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2 years ago

What ions exist in base solutions

Physics

1 answer:

nata0808 [166]2 years ago

7 0

Answer:

In this reaction, sodium hydroxide (NaOH) disassociates into sodium (Na+) and hydroxide (OH-) ions when dissolved in water, thereby releasing OH- ions into solution. Arrhenius acids are substances which produce hydrogen ions in solution. Arrhenius bases are substances which produce hydroxide ions in solution.

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se lanza un cuerpo desde el origen con velocidad horizontal de 40 m/s, y con un ángulo de 60º. calcular la máxima altura y el al

EastWind [94]

Answer:

1. $h = 244.8 m$

2. $x = 564.8 m$

Explanation:

1. La altura máxima se puede calcular usando la siguiente ecuación:

$v_{f}^{2} = v_{0}^{2} - 2gh$ (1)

Where:

$v_{f_{y}}$ : es la velocidad final = 0 (en la altura máxima)

$v_{0_{y}}$ : es la velocidad inicial horizontal en "y"

g: es la gravedad = 9.81 m/s²

h: es la altura máxima =?

La velocidad incial en "y" se puede calcular de la siguiente manera:

$tan(\theta) = \frac{v_{0_{y}}}{v_{0_{x}}}$

$v_{0_{y}} = tan(60)*40 m/s = 69.3 m/s$

Resolviendo la ecuación (1) para "h" tenemos:

$h = \frac{v_{0_{y}}^{2}}{2g} = \frac{(69.3 m/s)^{2}}{2*9.81 m/s^{2}} = 244.8 m$

2. Para calcular el alcance horizontal podemos usar la ecuación:

$x = v_{x}*t$

Primero debemos encontrar el tiempo cuando la altura es máxima ( $v_{f_{y}}$ = 0).

$v_{f_{y}} = v_{0_{y}} - gt$

$t = \frac{v_{0_{y}}}{g} = \frac{69.3 m/s}{9.81 m/s^{2}} = 7.06 s$

Ahora, como el tiempo de subida es el mismo que el tiempo de bajada, el tiempo máximo es:

$t_{m} = 2*7.06 s = 14.12 s$

Finalmente, el alcance horizontal es:

$x = 40 m/s*14.12 s = 564.8 m$

Espero que te sea de utilidad!

7 0

2 years ago

If a ball is given a push so that it has an initial velocity of 5 m/s down a certain inclined plane, then the distance it has ro

iren [92.7K]

Answer:

v= 13 m/s

Explanation:

Velocity is defined as the derivative of displacement with respect to time

v= ds/dt

Known data

s(t) = 5t + 2t² : distance that the ball has rolled after t seconds

vi= 5 m/s : initial velocity

t= 2 s

Problem develoment

s(t) = 5t + 2t²

v= ds/dt= 5 + 4t : velocity of the ball in function of the time

We replace t =2 s in the equation of velocity

v= 5 + 4(2)

v= 13 m/s : velocity after 2 seconds

4 0

3 years ago

Fill in the blank.<br> You don’t ____ me thass fine _ ___<br><br> -Man Tiffany

Jet001 [13]

Answer:

like, k bye

Hope this helps!

5 0

2 years ago

Read 2 more answers

A car is moving at a velocity of 25 km/h and increases velocity to 1200 km/h in 2 min. what is the acceleration?

OlgaM077 [116]

Answer:

a = 2.72 [m/s2]

Explanation:

To solve this problem we must use the following kinematics equation:

$v_{f} =v_{o} + a*t$

where:

Vf = final velocity = 1200 [km/h]

Vo = initial velocity = 25 [km/h]

t = time = 2 [min] = 2/60 = 0.0333 [h]

1200 = 25 + (a*0.0333)

a = 35250.35 [km/h2]

if we convert these units to units of meters per second squared

$35250.35[\frac{km}{h^{2} }]*(\frac{1}{3600^{2} })*[\frac{h^{2} }{s^{2} } ]*(\frac{1000}{1} )*[\frac{m}{km} ] = 2.72 [\frac{m}{s^{2} } ]$

3 0

2 years ago

A transformer has a secondary voltage of 140 volts and a secondary current of 3.5 amps. if the primary current is 10 amps, what

Lynna [10]

For an ideal transformer power loss is assumed to be zero

i.e. the power in primary coil due to input voltage must be equal to power in secondary coil due to output voltage

this can be written in form of equation

$V_1 i_1 = V_2 i_2$

here we know that

$V_2 = 140 volts$

$i_2 = 3.5 A$

$i_1 = 10 A{/tex]now we will use above equation[tex]140*3.5 = 10 * V_1$

$V_1 = 49 volts$

So primary coil voltage is 49 Volts

7 0

2 years ago