WHAT IS THE NET FORCE (CENTRIPETAL FORCE) FOR A 1500 KG

1 answer:

5 0

A :-) Given - m = 1500 kg
v = 5 m/s
r = 10 m
Solution -
F = mv^2 by r
F = 1500 x 5 x 5 by 10
( cut 5 and 10 because 5 x 2 = 10 )
F = 1500 x 5 by 2
( cut 2 and 1500 because
2 x 750 = 1500 )
F = 750 x 5
F = 3750 N
.:. The centripetal force is 3750 N.

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The wheels of the locomotive push back on the tracks with a constant net force of 7.50 × 105 N, so the tracks push forward on th

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Answer:

The freight train would take 542.265 second to increase the speed of the train from rest to 80.0 kilometers per hour.

Explanation:

Statement is incomplete. Complete description is presented below:

A freight train has a mass of $1.83\times 10^{7}\,kg$ . The wheels of the locomotive push back on the tracks with a constant net force of $7.50\times 10^{5}\,N$ , so the tracks push forward on the locomotive with a force of the same magnitude. Ignore aerodynamics and friction on the other wheels of the train. How long, in seconds, would it take to increase the speed of the train from rest to 80.0 kilometers per hour?

If locomotive have a constant net force ( $F$ ), measured in newtons, then acceleration ( $a$ ), measured in meters per square second, must be constant and can be found by the following expression:

$a = \frac{F}{m}$ (1)

Where $m$ is the mass of the freight train, measured in kilograms.

If we know that $F = 7.50\times 10^{5}\,N$ and $m = 1.83\times 10^{7}\,kg$ , then the acceleration experimented by the train is:

$a = \frac{7.50\times 10^{5}\,N}{1.83\times 10^{7}\,kg}$

$a = 4.098\times 10^{-2}\,\frac{m}{s^{2}}$

Now, the time taken to accelerate the freight train from rest ( $t$ ), measured in seconds, is determined by the following formula:

$t = \frac{v-v_{o}}{a}$ (2)

Where:

$v$ - Final speed of the train, measured in meters per second.

$v_{o}$ - Initial speed of the train, measured in meters per second.

If we know that $a = 4.098\times 10^{-2}\,\frac{m}{s^{2}}$ , $v_{o} = 0\,\frac{m}{s}$ and $v = 22.222\,\frac{m}{s}$ , the time taken by the freight train is: