How much force does it take to bring a 1,050 N car from rest to a velocity of 42 m/s in 13 seconds?

Physics

1 answer:

frozen [14]3 years ago

5 0

Answer:

F = 339.23 N

Explanation:

Weight of a car, W = 1050 N

Initial velocity, u = 0

Final velocity, v = 42 m/s

Time, t = 13 s

The weight of an object is given by :

W = mg

g is the acceleration due to gravity

$m=\dfrac{W}{g}\\\\m=\dfrac{1050}{10}\\\\m=105\ kg$

The force acting on car is :

$F=ma\\\\F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{105\times (42-0)}{13}\\\\F=339.23\ N$

So, the force acting on the car is 339.23 N.

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3 years ago

The period of a pendulum is measured 16 times. The average value of the period over these 16 trials is calculated to be 1.50 sec

marin [14]

Answer:

The additional trials needed is 48 trials

Explanation:

Given;

initial number of trials, n = 16 trials

the standard deviation, σ = 0.24 s

initial standard error, ε = 0.06 s

The standard error is given by;

$\epsilon = \frac{\sigma}{\sqrt{n} }$

To reduce the standard error to 0.03 s, let the additional number of trials = x

$0.03= \frac{0.24}{\sqrt{n+x} } \\\\0.03= \frac{0.24}{\sqrt{16+x} }\\\\0.03\sqrt{16+x} = 0.24\\\\\sqrt{16+x} = \frac{0.24}{0.03} \\\\\sqrt{16+x} = 8\\\\16+x = 8^2\\\\16+x = 64\\\\x = 64 -16\\\\x = 48 \ trials$