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2 years ago

Who knows the egg drop challenge

1 answer:

7 0

I know. If you're asking for ideas, I once did one and it did not break. I just took a cardboard box, filled it with thins like packing peanuts, crumpled newspaper, and put the egg inside a few layers of socks in the middle, and taped the box tightly shut. Our science teacher dropped from the roof of the school, and there wasn't even a crack on mine!

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An object weigh 40N in air ,weigh 20N when submerged in water,and 30N when submerged in a liquid of unknown liquid density.what

muminat

Answer:

The density is $\rho_u =500 kg /m^3$

Explanation:

From the question we are told that

The weight in air is $W_a = 40 \ N$

The weight in water is $W_w = 20 \ N$

The weight in a unknown liquid is $W_u = 30 \ N$

Now according to Archimedes principle the weight of the object in water is mathematically represented as

$W_w = W_a -m _w g$

Where $m_w$ is he mass of the water displaced

substituting value

$m_w g = 40 -20$

$m_w g = 20 \ N --- (1)$

Now according to Archimedes principle the weight of the object in unknown is mathematically represented as

$W_u = W_a -m _u g$

Where $m_u$ is he mass of the unknown liquid displaced

substituting value

$m_u g = 40 -30$

$m_u g = 10 \ N ---(2)$

dividing equation 2 by equation 1

$\frac{m_ug}{m_wg} = \frac{10}{20}$

$\frac{m_u}{m_w} = \frac{1}{2}$

=> $m_u = 0.5 m_w$

Now since the volume of water and liquid displaced are the same then

$\rho _u = 0.5 \rho_w$

This because

$density = \frac{mass}{volume}$

So if volume is constant

mass = constant * density

Where $\rho_u$ is the density of the liquid

and $\rho_ w$ is the density of water which is a constant with a value $\rho_w = 1000 kg/m^3$

$\rho_u = 1000*0.5$

$\rho_u =500 kg /m^3$

7 0

3 years ago

PLEASE ANSWER NEED HELP!!!!!!!! PLEASE THE CORRECT ANSWER!!!!!!

Leya [2.2K]

Answer:

Explanation:

5 0

2 years ago

1. In Newton’s ring experiment, the diameter of the 5th ring is 0.30 cm and diameter of 15th the ring is 0.62 cm. Find the diame

IgorC [24]

Answer:

Diameter of Newton’s 5th ring = 0.30 cm

Diameter of Newton’s 15th ring = 0.62 cm

Diameter of Newton’s 25th ring = ?

From Newton’s rings experiment we infer that

D2n+m − D2n = 4λmR

For the 5th and 15th rings we have

D215 − D25 = 4λ * 10 * R _______ (1) (m = 10)

For 15th and 25th rings

D225 − D215 = 4λ * 10 * R _______ (2) (m = 10)

We equate the two derivatives

Equation (2) = Equation (1)

D225 − D215 = D215 − D25

D225 = 2D215 – D25

Substituting the values into the equation

D225 = 2 * 0.62 * 0.62 – 0.3 * 0.3 =0.6788 cm2

D25 = 0.8239 cm

4 0

2 years ago

Billy drops a water balloon from the roof of his house. Since the balloon began with an original velocity of zero how far above

VashaNatasha [74]

T=2,5s
a=g≈10m/s²

h=s=?

$s= \frac{1}{2} at^2 \Leftrightarrow h=\frac{1}{2}gt^2=\frac{1}{2}*10m/s^2 *(2,5s)^2=5m/s^2* \frac{25}{4} s^2=\boxed{31,25m}$

"Non nobis, Domine, non nobis, sed Nomini tuo da gloriam."

Regards M.Y.

7 0

3 years ago

When distances were carefully measured from the Earth to globular clusters above and below the Milky Way plane (where our view o

uranmaximum [27]

Answer:

A) Spherically symmetric about a point in the constellation Sagittarius and concentrated in that direction

Explanation:

The globular clusters are present mainly in the direction of Sagittarius with the center of the system of globular cluster being measured as a spherical cluster cloud such that the center of the Milky Way can be taken as being in the Sagittarius constellation

4 0

3 years ago