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3 years ago

Which Salt is least soluble at 50 C

Physics

1 answer:

Anvisha [2.4K]3 years ago

4 0

The salt $c e_{2}\left(s o_{4}\right)_{3}$ is the least soluble salt at 50 $^{\circ} \mathrm{C}$ .

Answer: Option 1

<u>Explanation: </u>

As it is clear that $c e_{2}\left(s o_{4}\right)_{3}$ , a gas so the solubility of it decreases on increasing the temperature of water. Because gas molecule starts to evaporate on attaining heat with the increase of temperature thus leading to reduction in the solubility of them.

As most of the gas molecules get vaporised the solubility in water gradually decreases for this compound. It is also clear from the solubility graphs of the various salts.

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Consider three identical electric bulbs of power P. Two of bulbs are connected in series and the third one is connected in paral

Tamiku [17]

Answer: 3P/2

Explanation: Let the resistance of the bulbs be R.

now lets consider a Voltage V is supplied to the parallel circuit such that

$P=VI=V^2/R$

V=IR

both single bulb( bulb 3) and the two bulbs ( bulb 1 and bulb 2) are provided the same Voltage

( as the voltage remains same in parallel circuit)

we can calculate the Current across both circuits

At Bulb 3

Current 1=V/R

Power1=Voltage * Current1

Power1=V*V/R

Power1=P

At Bulb 1 and Bulb 2

Total Resistance= R+R=2R

$Current2=\frac{V}{2R}$

Power2=Voltage * Current2

$Power2=V*\frac{V}{2R} \\Power2=\frac{V^2}{2R} \\Power2=P/2$

$TotalPower=Power1+Power2\\TotalPower=P+P/2\\TotalPower=\frac{3P}{2}$

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2 years ago

Does heat always give of heat

iren2701 [21]

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2 years ago

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What is the wavelength of a wave that has a speed of 26 m/s and a frequency of 49? ? Show your work

nirvana33 [79]

Answer:

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Let me know if this helps!

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2 years ago

kramer

Answer:

60.2 J

Explanation:

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3 years ago

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Consider a father pushing a child on a playground merry-go-round. The system has a moment of inertia of 84.4 kg.m^2. The father

Sophie [7]

Answer:

Explanation:

Given that:

the initial angular velocity $\omega_o = 0$

angular acceleration $\alpha$ = 4.44 rad/s²

Using the formula:

$\omega = \omega_o+ \alpha t$

Making t the subject of the formula:

$t= \dfrac{\omega- \omega_o}{ \alpha }$

where;

$\omega = 1.53 \ rad/s^2$

∴

$t= \dfrac{1.53-0}{4.44 }$

t = 0.345 s

Using the formula:

$\omega ^2 = \omega _o^2 + 2 \alpha \theta$

here;

$\theta$ = angular displacement

∴

$\theta = \dfrac{\omega^2 - \omega_o^2}{2 \alpha }$

$\theta = \dfrac{(1.53)^2 -0^2}{2 (4.44) }$

$\theta =0.264 \ rad$

Recall that:

2π rad = 1 revolution

Then;

0.264 rad = (x) revolution

$x = \dfrac{0.264 \times 1}{2 \pi}$

x = 0.042 revolutions

Here; force = 270 N

radius = 1.20 m

The torque = F * r

$\tau = 270 \times 1.20 \\ \\ \tau = 324 \ Nm$

However;

From the moment of inertia;

$Torque( \tau) = I \alpha \\ \\ Since( I \alpha) = 324 \ Nm. \\ \\ Then; \\ \\ \alpha= \dfrac{324}{I}$

given that;

I = 84.4 kg.m²

$\alpha= \dfrac{324}{84.4} \\ \\ \alpha=3.84 \ rad/s^2$

For re-tardation; $\alpha=-3.84 \ rad/s^2$

Using the equation

$t= \dfrac{\omega- \omega_o}{ \alpha }$

$t= \dfrac{0-1.53}{ -3.84 }$

$t= \dfrac{1.53}{ 3.84 }$

t = 0.398s

The required time it takes= 0.398s

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2 years ago