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2 years ago

What happened to the balloon when it was placed on the bottle with the baking soda and vinegar and why.

Physics

1 answer:

user100 [1]2 years ago

5 0

Answer:

The ballon would be inflated. The reason is that the sodium bicarbonate in baking soda reacts with acetic acid in vinegar to produce gas.

Explanation:

The main component of baking soda is sodium bicarbonate, ${\rm Na_{2}CO_{3}}$ .

Vinegar is mostly a solution of acetic acid ${\rm CH_{3}COOH}$ in water.

Acids such as acetic acid react with carbonate salts. One of the products of such reactions is carbon dioxide ${\rm CO_{2}}$ , a gas.

In this question, when the acetic acid in vinegar reacts with sodium bicarbonate in the baking soda, the following reaction would occur:

$\begin{aligned}& {\rm Na_{2}CO_{3}}\, (aq) + 2\, {\rm CH_{3}COOH}\, (aq) \\ &\to 2\, {\rm CH_{3}COONa}\, (aq) + {\rm CO_{2}}\, (g)\end{aligned}$ .

The ${\rm CO_{2}}$ produced would then inflate the ballon placed on the opening of the bottle.

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A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t1 = 11.9 seconds to get up to it

Degger [83]

Answer:

a) $\omega_1=8.168\,rad.s^{-1}$

b) $n_1=7.735 \,rev$

c) $\alpha_1 =0.6864\,rad.s^{-2}$

d) $\alpha_2=4.1454\,rad.s^{-2}$

e) $t_2=1.061\,s$

Explanation:

Given that:

initial speed of turntable, $N_0=0\,rpm\Rightarrow \omega_0=0\,rad.s^{-1}$

full speed of rotation, $N_1=78 \,rpm\Rightarrow \omega_1=\frac{78\times 2\pi}{60}=8.168\,rad.s^{-1}$

time taken to reach full speed from rest, $t_1=11.9\,s$

final speed after the change, $N_2=120\,rpm\Rightarrow \omega_2=\frac{120\times 2\pi}{60}=12.5664\,rad.s^{-1}$

no. of revolutions made to reach the new final speed, $n_2=11\,rev$

(a)

∵ 1 rev = 2π radians

∴ angular speed ω:

$\omega=\frac{2\pi.N}{60}\, rad.s^{-1}$

where N = angular speed in rpm.

putting the respective values from case 1 we've

$\omega_1=\frac{2\pi\times 78}{60}\, rad.s^{-1}$

$\omega_1=8.168\,rad.s^{-1}$

(c)

using the equation of motion:

$\omega_1=\omega_0+\alpha . t_1$

here α is the angular acceleration

$78=0+\alpha_1\times 11.9$

$\alpha_1 = \frac{8.168 }{11.9}$

$\alpha_1 =0.6864\,rad.s^{-2}$

(b)

using the equation of motion:

$\omega_1\,^2=\omega_0\,^2+2.\alpha_1 .n_1$

$8.168^2=0^2+2\times 0.6864\times n_1$

$n_1=48.6003\,rad$

$n_1=\frac{48.6003}{2\pi}$

$n_1=7.735\, rev$

(d)

using equation of motion:

$\omega_2\,^2=\omega_1\,^2+2.\alpha_2 .n_2$

$12.5664^2=8.168^2+2\alpha_2\times 11$

$\alpha_2=4.1454\,rad.s^{-2}$

(e)

using the equation of motion:

$\omega_2=\omega_1+\alpha_2 . t_2$

$12.5664=8.168+4.1454\times t_2$

$t_2=1.061\,s$

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3 years ago

A woman 5.5 ft walks at a rate of 6 ft/sec towards a street light that is 22 ft above the ground. At what rate is the length of

Soloha48 [4]

Answer:

The length of her shadow is changing at the rate -2 m/s

Explanation:

Let the height oh the street light, h = 22 ft

Let the height of the woman, w = 5.5 ft

Horizontal distance to the street light = l

length of shadow = x

h/w = (l + x)/x

22/5.5 = (l + x)/x

4x = l + x

3x = l

x = 1/3 l

taking the derivative with respect to t of both sides

dx/dt = 1/3 dl/dt

dl/dt = -6 ft/sec ( since the woman is walking towards the street light, the value of l is decreasing with time)

dx/dt = 1/3 * (-6)

dx/dt = -2 m/s

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3 years ago

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13. A transformer has a primary coil with 600 turns and a secondary coil with 300 turns. If the output voltage is 320 volts, wha

vivado [14]

His is a step down transformer since n(primary) is greater than n(seconcary). You relate the input voltage with the ouput voltage with the following equation:

Vout = n2/n1*Vin (n2/n1 is essentially your 'transfer function' that dictates what a specified input would produce) 

Solving the equation: 

Vin = Vout*n1/n2 = (320V)*(600/300) = 640 V 

This is checked by seeing if Vin is greater than Vout, which it is for a step down transformer.

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