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3 years ago

The formula for the hydronium ion is

1 answer:

5 0

Hydronium is the commom name for the aqueous cation H 30+ the type of oxonium ion produced by protonation of water.. The Hydronium ion is H30+

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The earplug can reduce the sound level to about 18 decibels (dB). What percentage reduction is this intensity?

zlopas [31]

Answer:

1 x 10 -10 whisper at 1m distance.

Explanation:

Properly fitted ear plugs an reduce noise form 15-30db. Although they are better for low frequency

8 0

3 years ago

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Un puente de acero de 100 m de largo a 8° C aumenta su temperatura a 24°C ¿Cuánto medirá su longitud? Valor del coeficiente de d

BartSMP [9]

La longitud final del puente de acero es 100.018 metros.

Asumamos que la dilatación térmica experimentada por el puente de acero es pequeña, de modo que podemos emplear la siguiente aproximación lineal para determinar la longitud final del puente de acero ( $L$ ), en metros:

$L = L_{o}\cdot [1+\alpha\cdot (T_{f}-T_{o})]$ (1)

Donde:

$L_{o}$ - Longitud inicial del puente, en metros.
$\alpha$ - Coeficiente de dilatación, sin unidad.
$T_{o}$ - Temperatura inicial, en grados Celsius.
$T_{f}$ - Temperatura final, en grados Celsius.

Si tenemos que $L_{o} = 100\,m$ , $\alpha = 11.5\times 10^{-6}$ , $T_{o} = 8\,^{\circ}C$ y $T_{f} = 24\,^{\circ}C$ , entonces la longitud final del puente de acero es:

$L = (100\,m)\cdot [1+(11.5\times 10^{-6})\cdot (24\,^{\circ}C - 8\,^{\circ}C)]$

$L = 100.018\,m$

La longitud final del puente de acero es 100.018 metros.

Para aprender más sobre dilatación térmica, invitamos cordialmente a ver esta pregunta verificada: brainly.com/question/24953416

5 0

2 years ago

3.) An engineer is designing the runway for an airport. Of the planes that will use the airport,

scoray [572]

Answer: 704

Explanation:Vi = 0 m/s

vf = 65 m/s

a = 3 m/s2

d = ??

vf2 = vi2 + 2*a*d

(65 m/s)2 = (0 m/s)2 + 2*(3 m/s2)*d

4225 m2/s2 = (0 m/s)2 + (6 m/s2)*d

(4225 m 2/m2)/(6 m/s2) = d

d = 704 m

5 0

2 years ago

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In 1932 Albert Dremel of Racine, Wisconsin, created his rotary tool that has come to be known as a dremel.

kramer

Answer:

3054.32618 rad/s²

-431.1989 rad/s²

29080

Explanation:

Converting angular speed to rad/s

$\omega=35000\times \frac{2\pi}{60}=3665.19142\ rad/s$

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{3665.19142-0}{1.2}\\\Rightarrow a=3054.32618\ rad/s^2$

The average acceleration while speeding up is 3054.32618 rad/s²

The number of turns in the 1.2 seconds

$\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow \theta=0\times t+\frac{1}{2}\times 3054.32618\times 1.2^2\\\Rightarrow \theta=2199.11484\ rad=\frac{2199.11484}{2\pi}=349.99\ rotations$

The number of rotations in the 1.2 seconds is 349.99

Number of rotations in the 45 seconds

$\frac{35000}{60}\times 45=26250\ rotations$

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{0-3665.19142}{8.5}\\\Rightarrow a=-431.1989\ rad/s^2$

Average angular acceleration while slowing down -431.1989 rad/s²

$\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \theta=\frac{\omega_f^2-\omega_i^2^2}{2\alpha}\\\Rightarrow \theta=\frac{0^2-3665.19142^2}{2\times -431.1989}\\\Rightarrow \theta=15577.0668\ rad=\frac{15577.0668}{2\pi}\\ =2479.16718\ rotations$

Number of rotations while slowing down is 2479.16718

Total number of rotations is 349.99+26250+2479.16718 = 29079.15718 = 29080

3 0

2 years ago

NEED HELP ASAP WILL MARK BRAINLIEST

MissTica

Answer:

A. Atoms of the same element but ith different mass.

Explanation:

8 0

2 years ago

Read 2 more answers