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devlian [24]
3 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]3 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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S the statement, "You run faster than I do," qualitative or quantitative?
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"A 3 kg crate slides down a ramp. The ramp is 1 m in length and inclined at an angle of 30 degrees. The crate starts from rest a
OlgaM077 [116]

Answer:

v = 2.57 m /sec

Explanation: See Annex Free Body Diagram

From free body diagram and Newton´s second law we have

There is not movements in the y axis direction

cos 30°  =  √3/2       sin 30°  =  1/2

We have  P  = mg   =  3 Kg  *  9.8 m/sec²

P  =   29.4  Kg*m/ sec²      P  =  29.4 [N]

Py  =  P * cos 30°    Py =  29.4 [N] * √3/2   ⇒    Py = 25.43 [N]

Px  =   P * sin 30°    Px =  29.4 [N] * 1/2      ⇒     Px = 14.7  [N]

∑ F   =  m* a         ⇒    ∑ Fy   =  0       ∑ Fx  =  m *a

∑ Fy   =  Fn  -  Py   =  0         Py   = P*cos30°       Py = 25.43 [N]

Fn  =  25.43 [N]

Fr  =  μk * Fn      ⇒   Fr  =  0.19 * 25.43   ⇒ Fr  =  4.83 [N]    

Now

∑ Fx  =  m *a       mg sin30° - Fr =  m*a    ⇒   Px  - Fr  = m*a

14.7 [N]   -  4.83 [N]   =  3 [kg] * a       ⇒   9.87 /3   = a [m /sec²]

a = 3.29 [m/sec²]

From uniformly accelerated movement

distance  =  x₀  + V₀*t ± at²/2     but  x₀   and  V₀    =  0

Then

d = ( 1/2 )*a*t²     ⇒  1 [m]  * 2  =  3.29 [m/sec²] * t²

t  =  0.78 sec

And finally

v =  a*t      ⇒   v  =  3.29 *(.78)     ⇒   v = 2.57 m /sec

5 0
3 years ago
Sheila (m=56.8 kg) is in her saucer sled moving at 12.6 m/s at the bottom of the sledding hill near Bluebird Lake. She approache
FromTheMoon [43]

Answer:

y = 54.9 m

Explanation:

For this exercise we can use the relationship between the work of the friction force and mechanical energy.

Let's look for work

      W = -fr d

The negative sign is because Lafourcade rubs always opposes the movement

On the inclined part, of Newton's second law

Y Axis  

      N - W cos θ  = 0

The equation for the force of friction is

      fr = μ N

      fr = μ mg cos θ

We replace at work

     W = - μ m g cos θ  d

Mechanical energy in the lower part of the embankment

      Em₀ = K = ½ m v²

The mechanical energy in the highest part, where it stopped

     Em_{f} = U = m g y

     W = ΔEm =  Em_{f} - Em₀

    - μ m g d cos θ = m g y - ½ m v²

Distance d and height (y) are related by trigonometry

     sin θ = y / d

     y = d sin θ

   

    - μ m g d cos θ = m g d sin θ - ½ m v²

We calculate the distance traveled

     d (g syn θ + μ g cos θ) = ½ v²

     d = v²/2 g (sintea + myy cos tee)

     d = 9.8 12.6 2/2 9.8 (sin16 + 0.128 cos 16)

     d = 1555.85 /7.8145

     d = 199.1 m

Let's use trigonometry to find the height

      sin 16 = y / d

      y = d sin 16

      y = 199.1 sin 16

      y = 54.9 m

8 0
3 years ago
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