What average net force is needed to accelerate a 1280 kg car to a speed of 29 m/s in a time of 7 s?

Answer 1

1280 kg has 3 sigfigs (the trailing 0 is ignored)

29 m/s has 2 sigfigs

7 s only has 1 sigfig

Be careful when writing your numbers as many of us are very careful when answering in the correct accuracy.

The answer would be totally different if you used:

1280 kg has 3 sigfigs (the trailing 0 is ignored)29.0 m/s has 3 sigfigs7.00 s has 3 sigfigs

Answer 2

Given,

Mass (m) = 1280kgSpeed (v) = 29m/secTime (t) = 7secNet average force (F) = ?We know that:-Force (F) = mass (m) x acceleration (a)= mass (m) x speed (v) / time (t)= 1280kg x 29m/sec / 7sec= 5.3 x 10³ x kg.m/sec²∴ Net average force (F) = 5.3 x 10³N.

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Answer 3

Assuming that the car starts at a speed of zero, the acceleration is 29/7 = 4.1429 m/s^2.  From basic physics, we know that force = mass * acceleration, so force = 1280 kg * 4.1429 m/s^2 = 5302.8 newtons.  This is best rounded off to two digits, so we get 5300 newtons.

The "7 seconds" is not very accurate. It might be 6.5, it might be 7.5 seconds. So the force might be 4862 newtons, or 5811 newtons, or anything inbetween.

Answer 4

F = m*V/t = 1280*29/7 = 5300 N

Answer 5

F = ma = mv/t = around 5 kN but use a calculator.

Answer 6

F = m*a = 1280 * ΔV/Δt = 1280 * 29m/7s² = 5303N or about 5300N

Answer 7

Too many undefined parameters. From what speed? Nuclear? Chemical? Expressed how? Volts, Newtons, Joules, Ergs?