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3 years ago

What quantity of electricity

Physics

1 answer:

ki77a [65]3 years ago

6 0

Answer:

96500 C

Explanation:

The equation of the electrolytic process is;

4OH^-(aq) -----> 2H2O(l) +O2(g) + 4e

From the reaction equation; one mole of oxygen is liberated by 4 moles of electrons.

If one mole of electrons= 96500 Cmol-1, then;

1 mole of oxygen is liberated by 4×96500 Cmol-1

0.25 moles of oxygen will be liberated by 0.25 × 4×96500/1 = 96500 C

Therefore, 0.25 moles of oxygen is liberated by 96500 C

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If you are driving 80 km/h along a straight road and you look to the side for 2.1 s , how far do you travel during this inattent

ella [17]

Velocity = displacement / time

First, Always make sure that your units for things are the same. In this question, we are given time in seconds and hours. So, we need to make them all use 1 unit.

I will do seconds.

To convert 80km/h to m/s, divide by 3.6.

22.2222…. m/s = displacement / 2.1s

Multiply by 2.1s on both sides

46.6666……m = displacement

You travelled 47 meters.

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2 years ago

a car traveling at 11.2 m/s increases speed to 24.6 m/s over 4.4 seconds. what is the acceleration of the car

Sveta_85 [38]

Answer:

24.6 - 11.2 / 4.4 = 3.045m/s

Explanation:

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4 years ago

An object is located in air, 25 cm from the vertex of the concave surface of a block of glass (as viewed from the air side of th

Marizza181 [45]

Your question is missing one part as "magnification"

i have completed the missing part below

Answer:

a. $d_{i}=-0.0566cm$

b. $M=441.69$

Explanation:

For this type of numerical we will use the following formulas

$\frac{n1}{d_{o} }+\frac{n_{2} }{d_{i} }=\frac{n_{2}-n_{1} }{R}$ ......... Eq1

where,

$n_{1}$ $=$ refractive index of the medium surrounding refracting surface/object

i.e. $n_{1}$ = $n_{air}$

$n_{2}$ $=$ refractive index of the refracting surface/object

i.e. $n_{2}=n_{glass}$

$d_{0}=$ distance of object from the vertex of the refracting surface

$d_{i}=$ distance of image from the vertex of the refracting surface

$R=$ radius of curvature of the refracting surface

$M=\frac{d_{0} }{d_{i} }$ ........... Eq2

where,

$M=$ magnification

Convention:

$R>0\for\ the\ convex\ refractive\ surface\ of\ curvature\\\ R$

Given:

$n_{1}$ = $n_{air}$ $=1.0$

$n_{2}=n_{glass}$ $=1.5$

$d_{0}=$ $25cm$

$R=-11$ $cm$ because refraction surface is concave

Required:

a. $d_{i}=$ $?$

b. $M=?$

Solution:

a. putting values in eq1, we get

$\frac{1.0}{25}+\frac{1.5}{d_{i} }=\frac{1.5-1.0}{-11}$

$\frac{1.5}{d_{i} }=-0.045-0.040$

$d_{i}=(-0.085)(\frac{1}{1.5} )$

$d_{i}=-0.0566$ cm

b. $M=\frac{25}{-0.05665}$

$M=441.69$

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4 years ago

What happens to an object after being transported to three planets in the solar system?

alukav5142 [94]

The weight changes but the mass will stay the same.

6 0

3 years ago

a cannonball is fired with a speed of 76 m/s from the top of a cliff. It strikes the plane below with a speed of 89 m/s. if we n

galina1969 [7]

Assuming the cannonball is fired horizontally, its horizontal velocity stays at a constant 76 m/s. At the point it hits the ground, it has a speed of 89 m/s, so if its vertical velocity at that moment is $v_y$ , we have

$89\dfrac{\rm m}{\rm s}=\sqrt{\left(76\dfrac{\rm m}{\rm s}\right)^2+{v_y}^2}\implies{v_y}^2\approx2145\dfrac{\mathrm m^2}{\mathrm s^2}$

(taking the negative square root because we take the downward direction to be negative)

Recall that

${v_f}^2-{v_i}^2=2a\Delta x$

where $v_i$ and $v_f$ are the initial and final velocities, respectively; $a$ is the acceleration; and $\Delta x$ is the change in position of a body. In the cannonball's case, it starts with 0 vertical velocity and is subject to a downward acceleration with magnitude $g=9.80\frac{\rm m}{\mathrm s^2}$ . So we have

$2145\dfrac{\mathrm m^2}{\mathrm s^2}-0=-2g\Delta y\implies\Delta y\approx-109.44\,\mathrm m$

(which is negative because we take the cannonball's starting position at the top of the cliff to be the origin) so the cliff is about 109 m high.

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3 years ago