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3 years ago

Why do adults make bigger splashes when they jump into swimming pools than small children?

Physics

1 answer:

Semenov [28]3 years ago

5 0

Answer:

it bc the adults are bigger then us kid so when they dip in the pool it makes bigger splashes

Explanation:

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Que hace falta para descubrir hasta que punto todas y todos somos diferentes y ver que eso, más bien es valioso y positivo

lara [203]

Answer:

Para descubrir que las diferencias entre las personas son valiosas y positivas para la creación de sociedades más diversas y enriquecidas culturalmente, es necesario valorar las condiciones personales de cada persona de modo positivo, abandonando posturas discriminatorias como el racismo o la xenofobia, para así aceptar la diversidad social. Además, debe fomentarse el respeto por la opinión y la visión del otro, para que las personas no tomen al que piensa diferente como un adversario al que hay que confrontar. Solo así podrá lograrse una sociedad más diversa y respetuosa de las diferencias, sean culturales, étnicas o intelectuales.

5 0

3 years ago

Which statement is true about metals?

CaHeK987 [17]

B. Metals are usually defined to be from group 1 to group 13 in the periodic table which has 1-3 electrons in the outermost shell. They easily give them up which explains their conductivity.

7 0

3 years ago

A charge of 50 µC is placed on the y axis at y = 3.0 cm and a 77-µC charge is placed on the x axis at x = 4.0 cm. If both charge

zheka24 [161]

Answer:

The acceleration of an electron is $1.2\times10^{20}\ m/s^2$

Explanation:

Given that,

One Charge = 50 μC

Distance on y axis = 3.0 cm

Second charge = 77 μC

Distance on x axis = 4.0 cm

We need to calculate the force on electron due to q₁

Using formula of force

$F_{1}=\dfrac{kq_{1}q}{r^2}$

Here, q = charge of electron

Put the value into the formula

$F_{1}=\dfrac{9\times10^{9}\times50\times10^{-6}\times1.6\times10^{-19}}{(3\times10^{-2})^2}$

$F_{1}=8\times10^{-11}j\ \ N$

We need to calculate the force on electron due to q₂

Using formula of force

$F_{2}=\dfrac{kq_{2}q}{r^2}$

Here, q = charge of electron

Put the value into the formula

$F_{2}=\dfrac{9\times10^{9}\times77\times10^{-6}\times1.6\times10^{-19}}{(4\times10^{-2})^2}$

$F_{2}=6.93\times10^{-11}i\ \ N$

We need to calculate the net force

Using formula of net force

$F=F_{1}+F_{2}$

Put the value into the formula

$F=8\times10^{-11}j+6.93\times10^{-11}i$

The magnitude of the net force

$F=\sqrt{(8\times10^{-11})^2+(6.93\times10^{-11})^2}$

$F=1.058\times10^{-10}\ N$

We need to calculate the acceleration of an electron

Using formula of force

$F = ma$

$a=\dfrac{F}{m}$

Put the value into the formula

$a=\dfrac{1.058\times10^{-10}}{9.1\times10^{-31}}$

$a=1.2\times10^{20}\ m/s^2$

Hence, The acceleration of an electron is $1.2\times10^{20}\ m/s^2$

3 0

4 years ago

Almost all jobs require that you fill out an _____.

zvonat [6]

Almost all jobs require that you fill out an _____.;

D. Application

3 0

3 years ago

Read 2 more answers

A 2.00-kg ball is moving at 2.20 m/s toward the right. It collides elastically with a 4.00-kg ball that is initially at rest. 1)

loris [4]

Elastic collision is when kinetic energy before = kinetic energy after

Ek= 1/2mv^2

total before
Ek=1/2(2)(2.2^2) = 4.84 J

total after
Ek= 1/2(2+4)(v^2) = 3v^2

Before = after
4.84=3v^2 | divide by 3
121/75 = v^2 | square root both sides
v=1.27 m/s

7 0

3 years ago