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3 years ago

What kind of eruption is likely to occur in a volcano having magma that is low in silica

Physics

2 answers:

balu736 [363]3 years ago

5 0

Answer:

A slow eruption is likely to occur in a volcano having magma that is low in silica.

Explanation:

The magma flows out at a slow speed and with ease when the content of Silica is very low in it. The consistency of magma is made up of oxygen and silicon. On the other hand, if the content of silica is very high in the magma, the volcano erupts very explosively. Hence, it is to be noted that the amount and content of silica in the magma determine the flow of the magma.

Masja [62]3 years ago

3 0

A volcano erupts quietly if it's magma is low on silica. Low silica magma will be very thin, runny, and will flow easily, so the gases in the magma will bubble out gently. Hope this helps!:)

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Temka [501]

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Explanation:

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3 years ago

A gold wire that is 1.8 mm in diameter and 15 cm long carries a current of 260 mA. How many electrons per second pass a given cr

Musya8 [376]

Answer:

162500000.

Explanation:

Given that

Diameter of the wire , d= 1.8 mm

The length of the wire ,L = 15 cm

Current ,I = 260 m A

The charge on the electron ,e= 1.6 x 10⁻¹⁹ C

We know that Current I is given as

$I=\dfrac{q}{t}$

I=Current

q=Charge

t=time

q= I t

q= 260 m t

The total number of electron = n

q= n e

$n=\dfrac{260\times 10^{-3}\ t}{1.6\times 10^{-9}}$

n=162500000 t

$\dfrac{n}{t}=16250000$

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4 0

3 years ago

Una prenda de 320gramos de ropa gira en el interior de una lavadora si dicha lavadora tiene 40 cm y gira con una frecuencia de 4

Nitella [24]

Answer:

Período del tambor: $T = 0.25\,s$ , fuerza sobre la prenda: $F \approx 80.852\,N$ , velocidad lineal del tambor: $v \approx 10.053\,\frac{m}{s}$ , velocidad angular del tambor: $\omega \approx 25.133\,\frac{rad}{s}$ .

Explanation:

La expresión tiene un error por omisión, su forma correcta queda descrita a continuación:

"Una prenda de 320 gramos de ropa gira en el interior de una lavadora si dicha lavadora tiene un radio de 40 centímetros y gira con una frecuencia de 4 hertz. Halle a) el período, b) la velocidad angular, c) la fuerza con la que gira la prenda y d) la velocidad lineal de la lavadora."

El tambor gira a velocidad angular constante ( $\omega$ ), en radianes por segundo, lo cual significa que la prenda experimenta una aceleración centrífuga ( $a$ ), en metros por segundo al cuadrado. En primer lugar, calculamos el período de rotación del tambor ( $T$ ), en segundos:

$T = \frac{1}{f}$ (1)

Donde $f$ es la frecuencia, en hertz.

( $f = 4\,hz$ )

$T = \frac{1}{4\,hz}$

$T = 0.25\,s$

Ahora determinamos la fuerza aplicada sobre la prenda ( $F$ ), en newtons:

$F = m\cdot a$ (2)

$F = \frac{4\pi^{2}\cdot m \cdot r}{T^{2}}$ (2b)

Donde:

$m$ - Masa de la prenda, en kilogramos.

$r$ - Radio interior del tambor, en metros.

( $m = 0.32\,kg$ , $r = 0.4\,m$ , $T = 0.25\,s$ )

$F = \frac{4\pi^{2}\cdot (0.32\,kg)\cdot (0.4\,m)}{(0.25\,s)^{2}}$

$F \approx 80.852\,N$

La velocidad lineal de la lavadora es:

$v = \frac{2\pi\cdot r}{T}$ (3)

( $r = 0.4\,m$ , $T = 0.25\,s$ )

$v = \frac{2\pi\cdot (0.4\,m)}{0.25\,s}$

$v \approx 10.053\,\frac{m}{s}$

Y la velocidad angular del tambor de la lavadora:

$\omega = \frac{2\pi}{T}$

( $T = 0.25\,s$ )

$\omega = \frac{2\pi}{0.25\,s}$

$\omega \approx 25.133\,\frac{rad}{s}$

7 0

2 years ago

Block A has a mass of 0.5kg, and block B has a mass of 2kg. Block is is released at a height of 0.75 meters above B. The coeffic

VikaD [51]

Answer:

0.075 m

Explanation:

The picture of the problem is missing: find it in attachment.

At first, block A is released at a distance of

h = 0.75 m

above block B. According to the law of conservation of energy, its initial potential energy is converted into kinetic energy, so we can write:

$m_Agh=\frac{1}{2}m_Av_A^2$

where

$g=9.8 m/s^2$ is the acceleration due to gravity

$m_A=0.5 kg$ is the mass of the block

$v_A$ is the speed of the block A just before touching block B

Solving for the speed,

$v_A=\sqrt{2gh}=\sqrt{2(9.8)(0.75)}=3.83 m/s$

Then, block A collides with block B. The coefficient of restitution in the collision is given by:

$e=\frac{v'_B-v'_A}{v_A-v_B}$

where:

e = 0.7 is the coefficient of restitution in this case

$v_B'$ is the final velocity of block B

$v_A'$ is the final velocity of block A

$v_A=3.83 m/s$

$v_B=0$ is the initial velocity of block B

Solving,

$v_B'-v_A'=e(v_A-v_B)=0.7(3.83)=2.68 m/s$

Re-arranging it,

$v_A'=v_B'-2.68$ (1)

Also, the total momentum must be conserved, so we can write:

$m_A v_A + m_B v_B = m_A v'_A + m_B v'_B$

where

$m_B=2 kg$

And substituting (1) and all the other values,

$m_A v_A = m_A (v_B'-2.68) + m_B v_B'\\v_B' = \frac{m_A v_A +2.68 m_A}{m_A + m_B}=1.30 m/s$

This is the velocity of block B after the collision. Then, its kinetic energy is converted into elastic potential energy of the spring when it comes to rest, according to

$\frac{1}{2}m_B v_B'^2 = \frac{1}{2}kx^2$