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Vedmedyk [2.9K]
2 years ago
11

Electrical energy is sold by the kilowatt hour How much would it cost to heat a hot tub from 15 to 43

Physics
1 answer:
Alexeev081 [22]2 years ago
7 0

Answer:

hot tub is hot

Explanation:

dont get fried

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(d) What type of transformation is done by the following things?
mafiozo [28]

Answer:

human body is answer according to our studies

7 0
3 years ago
Why must mine tailings be stored and disposed of carefully?
gavmur [86]
After thorough researching, the mine tailings must be stored and disposed of carefully because they have lots of chemical and various toxic materials. They can also leach to the aquifers. The correct answer to the following given statement above is they have chemicals which are dangerous.
7 0
3 years ago
Suppose that a rectangular toroid has 1,500 windings and a self-inductance of 0.060 H. If the height of the rectangular toroid i
V125BC [204]

Answer:

Current in the toroid will be I=17.32\times 10^{-3}A

Explanation:

We have given number of winding in rectangular toroid N = 1500

Self inductance of toroid L = 0.06 H

Magnetic energy stored in toroid E=9\times 10^{-16}J

We have to find the current in the toroid

Magnetic energy stored is equal to E=\frac{1}{2}Li^2

9\times 10^{-6}=\frac{1}{2}\times 0.06\times I^2

I=17.32\times 10^{-3}A

So current in the toroid will be I=17.32\times 10^{-3}A

5 0
4 years ago
How much energy (in Joules) is released when 12.0 g of water cools from 20.0 °C to 11.0 °C? This is a grade 10 question from the
KATRIN_1 [288]

Answer: - 452.088joule

Explanation:

Given the following :

Mass of water = 12g

Change in temperature(Dt) = (11 - 20)°C = - 9°C

Specific heats capacity of water(c) = 4.186j/g°C

Q = mcDt

Where Q = quantity of heat

Q = 12g × 4.186j/g°C × - 9°C

Q = - 452.088joule

7 0
3 years ago
Please answer the number 5 6 and 7. I don't know what to do its our hw in physics. (new lesson as well)
Aleks04 [339]

From the picture, I see that you had no trouble at all with #4.
Well, #5, 6, and 7 are easily handled in exactly the same way.

Just as you did with #4, please sketch these on paper
as I walk you through the solutions.  That'll help you see
immediately what's going on.

#5.b).
Traveling east at 3 m/s for 4 seconds,
he covers  (3 m/s) x (4 sec) = 12 meters.

Traveling south at 5 m/s for 2 seconds,
he covers (5 m/s) x (2 sec) = 10 meters.

The total distance he covers is  (12m + 10m) = 22 meters.

#5.c).
Average speed (scalar)

                           = (distance covered)/(time to cover the distance)

                           = (22 meters)/(6 sec) = 3-2/3 m/s .

#5.d).
Displacement (vector)

                       = distance between the start-point and the end-point,
                          regardless of the route traveled,
                      
  in the direction from the start-point to the end-point.

Distance from the start-point to the end-point =

               √(12² + 10²) = √(144 + 100) = √(244) = 15.62 meters

in the direction of  arctan(10/12) south of east

                             =  39.8° south of east.
 
#5.e).
Average velocity (vector) =

             (displacement vector) / (time)

         =  15.62 meters directed 39.8° south of east / 6 seconds

         =  2.603 m/s directed 39.8° south of east.

 #6).
Magnitude = √(5.2² + 2.1²) = √(27.04 + 4.41) = √31.45 = 5.608 km.

Direction = arctan(5.2/2.1) south of east

                =   68° south of east  =  158° bearing . 

#7).
Magnitude = √(39² + 57²) = √(1521 + 3249) = √( 4770)

                                                                         =  69.07 m/s .

Direction = arctan (57/39) south of west

               =   55.6° south of west

                    Bearing = 214.4°

                    Compass: 0.65° past "southwest by south".  


I'm grateful for the privilege and opportunity to practice my math,
and I shall cherish the bounty of 5 points that came with it.

8 0
4 years ago
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