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2 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.

2 answers:

lara [203]2 years ago

7 0

Have it land straight up

Ugo [173]2 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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2 years ago

PLEASE HELP ME WITH THIS ONE QUESTION

Vlad [161]

Answer: $7.95^{\circ}C$

Explanation:

Given

Mass of water is $m_1=1\ kg$

mass of ice is $m_2=0.1\ kg$

Latent heat of fusion $L=3.33\times 10^5\ kJ/kg$

The heat capacity of water is $c_{w}=4186\ J/kg^{\circ}C$

Suppose water is at $T^{\circ} C$ and it reaches to $0^{\circ}C$ to melt the ice

the heat released by water must be equivalent to heat absorbed by the ice

$\therefore \quad m_1c_w(T-0)=m_2\times L\\\Rightarrow 1\times 4186\times T=0.1\times 3.33\times 10^5\\\Rightarrow T=7.95^{\circ}C$

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2 years ago

Sickle cell anemia is known to run in a family. A pedigree chart for this family is shown below. The parents are shown at the to

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3 years ago

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A child\'s top is held in place, upright on a frictionless surface. The axle has a radius of r = 2.96 mm. Two strings are wrappe

sladkih [1.3K]

Answer:

Explanation:

Given

radius r=2.96 mm

Tension T=2.4 N

time taken=0.74 s

Let $\alpha$ be the angular acceleration

$2 T\times r=I\times \alpha$

$2\times 2.4\times 2.96\times 10^{-3}=0.5\times m\times (2.96\times 10^{-3})^2\times \alpha$

$\alpha =\frac{4\times 2.4}{m\times 2.96\times 10^{-3}}$

$\alpha =\frac{3.24\times 10^3}{m} rad/s^2$

$\omega =\omega _0+\alpha \cdot t$

$\omega =0+\frac{3.24\times 10^3}{m}\times 0.74$

$\omega =\frac{2.4\times 10^3}{m} rad/s$

Angular momentum

$L=I\omega$

$L=0.5\times mr^2\times \omega$

$L=0.5\times m\times (2.96\times 10^{-3})^2\times \frac{2.4\times 10^3}{m}$

$L=0.01051 kg-m^2/s$

4 0

2 years ago

The cylinder in the picture is rotating at 500 RPMs. The friction coefficients between the cylinder and block B are static=0.5 a

Tju [1.3M]

Coefficient of static friction = 0.5
Coefficient of Kinetic friction = 0.3
Angular velocity = 500 RPMs

<h3>The Radius of the System</h3>

Let R be the radius of cylinder

$m_a + m_b = 4 + 3 = 7kg$

The angular velocity is 500 RPMs

$\omega ^2 = \frac{500 * 2\pi}{60} rad/s\\N = (M_a + M_b)\omega ^2 R$

The normal force

$f = \mu N = (M_a + M_b) g\\\mu (M_a + M_b) \omega ^2 R = M_a + M_b\\R = \frac{1}{\mu \omega ^2 R}\\\mu_s = 0.5\\R = \frac{1}{0.5 * (\frac{500 * 2\pi}{60})^2 }\\R = 0.0073m\\R = 7.3mm$

Since the radius is very little for two block to execute circular motion so system will slide down.

Learn more on coefficient of static friction here;

brainly.com/question/11841776

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brainly.com/question/26400616

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1 year ago