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devlian [24]
4 years ago
7

A moderate force will break an egg. However, an egg droppedon the road usually breaks while one dropped on the grass usuallydoes

nt break. THis is because the egg dropped on thegrass............???
A. has a greater change in momentum
B. has a lesser change in momentum
C. the time interval for stoppin is greater
D. the time interval for stoppin is less
Physics
1 answer:
Fed [463]4 years ago
6 0

Answer:

C. the time interval for stopping is greater.

Explanation:

As the egg falls onto the grass, it takes a a greater amount of time for it to stop, and thus the force that is being applied to it is in increments; there is never enough force applied on the egg for it to break. That's why the egg doesn't break when it lands on the grass.

In contrast, when the egg is dropped on the road, <em>all of the force that is being applied by the road on the egg is in the tiny interval when the egg touches the road</em>, That force is large enough to break the egg because it is being applied in a tiny amount of time. That's why the egg dropped on the road breaks.

<em>So here is the rule of thumb: if you don't want to break your things but still want to drop them, drop them such that it takes  some for them to stop—because force will applied to them gradually. </em>

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A ball is thrown from a rooftop with an initial downward velocity of magnitude vo = 2.9 m/s. The rooftop is a distance above the
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Answer:

a) The velocity of the ball when it hits the ground is -20.5 m/s.

b) To acquire a final velocity of 27.3 m/s, the ball must be thrown from a height of 38 m.

Explanation:

I´ve found the complete question on the web:

<em />

<em>A ball is thrown from a rooftop with an initial downward velocity of magnitude v0=2.9 m/s. The rooftop is a distance above the ground, h= 21 m. In this problem use a coordinate system in which upwards is positive.</em>

<em>(a) Find the vertical component of the velocity with which the ball hits the ground.</em>

<em>(b) If we wanted the ball's final speed to be exactly 27, 3 m/s from what height, h (in meters), would we need to throw it with the same initial velocity?</em>

<em />

The equation of the height and velocity of the ball at any time "t" are the following:

h = h0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

h = height of the ball at time t.

h0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

v = velocity of the ball at a time "t".

First, let´s find the time it takes the ball to reach the ground (the time at which h = 0)

h = h0 + v0 · t + 1/2 · g · t²

0 = 21 m - 2.9 m/s · t - 1/2 · 9.8 m/s² · t²

Solving the quadratic equation using the quadratic formula:

t = 1.8 s  ( the other solution of the quadratic equation is rejected because it is negative).

Now, using the equation of velocity, let´s find the velocity of the ball at

t = 1.8 s:

v = v0 + g · t

v = -2.9 m/s - 9.8 m/s² · 1.8 s

v = -20.5 m/s

The velocity of the ball when it hits the ground is -20.5 m/s.

b) Now we have the final velocity and have to find the initial height. Using the equation of velocity we can obtain the time it takes the ball to acquire that velocity:

v = v0 + g · t

-27.3 m/s = -2.9 m/s - 9.8 m/s² · t

(-27.3 m/s + 2.9 m/s) / (-9.8 m/s²) = t

t = 2.5 s

The ball has to reach the ground in 2.5 s to acquire a velocity of 27.3 m/s.

Using the equation of height, we can obtain the initial height:

h = h0 + v0 · t + 1/2 · g · t²

0 = h0 -2.9 m/s · 2.5 s - 1/2 · 9.8 m/s² · (2.5 s)²

-h0 = -2.9 m/s · 2.5 s - 1/2 · 9.8 m/s² · (2.5 s)²

h0 = 38 m

To acquire a final velocity of 27.3 m/s, the ball must be thrown from a height of 38 m.

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Answer:

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Explanation:

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