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svp [43]
3 years ago
10

Can I have some tips, tricks, and ideas for dropping an egg 12 feet in the air onto a hard surface without it breaking. It HAS t

o be raw. No jelly or peanut butter, the egg cannot be taped, no plastic or glass, it can not be rubber banded either. No parachutes and it has to be smaller than a soccer or the same size.
Physics
2 answers:
lara [203]3 years ago
7 0
Have it land straight up
Ugo [173]3 years ago
4 0
Ok I Did Something Like This In Science My Idea Get a cardboard box put marshmallows around it put a container like i used a small cereal container i put it in the middle of the box put the egg in the container and surround the egg with marsh mellows and tape the top of my box and it survived 

Hoped this is useful if not im sorry at least i tried :)
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9. Captain America is chasing Red Skull. He plans to throw his shield to knock down Red Skull but needs to know how fast Red Sku
Sedaia [141]

Red Skull's relative velocity to Captain America, towards the left front of the

truck is approximately <u>33.23 m/s</u> in a direction from the North of

approximately <u>9.18°</u>.

Reasons:

Assumptions;

Taking the north direction as positive.

The activity takes place on the trucks.

The trucks are moving towards each other.

Solution:

Vector form of net speed of Red Skull, is given as follows;

  • v₁ = -(\frac{\sqrt{2} }{2} × 3.5)·i + (\frac{\sqrt{2} }{2} × 3.5 + 12.5)·j

Vector form of the net speed of Captain America is given as follows;

  • v₂ =  (\frac{\sqrt{2} }{2} × 4.0)·i - (\frac{\sqrt{2} }{2} × 4.0 + 15)·j

Relative velocity, v₁₂ = v₁ - v₂

∴ v₁₂ = (-(\frac{\sqrt{2} }{2} × 3.5) - (\frac{\sqrt{2} }{2} × 4.0))·i + ((\frac{\sqrt{2} }{2} × 3.5 + 12.5) + (\frac{\sqrt{2} }{2} × 4.0 + 15))·j

  • v₁₂ = -\frac{ 15 \cdot \sqrt{2} }{4}·i + \frac{ 110 + 15 \cdot \sqrt{2} }{4}·j

Red Skull's velocity relative to Captain America,  v₁₂ = -\frac{ 15 \cdot \sqrt{2} }{4}·i + \frac{ 110 + 15 \cdot \sqrt{2} }{4}·j

  • v₁₂ ≈ -5.3·i + 32.8·j

Therefore;

  • Red Skull appears to be moving West at <u>5.3 m/s</u> and North at <u>32.8 m/s</u>

  • The direction is arctan \left(\frac{32.8}{-5.3} \right) \approx -80.2^{\circ}

Therefore;

  • Red Skull appear to be moving at 90° - 80.2° ≈ 9.18° towards the left front end of the truck moving North

The magnitude of the velocity, |v₁₂|, is given as follows;

  • |v_{12}| = \sqrt{\left(-\frac{ 15 \cdot \sqrt{2} }{4}\right)^2 + \left(\frac{ 110 + 15 \cdot \sqrt{2} }{4}\right)^2} = \dfrac{ 5 \cdot \sqrt{130+33 \cdot\sqrt{2} } }{2} \approx 33.23·

The magnitude of Red Skull's velocity relative to Captain America is,

therefore;

|v₁₂| ≈ <u>33.23 m/s</u>

Learn more here:

brainly.com/question/24430414

6 0
3 years ago
Two tiny conducting spheres are identical and carry charges of -18.8 µC and +46.5 µC. They are separated by a distance of 2.47 c
sergiy2304 [10]

Answer:

F=-12896N, attractive.

Explanation:

For calculating this force we use the Coulomb Law:

F=\frac{kq_1q_1}{r^2}

Where k=9\times10^9Nm^2/C^{-2} is the Coulomb's constant, q_1 and q_2 the values of each charge and r the distance between them.

Since the Coulomb's constant as I wrote it is in S.I. we have to write all the magnitudes in that system of units, and substitute:

F=\frac{(9\times10^9Nm^2/C^{-2})(-18.8\times10^{-6}C)(46.5\times10^{-6}C)}{(0.0247m)^2}=-12896N

This force is attractive since both charges are of opposite sign.

3 0
3 years ago
Q 1: Calculate the pressure.
Oksana_A [137]

Answer:

1: 300pa

2: 10cm2

3: 430pa

4: a: 1.6667pa

b:2.5m2

c:20pa

Explanation:

7 0
4 years ago
How do you find period when you have frequency
nydimaria [60]

Period and frequency are mutual reciprocals.

Period = 1 / frequency .

Frequency = 1 / period

(Frequency) x (Period) = 1


8 0
3 years ago
Calculate the limit of resolution for the oil lens of your microscope. assume an average wavelength of 500 nm
natima [27]
The resolution of a microscope is the distance with the shortest measurement between two different points given a specimen  with the premise that it can still be seen clearly or distinguished by the one looking through the microscope. It can be calculate from the ratio of the wavelength of the light and twice the numerical aperture or the refractive index of the lens. Most of the microscopes have a numerical aperture ranging from 1.2 to 1.4. Resolution and the numerical aperture are indirectly proportional so that as the aperture increases the resolution would decrease. We calculate as follows:

<span>Resolution = wavelength / ((2) (numerical aperture))
Resolution = 500 nm / (2 ) ( 1.25) = 200 nm  = 0.2 um</span>
3 0
4 years ago
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