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3 years ago

Why would an egg break immediately when it hits the ground?

Physics

2 answers:

iogann1982 [59]3 years ago

7 0

All I can say is "Gravity"

Elza [17]3 years ago

6 0

As an egg falls towards the floor, it begins to travel faster and faster. When it slams into the floor, the egg is stopped almost immediately. This force of the floor against the eggshell is too large, so it breaks.

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A string is wrapped around a uniform disk of mass M = 1.2 kg and radius R = 0.07 m. (Recall that the moment of inertia of a unif

pentagon [3]

Explanation:

Given that,

Mass of disk = 1.2 kg

Radius = 0.07 m

Radius of rod = 0.11 m

Mass of small disk = 0.5 kg

Force = 29 N

Time t = 0.022 s

$\theta=0.023\ m$

Distance d= 0.039 m

(I). We need to calculate the speed of the apparatus

Using work energy theorem

$W=\Delta K.E$

$Fd=\dfrac{1}{2}mv^2$

$v=\sqrt{\dfrac{2Fd}{M+4m}}$

Where, m = total mass

v = velocity

F = force

d = distance

Put the value into the formula

$v=\sqrt{\dfrac{2\times29\times0.039}{1.2+4\times0.5}}$

$v=0.840\ m/s$

(b). We need to calculate the angular speed of the apparatus

Using formula of torque

$\tau=I\alpha$

$F\times=(\dfrac{1}{2}MR^2+4mb^2)\alpha$

$29\times0.07=(\dfrac{1}{2}\times1.2\times0.07^2+4\times0.5\times0.11^2)\alpha$

$\alpha=\dfrac{29\times0.07}{0.02714}$

$\alpha=74.79\ rad/s^2$

We need to calculate the angular speed of the apparatus

Using equation of angular motion

$\omega=\omega_{0}+\alpha t$

Put the value into the formula

$\omega=0+74.79\times0.022$

$\omega=1.645\ rad/s$

(c). We need to calculate the angular speed of the apparatus

Using equation of angular motion

$\omega_{0}^2=\omega^2+2\alpha t$

Put the value into the formula

$\omega_{0}^2=1.645^2+2\times74.79\times0.022$

$\omega=2.44\ rad/s$

Hence, This is required equation.

7 0

3 years ago

Which of the following scenarios would create the most friction between the two surfaces?

Tcecarenko [31]

The Answer to this question would Be A

Hope this helps

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5 0

3 years ago

Read 2 more answers

What unit is used for the value of G in Newton's Law of Universal<br> Gravitation?<br> *

Helga [31]

Answer:

In SI units, its value is approximately 6.674×10−11 m3⋅kg−1⋅s−2. The modern notation of Newton's law involving G was introduced in the 1890s by C. V. Boys. The first implicit measurement with an accuracy within about 1% is attributed to Henry Cavendish in a 1798 experiment.

Explanation:

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3 0

2 years ago

If the earth shrank until its radius were only one-quarter its present size without changing its mass what would a 20 n object w

Dahasolnce [82]

Basing on the information given, we can calculate the new weight of the object by the following given:current weight = 20 Ng = 10m/s2

20N/4 = 5N

Thank you for your question. Please don't hesitate to ask in Brainly your queries.

5 0

3 years ago

Un movil viaja a 40km/h y comienza a reducir su velocidad a partir del instante t=0. Al cabo de 6 segundo se detiene completamen

aleksklad [387]

Answer:

1,85 m / s²

Explanation:

De la pregunta anterior, se obtuvieron los siguientes datos:

Velocidad inicial (u) = 40 km / h

Hora inicial (t₁) = 0

Tiempo final (t₂) = 6 s

Velocidad final (v) = 0

Aceleración (a) =?

A continuación, convertiremos 40 km / ha m / s. Esto se puede obtener de la siguiente manera:

1 km / h = 0,2778 m / s

Por lo tanto,

40 km / h = 40 km / h × 0,2778 m / s / 1 km / h

40 km / h = 11,11 m / s

Por tanto, 40 km / h equivalen a 11,11 m / s.

Finalmente, determinaremos la aceleración del móvil durante el período en el que desaceleró. Esto se puede obtener de la siguiente manera:

Velocidad inicial (u) = 11,11 m / s

Hora inicial (t₁) = 0

Tiempo final (t₂) = 6 s

Velocidad final (v) = 0

Aceleración (a) =?

a = (v - u) / (t₂ - t₁)

a = (0 - 11,11) / (6 - 0)

a = - 11,11 / 6

a = –1,85 m / s²

Por tanto, la aceleración del móvil durante el período en el que se ralentizó es de –1,85 m / s²

6 0

3 years ago