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/s

Question

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

in progress 0
Thiên Hương 5 years 2021-07-16T09:48:15+00:00 2 Answers 54 views 1

Answers ( )

  1. Calantha
    0
    2021-07-16T09:49:22+00:00
    Reply

    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 N
    • Mass 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 !

  2. thanhha
    0
    2021-07-16T09:49:22+00:00
    Reply

    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)

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )