Physics, 04.10.2020 01:01 malaysiae6321

The net fedrce on a 12 kg cart is 3.2N backward. Ignoring friction, what is the acceleration of the car?
0.27 m/s2
3.8 m/s2
8.2 m/s2
18 m/s2
38m/s2

Answers: 2

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Answer from: angelinaranee15

We have,

Mass,m = 12 kg

Force,F = 3.2 N

To be calculated:-

Calculate the acceleration ( a ) .

Solution:-

We know that,

$\bf \: Force = Mass × Acceleration$

$\sf\implies \: 3.2 = 12 \: \times a$

$\sf \implies \: a = \dfrac{3.2}{12}$

$\sf \implies \: a = \dfrac{32}{12 \times 10}$

$\sf \implies \: a \: = \dfrac{8}{3 0}$

$\sf \implies \: a = 0.266$

$\sf \implies \: a = 0.27 \: m {s}^{ - 2} \: (approx)$

Answer from: matt16913

We are given with the amount of force applied and mass of a cart and we have to find the acceleration of the box.

Force applied = 3.2 NMass of the cart = 12 kg

According to Netwon's 2nd law,

After deriving, we will get that

Force = Mass × Accleration

Plugging the given values to get mass,

➝ 3.2 N = 12 kg × accleration

➝ Accleration = 3.2 / 12 m/s²

➝ Accleration = 0.27 m/s² (approx.)

Thus, the required answer is Option A.

Carry On Learning

Answer from: Quest

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Answer from: Quest

answer: at 24 february, 2014 the population gets twice of its initial population.

explanation:

since we have given that

the population for alpha city, t years after january 1, 2004 is given as

$p(t)=0.3t^2+6t+80$

first we find out the initial population, i.e. t=0,

so, our quadratic equation becomes,

$p(0)=0+0+80=80$

according to question, we have said that if the population has twice its initial population, then it becomes

$p(t)=2\times 80=160$

so, time taken to reach this above mentioned population is given by

$160=0.3t^2+6t+80\\\\160-80=0.3t^2+6t\\\\80=0.3t^2+6t\\\\0.3t^2+6t-80=0\\\\\text{using quadratic formula ,we get }\\\\t_{1,\: 2}=\frac{-60\pm \sqrt{60^2-4\cdot \: 3\cdot \: 800}}{2\cdot \: 3}\\\\t=-29.14\ and\ t=9.149$

now, we know that time can't be negative so, we take t=9.149 years.

hence, at 24 february, 2014 the population gets twice of its initial population.

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The net fedrce on a 12 kg cart is 3.2N backward. Ignoring friction, what is the acceleration of the...

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