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sertanlavr [38]
3 years ago
11

The standard molar heat of vaporization for water is 40.79 kj/mol. How much energy would be required to vaporize 4.72 x 10^10 mo

lecules of water
Physics
1 answer:
stellarik [79]3 years ago
7 0
(5.81 x 10^10 molecules) / (6.022 x 10^23 molecules/mol) x (40.79 kj/mol) = 3.94 x 10^-12 kj
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pochemuha

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A voltage of 220V (rms) line to netralvoltage is applied across a three phase Y connected resistive load. Each phase of the Y co
mrs_skeptik [129]

Answer:

Current through each phase Vp = 2.2A

Total three phase power Pt= 1.45kW

Power factor of the load pf = 1

Explanation:

i) Find current through each phase

Vp =220V (rms)

Z =100 Ω

I = Vp/Z

 = 220/100

 = 2.2A

ii) Find the total three phase power

for a resistive load, Power, P = VI

Power for each phase is given as:

P = 220 * 2.2

  = 484 W

Total power TP =3* P

                          =484*3

                          = 1452W

                           =1.45kW

 iii) Find the power factor of the load

Phase angle for a resistive load is 0.

α= 0

Hence, power factor of load = cos α

                                          pf      = cos 0

                                             pf    = 1

5 0
3 years ago
A fishbowl has a circular opening with a diameter of 13 cm. The fishbowl sits upright on a table in a magnetic field of 0.00110
azamat

Answer:

Did you ever get the answer?

Explanation:

5 0
3 years ago
A 0.0663 kg ingot of metal is heated to 241◦C
Westkost [7]

Answer:280.216j/kg°C

Explanation:

Mass of metal=0.0663kg

mass of water=0.395kg

Final temperature=27.4°C

Temperature of metal=241°C

Temperature of water=25°C

specific heat capacity of water=4186j/kg°C

0.0663xax(241-27.4)=0.395x4186x(27.4-25)

0.0663xax213.6=0.395x4186x2.4

14.16168a=3968.328

a=3968.328 ➗ 14.16168

a=280.216j/kg°C

4 0
3 years ago
A soccer ball is kicked horizontally off a bridge with a heigh of 36m. The ball travels 25m horizontally before it hits the pave
kow [346]

Answer:

The initial velocity was U=22.14m/s

Explanation:

Step one :

Applying the third equation of motion

v² = u²+ 2as

Where v= Final velocity

U =initial velocity

a= acceleration due to gravity

S= distance or displacement

Step two :

V= 0

a= 9.81m/s²

S=25m

U=?

Step three :

Substituting into the equation we have

0²=U²+2*9.81*25

0=U²+490.5

U²=-490.5

U=√490.5

U=22.14m/s

5 0
3 years ago
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