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2 years ago

A 325 N force applied to an object for 5.5 s. What’s the impulse?

1 answer:

4 0

According to Newton second law of motion, the resultant force is directly proportional to the rate of change in momentum while maintaining other factors constant. Therefore, F = (mv-mu)/t where F is the resultant force , m is the mass of the object, v is the final velocity and u is the initial velocity.
Hence, Ft = mv-mu, but impulse is given by force multiplied by time, thus, impulse is equivalent to the change in momentum.
Impulse = Ft
= 325 × 2.2 sec
= 715 Ns

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The dragster has a mass of 1.3 Mg and a center of mass at G. A parachute is attached at C provides a horizontal braking force of

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Answer:

The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²

Explanation:

The additional information to the question is embedded in the diagram attached below:

The height between the dragster and ground is considered to be 0.35 m since is not given ; thus in addition win 0.75 m between the dragster and the parachute; we have: (0.75 + 0.35) m = 1.1 m

Balancing the equilibrium about point A;

F(1.1) - mg (1.25) = $ma_a (0.35)$

$1.8v^2(1.1)$ - 1200(9.8)(1.25) = 1200a(0.35)

$1.8v^2(1.1)$ - 14700 = 420 a ------- equation (1)

$F_x = ma_x \\ \\ = 1.8v^2 = 1200 \ a$ --------- equation (2)

Replacing equation 2 into equation 1 ; we have :

${1.1 * 1200 \ a} - 14700 = 420 a$

1320 a - 14700 = 420 a

1320 a - 420 a =14700

900 a = 14700

a = 14700/900

a = 16.33 m/s²

The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²

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2 years ago

Write a hypothesis for Part II of the lab, which is about the relationship described by F = ma. In the lab, you will use a toy c

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Answer:

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If we apply force to a toy car then It will accelerate.

This is how Newton's second law of motion is verified.

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2 years ago

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Answer:

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