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Lyrx [107]
3 years ago
9

A runner is training and runs past meter mark 65 and starts a stopwatch. After the runner passes meter mark 15 the stopwatch sto

ps at 8 seconds.
What was the runners velocity?
Question options:
-6.25 m/s
6.25 m/s
8.13 m/s
-1.88 m/s
Physics
1 answer:
denis-greek [22]3 years ago
7 0

Answer:

6.25 m/s

Explanation:

Given parameters

Initial mark = 65

Final mark  = 15

time taken = 8s

Unknown:

Velocity of the runner  = ?

Solution:

Velocity is the displacement divided by time;

        Velocity  = \frac{displacement}{time}  

Displacement  = Initial mark - final mark  = 65 - 15 = 50m

Now, insert the parameters and solve;

       Velocity  = \frac{50}{8}   = 6.25m/s

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For every action, there is an equal and opposite reaction.

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How many carbon atoms are there in 5 molecules of CHCl3?
lukranit [14]

There are 5 carbon atoms present in the Chemical formula, CHCl_\\{3} .

<h3>What is a chemical formula?</h3>

A chemical formula is used to show the number and type of atoms present in a molecule, using the atomic symbols of the elements and numerical subscripts.

From the chemical formula;

We can deduce that we have the following elements and their corresponding number of atoms in one molecule of the compound;

Carbon = 1 atom

Hydrogen = 1 atom

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So, in 5 molecules of the compound;

Carbon = 1 x 5 = 5 atoms

Hydrogen = 1 x 5 = atoms

Chlorine = 3 x 5 atoms = 15 atoms

In conclusion, from the solution above, we can say that we have 5 atoms of carbon present in 5 molecules of CHCl_3.

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5 0
1 year ago
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3 years ago
(II) A 1200-kg car moving on a horizontal surface has speed v = 85 km/h when it strikes a horizontal coiled spring and is brough
Pie

Answer:

k = 138440 N/m

Explanation:

given,

Mass of the car = 1200 Kg

speed of the car = 85 km/h

                            = 85 x 0.278 = 23.63 m/s

KE of the car

KE = \dfrac{1}{2} mv^2

Using Spring Energy Formula

KE_{spring}=\dfrac{1}{2}kx^2

Using conservation of energy

\dfrac{1}{2} mv^2=\dfrac{1}{2}kx^2

k = \dfrac{mv^2}{x^2}

k = \dfrac{1200\times 23.63^2}{2.2^2}

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Spring stiffness constant of the spring is equal to k = 138440 N/m

7 0
3 years ago
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