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den301095 [7]
3 years ago
5

A car moving south speeds up from 10 m/s to 40 m/s in 15 seconds. What is the car’s acceleration? A. 2 m/s2 B.15 m/s2 C.30 m/s2

D.50 m/s2
Physics
2 answers:
Ainat [17]3 years ago
6 0
A. 2 m/s^2

40-10=30
30÷15 =2
Black_prince [1.1K]3 years ago
5 0

Answer:

Acceleration, a=2\ m/s^2

Explanation:

It is given that,

A car moving south speeds up from 10 m/s to 40 m/s in 15 seconds.

It means,

Initial velocity of the car, u = 10 m/s

Final velocity of the car, v = 40 m/s

Time taken, t = 15 seconds

We have to find the acceleration of the car. Mathematically, acceleration of the car is given by taking difference in velocities divided by time taken i.e.

a=\dfrac{v-u}{t}

a=\dfrac{40\ m/s-10\ m/s}{15\ s}

a=2\ m/s^2

Hence, this is the required solution.

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Answer:

\rm KE\pm \Delta KE = 17.6\pm 6.8\ J.

Explanation:

<u>Given:</u>

  • Mass, \rm m\pm\Delta m = 1.3\pm 0.4\ kg.
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where,

\rm \Delta m,\ \Delta v are the uncertainties in mass and velocity respectively.

The kinetic energy is given by

\rm KE = \dfrac 12 mv^2 = \dfrac 12 \times 1.3\times 5.2^2=17.576\approx 17.6\ J.

The uncertainty in kinetic energy is given as:

\rm \dfrac{\Delta KE}{KE}=\dfrac{\Delta m}{m}+\dfrac{2\Delta v}{v}\\\dfrac{\Delta KE}{17.6}=\dfrac{0.4}{1.3}+\dfrac{2\times 0.2}{5.2}\\\dfrac{\Delta KE}{17.6}=0.384\\\Rightarrow \Delta KE = 17.6\times 0.384 = 6.7854\ J\approx6.8\ J\\\\Thus,\\\\KE\pm \Delta KE = 17.6\pm 6.8\ J.

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