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3 years ago

I really need help!!

Physics

2 answers:

pishuonlain [190]3 years ago

4 0

-- Ice melts and becomes liquid when its temperature reaches 32°F.
The "warm and sunny day" is certainly warmer than that.

-- Table salt melts and becomes liquid when its temperature reaches 1474°F.
I'm pretty sure the day is not THAT warm and sunny.

Mars2501 [29]3 years ago

3 0

Omg! i just answered this <em>exact same</em> question last year!!! the salt doesnt melt cuz it is an ionic compound, which needs large amounts of energy to melt. it is simply sunny and warm outside, so it doesnt melt. hope i answered your question! ;)

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I WILL MARK YOU THE BRAINLIEST NO LINKS AND IF YOU DONT UNDERSTAND THE QUESTION DONT ANSWER

elena-s [515]

Answer:

Perform a simple test of the material in the pan to assess whether it is real gold. Raw gold appears brassy yellow and bright. If you think it is gold, place your hand between it and the sun to create shade over the gold. If it still appears bright in the pan, chances are that it is real gold.

Explanation:

6 0

3 years ago

Equations E = 1 2πε0 qd z3 and E = 1 2πε0 P z3 are approximations of the magnitude of the electric field of an electric dipole,

tamaranim1 [39]

Answer:

The ratio of $E_{app}$ and $E_{act}$ is 0.9754

Explanation:

Given that,

Distance z = 4.50 d

First equation is

$E_{act}=\dfrac{qd}{2\pi\epsilon_{0}\times z^3}$

$E_{act}=\dfrac{Pz}{2\pi\epsilon_{0}\times (z^2-\dfrac{d^2}{4})^2}$

Second equation is

$E_{app}=\dfrac{P}{2\pi\epsilon_{0}\times z^3}$

We need to calculate the ratio of $E_{act}$ and $E_{app}$

Using formula

$\dfrac{E_{app}}{E_{act}}=\dfrac{\dfrac{P}{2\pi\epsilon_{0}\times z^3}}{\dfrac{Pz}{2\pi\epsilon_{0}\times (z^2-\dfrac{d^2}{4})^2}}$

$\dfrac{E_{app}}{E_{act}}=\dfrac{(z^2-\dfrac{d^2}{4})^2}{z^3(z)}$

Put the value into the formula

$\dfrac{E_{app}}{E_{act}}=\dfrac{((4.50d)^2-\dfrac{d^2}{4})^2}{(4.50d)^3\times4.50d}$

$\dfrac{E_{app}}{E_{act}}=0.9754$

Hence, The ratio of $E_{app}$ and $E_{act}$ is 0.9754

8 0

3 years ago

Find the required angular speed, ω, of an ultracentrifuge for the radial acceleration of a point 1.80 cm from the axis to equal

Arturiano [62]

Answer:

The required angular speed ω of an ultra-centrifuge is:

ω = 18074 rad/sec

Explanation:

Given that:

Radius = r = 1.8 cm

Acceleration due to g = a = 6.0 x 10⁵ g

Sol:

We know that

Angular Acceleration = Angular Radius x Speed²

a = r x ω ²

Putting the values

6 x 10⁵ g = 1.8 cm x ω ²

Converting 1.8 cm to 0.018 m, also g = 9.8 ms⁻²

6 x 10⁵ x 9.8 = 0.018 x ω ²

ω ² = (6 x 10⁵ x 9.8) / 0.018

ω ² = 5880000 / 0.018

ω ² = 326666667

ω = 18074 rad/sec

7 0

3 years ago

___Fe + ___O2 →___Fe2O3

Anni [7]

D -
You basically keep varying the coefficient so until the number of atoms for each element is equal. I started with iron to show how you might have to change coefficients but it would be more efficient and quicker to start with oxygen since it changes the value or iron required.

4 0

3 years ago

Read 2 more answers

a water droplet falling through the air can oscillate with some angular frequency that depends on its surface tension, density,

Anna11 [10]

The oscillation angular frequency of a drop with half of the first drop's radius is 4ω

<h3>What is surface tension?</h3>

Surface tension is the tension force exerted on an object by the surface of a liquid.

<h3>What is angular frequency?</h3>

Angular frequency is the frequency of oscillation of a rotating object. It is given in rad/s.

<h3>What is the oscillation angular frequency of a drop with half of the first drop's radius?</h3>

Given that

the angular frequency of the drop is ω and
radius r.

Since the energy of the drop is conserved, using the law of conservation of angular momentum, we have

Iω = I'ω' where

I = initial rotational inertia of droplet = mr²
where m = mass of drop and
r = initial radius of droplet,
ω = initial angular frequency of droplet,
I' = initial rotational inertia of droplet = mr² where
m = mass of drop and
r' = final radius of droplet, and
ω = final angular frequency of droplet

So, Iω = I'ω'

Making ω' subject of the formula, we have

ω' = Iω/I'

ω' = mr²ω/mr'²

ω' = r²ω/r'²

Given that the drop is half of the first drop's radius, r' = r/2

So, ω' = r²ω/r'²

ω' = r²ω/(r/2)²

ω' = r²ω/r²/4

ω' = 4ω

So, the oscillation angular frequency of a drop with half of the first drop's radius is 4ω

Learn more about angular frequency here:

brainly.com/question/28036464

#SPJ1

8 0

2 years ago