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1 year ago

Question 1 of 20:

Physics

1 answer:

xxMikexx [17]1 year ago

4 0

Unlike the IQ, your emotional intelligence (EQ) may be improved over time by becoming more mindful

Emotional intelligence is the understanding and usage of one's own emotions in a positive way to relieve their stress. It also helps for effective communication, to overcome challenges, to empathize and to defuse any bad situation.

Option A is the correct for IQ ( Intelligence quotient ). Option C is wrong because EQ can be built in any stage of your life. EQ is basically emotions. So option D is incorrect.

Therefore, Unlike the IQ, your emotional intelligence (EQ) may be improved over time by becoming more mindful

To know more about emotional intelligence

brainly.com/question/13129837

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A typical car has 16 L of liquid coolant circulating at a temperature of 95 ∘C through the engine's cooling system. Assume that,

amm1812

Answer:

$\Delta V=0.0592\ L$

Explanation:

Given:

Initial temperature of the coolant, $T_f=95^{\circ}C$

final temperature of the coolant, $T_f=105^{\circ}C$

total volume of the coolant, $V_c=16\ L$

coefficient of volume expansion for coolant, $\beta_c=410\times 10^{-6}\ ^{\circ}C^{-1}$

volume of Al radiator, $V_a=2\ L$

volume of steel radiator, $V_s=14\ L$

We have:

coefficient of volume expansion for Aluminium, $\beta_a=75\times 10^{-6}\ ^{\circ}C^{-1}$

coefficient of volume expansion for steel, $\beta_a=35\times 10^{-6}\ ^{\circ}C^{-1}$

Now, change in volume of the coolant after temperature rises:

$\Delta V_c=V_c.\beta_c.\Delta T$

$\Delta V_c=16\times 410\times 10^{-6}\times (105-95)$

$\Delta V_c=0.0656\ L$

Now, volumetric expansion in Aluminium radiant:

$\Delta V_a=V_a.\beta_a.\Delta T$

$\Delta V_a=2\times 75\times 10^{-6}\times (105-95)$

$\Delta V_a=0.0015\ L$

Now, volumetric expansion in steel radiant:

$\Delta V_s=V_s.\beta_s.\Delta T$

$\Delta V_s=14\times 35\times 10^{-6}\times (105-95)$

$\Delta V_s=0.0049\ L$

∴Total extra accommodation volume created after the expansion:

$V_X=\Delta V_s+\Delta V_a$

$V_X=0.0049+0.0015$

$V_X=0.0064\ L$

Hence, the volume that will overflow into the small reservoir will be the volume of coolant that will be extra after the expanded accommodation in the radiator.

$\Delta V=\Delta V_c-\Delta V_X$

$\Delta V=0.0656-0.0064$

$\Delta V=0.0592\ L$

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