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Dmitriy789 [7]
3 years ago
14

why would a soft container protect an egg from a drop better than a hard container? (use physics terms!)​

Physics
1 answer:
jeyben [28]3 years ago
6 0

The average  force exerted on the egg is smaller for a soft container

Explanation:

When the egg falls down and reaches the ground, its change in momentum is given by

\Delta p = m\Delta v

where

m is the mass of the egg

\Delta v is the change in velocity

According to the impulse theorem, the change in momentum is equal to the impulse exerted by the floor on the egg:

\Delta p = I = F \Delta t

where

F is the average force exerted on the egg

\Delta t is the duration of the collision

The equation can be rewritten as

F=\frac{\Delta p}{\Delta t}

A soft container is able to increase the duration of the collision, \Delta t, compared to a hard container. Since the change in momentum of the egg, \Delta p, is the same (we assume the same initial conditions), this means that the ratio \frac{\Delta p}{\Delta t} is smaller for a soft container, and therefore the average force exerted on the egg is also smaller.

Learn more about impulse:

brainly.com/question/9484203

#LearnwithBrainly

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The famous astronomer, Kepler, determined that the planetary orbits were: 1. ellipses 2. circles 3. epicentric 4. geocentric
Yanka [14]

Answer:

geocentric

i hope this helps!!!!

Explanation:

3 0
3 years ago
You throw a baseball at an angle of 30.0∘∘ above the horizontal. It reaches the highest point of its trajectory 1.05 ss later. A
garri49 [273]

Answer:

The speed with which the baseball leaves the hand = 20.58 m/s

Explanation:

The time take to reach highest height during a projectile's flight is given by

t = (u sin θ)/g

u = initial velocity of the baseball = ?

θ = angle of throw above the horizontal

g = acceleration due to gravity = 9.8 m/s²

1.05 = (u sin 30)/9.8

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u = 20.58 m/s

7 0
3 years ago
Displacement is the difference between the initial position and the____ position of an object
SOVA2 [1]
<span>Displacement is the difference between the initial position and the FINAL position of an object.

Hope this helps!</span>
4 0
3 years ago
How much work (MJ) must a car produce to drive 125 miles if an average force of 306 N must be maintained to overcome friction?
baherus [9]

Answer:

The right solution is "61.557 MJ". A further explanation is given below.

Explanation:

The given values are:

Force,

F = 306 N

Drive,

D = 125 miles,

i.e.,

  = 201168

meters

As we know,

The work done will be:

= F\times S

On substituting the given values, we get

= 306\times 201168

= 61557408 \ J

On converting it in "MJ", we get

= 61.557\times 10^6 \ J

= 61.557 \ MJ

6 0
3 years ago
If the normal force exerted by the track on the car when it is at the top of the track (point B) is 6.00 N , what is the normal
Eddi Din [679]

Answer:

So, the force, F is the agent which provides the basic property of motion or rest to the body.While, the car is more obviously to have a mass, m and that the angle withe road or surface is also given which is normal to the road(i.e angle =90 degree). Then we say that lets say that the car is moving with the constant velocity of 20 m/sec and its kept unchanged by the car. So, we have the mass, m as 1 kg for the car and the value of the gravity we have the g=9.8 m/sec.

Now,

We have F=ma,

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so we can have another equation for it as,

Now, providing the required data  to, it;  ∴t =2 sec,

F=(1)×(20/2),

  • <u>F=10 N.</u>
  • So, the car would be acting the force,F of about 10 N  while the car is present on the lower region of the track.

4 0
3 years ago
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