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daser333 [38]
3 years ago
9

How much energy is used when a 110kw appliance is used for 3 hours

Physics
1 answer:
Bogdan [553]3 years ago
6 0
We could take the easy way out and just say

(110 kW) x (3 hours) = 330 kilowatt hours .

But that's cheap, and hardly worth even 5 points.
If we want to talk energy, let's use the actual scientific unit of energy.
________________________________________________

" 110 kw " means 110,000 watts = 110,000 joules/second .

(3 hours) x (3600 sec/hour) = 10,800 seconds.

(110,000 joules/second) x (10,800 seconds) = 1.188 x 10⁹ Joules

 That's

==>  1,188,000,000 joules

==>  1,188,000 kilojoules

==>  1,188 megajoules

==>  1.188 gigajoules

Atsa nawfulotta energy ! 
It goes back to that "110 kw appliance" that we started with. 
That's no common ordinary household appliance.  110 kw is something like
147 horsepower.  In order to bring 110 kw into your house, you'd need to
take 458 Amperes through the 240-volt line from the pole.  Most houses
are limited to 100 or 200 Amperes, tops.  And the TRANSFORMER on
the pole, that supplies the whole neighborhood, is probably a 50 kw unit.  
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The length of vector A is 93.8 meters and the length of Ay is 38.4 meters, then the length of A, must be
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85.556metres

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2 years ago
A space probe is fired as a projectile from the Earth's surface with an initial speed of 2.05 104 m/s. What will its speed be wh
Elanso [62]

Answer:

The value is  v  =  2.3359 *10^{4} \ m/s

Explanation:

From the question we are told that

  The  initial speed is u =  2.05 *10^{4} \  m/s

 Generally the total energy possessed by the space probe when on earth is mathematically represented as

             T__{E}} =  KE__{i}} +  KE__{e}}

Here  KE_i is the kinetic energy of the space probe due to its initial speed which is mathematically represented as

          KE_i =   \frac{1}{2}  *  m  *  u^2

=>       KE_i =   \frac{1}{2}  *  m  *  (2.05 *10^{4})^2

=>       KE_i =  2.101 *10^{8} \ \ m \ \ J

And  KE_e is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as

       KE_e =  \frac{1}{2}  *  m *  v_e^2

Here v_e is the escape velocity from earth which has a value v_e =  11.2 *10^{3} \  m/s

=>    KE_e =  \frac{1}{2}  *  m *  (11.3 *10^{3})^2

=>    KE_e =  6.272 *10^{7} \  \  m  \ \   J

Generally given that at a position that is very far from the earth that the is Zero, the kinetic energy at that position is mathematically represented as

        KE_p =  \frac{1}{2}  *  m *  v^2

Generally from the law energy conservation we have that

        T__{E}} =  KE_p

So

       2.101 *10^{8}  m  +  6.272 *10^{7}  m  =   \frac{1}{2}  *  m *  v^2

=>     5.4564 *10^{8} =   v^2

=>     v =  \sqrt{5.4564 *10^{8}}

=>     v  =  2.3359 *10^{4} \ m/s

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Answer:

A

Explanation:

8 0
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