-nickel alloy used in the electronics industry. A particular alloy with a density of

klemol [59]

<span>To find the volume of the plate without accounting for the hole firstly
V = (15.0 cm)(12.5 cm)(0.250 cm) = 46.875 cm^3
and the volume of the hole is
(pi)(1.25 cm)^2(0.250 cm) = 1.2272 cm^3
we will subtract the volume of the hole from the rest 45.648 cm^3
the multiply this by the density of the alloy to find the mass
(8.80 g/cm^3)(45.648 cm^3) = 401.701 g.
0.044% of this is Si, so (0.00044)(401.701 g) = 0.17675 g is silicon.
by the number of atoms and using average atomic mass of silicon and Avogadro's number to find the number of silicon atoms:
(0.17675 g)(1 mol/28.0855 g)(6.022E23 atoms/1 mol) =3.794E21atoms of Si
3.10% of these are Si-30:(0.0310)(3.794E18 atoms)=1.176E20 atoms of Si-30 and with two significant figures, 1.2E20 atoms.
hope this helps
</span>

4 0

3 years ago

At 298 K, the osmotic pressure of a glucose solution (C6H12O6 (aq)) is 12.1 atm. Calculate the freezing point of the solution. T

Anarel [89]

<u>Answer:</u> The freezing point of solution is -0.974°C

<u>Explanation:</u>

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

where,

$\pi$ = osmotic pressure of the solution = 12.1 atm

i = Van't hoff factor = 1 (for non-electrolytes)

M = molarity of solute = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the solution = 298 K

Putting values in above equation, we get:

$12.1atm=1\times M\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\M=\frac{12.1}{1\times 0.0821\times 298}=0.495M$

This means that 0.495 moles of glucose is present in 1 L or 1000 mL of solution

To calculate the mass of solution, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of solution = 1.034 g/mL

Volume of solution = 1000 mL

Putting values in above equation, we get:

$1.034g/mL=\frac{\text{Mass of solution}}{1000mL}\\\\\text{Mass of solution}=(1.034g/mL\times 1000mL)=1034g$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of glucose = 0.495 moles

Molar mass of glucose = 180.16 g/mol

Putting values in above equation, we get:

$0.495mol=\frac{\text{Mass of glucose}}{180.16g/mol}\\\\\text{Mass of glucose}=(0.495mol\times 180.16g/mol)=89.18g$

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

The equation used to calculate depression in freezing point follows:

$\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Freezing point of pure solution = 0°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal freezing point elevation constant = 1.86°C/m

$m_{solute}$ = Given mass of solute (glucose) = 89.18 g

$M_{solute}$ = Molar mass of solute (glucose) = 180.16 g/mol

$W_{solvent}$ = Mass of solvent (water) = [1034 - 89.18] g = 944.82 g

Putting values in above equation, we get:

$0-\text{Freezing point of solution}=1\times 1.86^oC/m\times \frac{89.18\times 1000}{180.16g/mol\times 944.82}\\\\\text{Freezing point of solution}=-0.974^oC$

Hence, the freezing point of solution is -0.974°C

8 0

3 years ago

How many molecules are in 68.0 g of H2S

Kryger [21]

The correct answer is 12.044 × 10²³ molecules.

The molecular mass of H₂S is 34 gram per mole.

Number of moles is determined by using the formula,

Number of moles = mass/molecular mass

Given mass is 68 grams, so no of moles will be,

68/34 = 2 moles

1 mole comprises 6.022 × 10²³ molecules, therefore, 2 moles will comprise = 6.022 × 10²³ × 2

= 12.044 × 10²³ molecules.

3 0

3 years ago