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Valentin [98]
3 years ago
9

A 2000kg car experiences a braking force of 10000N and skids to a stop in 6 seconds. The speed of the car just before the brakes

were applied was
A) 45 m/s
B) 15 m/s
C) 1.2 m/s
D) 30 m/s
E) none of these
So far I got a= 5m/s^2 and u= 30m/s but I think the answer is 45 m/s but I do not know where to go from there.
Physics
1 answer:
mylen [45]3 years ago
4 0

Answer:

D. 30m/s

Explanation:

According to the equation of motion

v=u +at

v = Final velocity

u = initial velocity

a = acceleration

t = time taken

Since we don't know the acceleration of the body, we will find its acceleration using newtons first law of motion F = ma

10000 =2000a

a = 5m/s²

initial velocity = 0m/s since the body starts from rest

v is what we are looking for

t = 6seconds

Substituting this datas in the formula

v = 0+ 5(6)

v = 30m/s (D)

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protons in an atomic nucleus are typically 10-15 m apart. what is the electric force of repulsion between nuclear protons?
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onsider 1000 mL of a 1.00 × 10-4 M solution of a certain acid HA that has a Ka value equal to 1.00 × 10-4. Water was added or re
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Answer:

The volume of the final solution, V = 0.0305L

Explanation:

Number of moles = Concentration * volume

Concentration of HA = 1.00 * 10⁻⁴M

Volume of HA = 1000mL = 1 L

Number of moles of HA =  1.00 * 10⁻⁴ * 1

Number of moles of HA =  1.00 * 10⁻⁴ mols

Equation of reaction:

HA → H⁺  +  A⁻

If 1 mol of HA produces 1 mol of H⁺  and  A⁻, 1.00 * 10⁻⁴ mol of HA will produce 1.00 * 10⁻⁴ mol of  H⁺  and  A⁻.

Since only 16% dissociation occurs = 0.16

Number of moles of  H⁺ produced = 0.16 *  1.00 * 10⁻⁴

Number of moles of  H⁺ produced = 1.6 * 10⁻⁵mols

Number of moles of  A⁻ produced = 0.16 *  1.00 * 10⁻⁴

Number of moles of  A⁻ produced = 1.6 * 10⁻⁵mols

Since 16% of HA dissociated into  H⁺  and  A⁻, 84% of HA is left

Number of mols of HA left = 0.84 *  1.00 * 10⁻⁴

Number of mols of HA left =  8.4 * 10⁻⁵mols

Concentration = num of moles/volume

Let the volume of the final solution be V

Conc of HA = 8.4 * 10⁻⁵/V

Conc of H⁺ = 1.6 * 10⁻⁵/V

Conc of A⁻ =  1.6 * 10⁻⁵/V

To calculate the dissociation constant

k_{a} = [H^{+} ][A^{-} ]/[HA]

k_{a}= [1.6 * 10^{-5} /V][1.6 * 10^{-5} /V]/[8.4 * 10^{-5} /V]\\k_{a}= 3.05 * 10^{-6} /V\\k_{a} = 1.00 * 10^{-4}\\ 1.00 * 10^{-4} = 3.05 * 10^{-6} /V\\V= 3.05 * 10^{-6}/ 1.00 * 10^{-4}\\V=3.05 * 10^{-2}\\V=0.0305 L

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