How to determine the parametric equation using Newtons second law

Physics

1 answer:

GenaCL600 [577]2 years ago

6 0

Answer:

Newton's second law of motion states that the acceleration of a system is directly proportional to and in the same direction as the net external force acting on the system, and inversely proportional to its mass. In equation form, Newton's second law of motion is a=Fnetm a = F net m .

Explanation:

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A car is behind a truck going 25 m/s on the highway. The car’s driver looks for an opportunity to pass, guessing that his car ca

user100 [1]

Answer:

No he should not attempt the pass

Explanation:

Let t be the time it takes for the car to pass the truck. The driver should ONLY attempt to pass when the distance covered by himself plus the distance covered by the oncoming car is less than or equal 400 m (a near miss)

At acceleration of 1m/s2 and a clear distance of 10 + 20 + 10 = 40 m, we can use the following equation of motion to estimate the time t in seconds

$s = at^2/2$

$40 = 1t^2/2$

$t^2 = 80$

$t = \sqrt{80} = 8.94 s$

Within this time frame, the first car would have traveled a total distance of the clear distance (40m) plus the distance run by the truck, which is

8.94 * 25 = 223.6m

So the total distance traveled by the first car is 223.6 + 40 = 263.6m

The distance traveled by the 2nd car within 8.94 s at rate of 25m/s is

8.94 * 25 = 223.6 m

So the total distance covered by both cars within this time frame

223.6 + 263.6 = 487.2m > 400 m

So no, he should not attempt the pass as we will not clear it in time.

8 0

3 years ago

A 34-kg child runs with a speed of 2.8 m/s tangential to the rim of a stationary merrygo-round. The merry-go-round has a momentu

tekilochka [14]

Answer: $0.43\ rad/s$

Explanation:

Given

Mass of child $m=34\ kg$

speed of child is $v=2.8\ m/s$

Moment of inertia of merry go round is $I=510\ kg.m^2$

radius $r=2.31\ m$

Conserving the angular momentum

$\Rightarrow mvr=I\omega \\\Rightarrow 34\times 2.8\times 2.31=510\times \omega\\\\\Rightarrow \omega=\dfrac{219.912}{510}\\\Rightarrow \omega=0.43\ rad/s$