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rosijanka [135]
3 years ago
12

Give 1 real life example of a scenario that takes advantage of the inverse relationship between force and time when impulse is c

onstant. Describe how it is an example of impulse and how force and time are involved.
Physics
1 answer:
OverLord2011 [107]3 years ago
7 0

Answer:

On real life example of a scenario that takes advantage of the inverse relationship between force and time when impulse is constant is when making a serve with a lawn tennis racket

How It is an example of impulse is that when a serve is made by moving the bat slowly, the lawn tennis player uses less force and the ball is in contact with the string for longer a period

When however, the lawn tennis player moves the racket faster, with the strings of the racket highly tensioned  he uses more force and the ball also spends less time on the racket to produce the same momentum

Explanation:

The impulse of a force, ΔP is given by the following formula;

ΔP = F × Δt

Where ΔP is constant, we have;

F ∝ 1/Δt

Therefore, for the same impulse, when the force is increased, the time of contact is decreases and vice versa.

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v = 7934.2 m/s

Explanation:

Here the total energy of the Asteroid and the Earth system will remains conserved

So we will have

-\frac{GMm}{r} + \frac{1}{2}mv_0^2 = -\frac{GMm}{R} + \frac{1}{2}mv^2

now we know that

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now from above formula

GMm(\frac{1}{R} - \frac{1}{r}) + \frac{1}{2}mv_0^2 = \frac{1}{2}mv^2

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2GM(\frac{1}{R} - \frac{1}{r}) + v_0^2 = v^2

now plug in all data

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A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 3.00 m/s, and
Leya [2.2K]

Answer:

A) 0.9844 s

B) x2 = 0.4587 m

C) v = 6.657 m/s

Explanation:

We are given;

Height of take off point above pool; x1 = 1.8 m

Initial take off velocity; u = 3 m/s

Final velocity at highest point before free fall; v = 0 m/s

B) To find the highest point above the board her feet reaches means the distance from take off to the top of the motion just before free fall.

Thus, we will be using equation of motion and we have;

v² = u² + 2gs

Now, let s = x2 which will be the distance between take off and the top before free fall.

So;

v² = u² + 2g(x2)

Now,since the motion is against gravity, g will be negative.

Thus;

v² = u² + 2(-9.81)(x2)

Plugging in the relevant values to give;

0² = 3² - (19.62x2)

19.62(x2) = 9

x2 = 9/19.62

x2 = 0.4587 m

A) We want to find how long her feet is in air.. It means we want to find out the time to get to a distance of x1 and also the time to achieve the distance (x1 + x2) on free-fall.

Thus, using equation of motion;

v = u + gt

Again, g = -9.81

Thus;

0 = 3 - 9.81t1

9.81t1 = 3

t1 = 3/9.81

t1 = 0.3058 s

Now, for the time taken to achieve the distance (x1 + x2) on free-fall, we will use the formula;

s = ut + ½gt²

Where s = (x1 + x2) = 1.8 + 0.4587 = 2.2587 m

And now, u = 0 m/s because the start of the free fall is from maximum height with velocity of 0 m/s. Again, g = - 9.81 m/s²

Thus;

2.2587 = 0 - ½(-9.81)(t2)²

2.2587 = 4.905(t2)²

(t2)² = 2.2587/4.905

(t2)² = 0.4605

t2 = √0.4605

t2 = 0.6786 s

Thus, total time of feet in air = t1 + t2 = 0.3058 + 0.6786 = 0.9844 s

C) Velocity when feet hit the water would be given by;

v = u + gt

Where u = 0 m/s and t = t2 = 0.6786

Since it's in direction of gravity, g = 9.81 m/s

v = 0 + (0.6786 × 9.81)

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