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2 years ago

What does the 2 mean in the formula 5Mg3(PO4)2?

Physics

2 answers:

dalvyx [7]2 years ago

8 0

D. There are two phosphate ions in a molecule of magnesium phosphate. Hope this helps!! :-)

stira [4]2 years ago

3 0

D. There are two phosphate ions in a molecule of magnesium phosphate

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Which of the following is not an application of Doppler technology?

allsm [11]

The correct answer to the question above is The third Option: C; ultrasound imaging of the liver. The ultrasound imaging of the liver is definitely not an application of Doppler technology.

Hope this helps! :)

8 0

3 years ago

Read 2 more answers

Un engrane que gira con una velocidad de 20 rad/s, es acelerado durante 5 segundos hasta alcanzar una velocidad de 35 rad/s

larisa [96]

Answer:

a) La aceleración angular es: $\alpha=2\: rad/s^{2}$

b) El engranaje gira 125 radianes.

c) El engranaje hara aproximadamente 20 revoluciones.

Explanation:

La aceleración angular se define como:

$\alpha=\frac{\Delta \omega}{\Delta t}$

Donde:

Δω es la diferencia de velocidad angular (en otras palabras ω(final)-ω(inicial))
Δt es el tiempo en el que occure el cambio de velocidad angular

$\alpha=\frac{35-25}{5}$

$\alpha=2\: rad/s^{2}$

El desplazamiento angular puede ser calculado usando la siguiente ecuación:

$\theta=\theta_{i}+\omega_{i}t+\frac{1}{2}\alpha t^{2}$

Aqui el angulo inicial es 0, por lo tanto.

$\theta=20(5)+\frac{1}{2}(2)(5)^{2}$

$\theta=125\: rad$

El engranaje gira 125 radianes.

Lo que debemos hacer aquí es convertir radianes a revoluciones.

Recordemos que 2π rad = 1 rev

Entonces:

$\theta=125\: rad \times \frac{1\: rev}{2\pi\: rad}=19.89\: rev$

Por lo tanto el engranaje hara aproximadamente 20 revoluciones.

Espero te haya sido de ayuda!

6 0

2 years ago

A thin walled spherical shell is rolling on a surface. What

ExtremeBDS [4]

Answer:

$=\frac{1/3}{5/6} = 0.4$

Explanation:

Moment of inertia of given shell $= \frac{2}{3} MR^2$

where

M represent sphere mass

R -sphere radius

we know linear speed is given as $v = r\omega$

translational $K.E = \frac{1}{2} mv^2 = \frac{1}{2} m(r\omega)^2$

rotational $K.E = \frac{1}{2} I \omega^2 = \frac{1}{2} \frac{2}{3} MR^2 \omega^2$

total kinetic energy will be

$K.E = \frac{1}{2} m(r\omega)^2 + \frac{1}{2} \frac{2}{3} MR^2 \omega^2$

$K.E =\frac{5}{6} MR^2 \omega^2$

fraction of rotaional to total K.E

$=\frac{1/3}{5/6} = 0.4$

8 0

3 years ago

What type of appeal does Gore mostly use to persuade the audience?

Mazyrski [523]

I thinks its He uses proof to show the evidence is relevant. But im not totally positive on it hope this helps

8 0

3 years ago

Read 2 more answers

A satellite of mass m is in a circular orbit of radius R2 around a spherical planet of radius R1 made of a material with density

Amiraneli [1.4K]

Answer:

a) K = 2/3 π G m ρ R₁³ / R₂ , b) U = - G m M / r

Explanation:

The law of universal gravitation is

F = G m M / r²

Part A

Let's use Newton's second law

F = m a

The acceleration is centripetal

a = v² / R₂

G m M / R₂² = m v² / R₂

v² = G M / R₂

They give us the density of the planet

ρ = M / V

V = 4/3 π R₁³

M = ρ V

M = ρ 4/3 π R₁³

v² = 4/3 π G ρ R₁³ / R₂

K = ½ m v²

K = ½ m (4/3 π G ρ R₁³ / R₂)

K = 2/3 π G m ρ R₁³ / R₂

Part B

Potential energy and strength are related

F = - dU / dr

∫ dU = - ∫ F. dr

The force was directed towards the center and the vector r outwards therefore there is an angle of 180º between the two cos 180 = -1

U- U₀ = G m M ∫ dr / r²

U - U₀ = G m M (- r⁻¹)

We evaluate for

U - U₀ = -G m M (1 / $r_{f}$ - 1 / $r_{i}$ )

They indicate that for ri = ∞ U₀ = 0

U = - G m M / r

6 0

3 years ago