Answer:
<u>The transformation of energy in a torch light is as follows:</u>
1) When the torch is turned ON, the chemical energy in the batteries is converted into electrical energy.
2) The electrical energy is converted into heat and light energy. (We feel the torch to be hot after some time and we can see the light energy)
Hope this helped!
<h2>~AnonymousHelper1807</h2>
Answer:
V = 10 km / 1 hr = 10 km/hr
V = -10 j km / hr if one were to use i, j, k as unit vectors with the usual orientation
<span>D is the correct answer. A Bourdon gage is a popular and commonly used kind of gauge for measuring pressure and vacuum. One use for a Bourdon gage is to indicate steam pressure.</span>
The solution to the problem is as follows:
<span>Average = 80
So Sum = 80 * 5 = 400
Mode = 88, so two results are 88 (if three results were 88, then the median would be 88).
Three results are 81, 88, and 88.
That leaves 143. We could still have one 81 score, so that leaves the lowest score as 62.
Greg is in a car at the top of a roller-coaster ride. The distance, d, of the car from the ground as the car descends is determined by the equation d = 144 - 16t2, where t is the number of seconds it takes the car to travel down to each point on the ride. How many seconds will it take Greg to reach the ground?
d = 144 - 16t2
0 = 144 - 16t2
16t^2=144
t^2=9
t=3</span>
Answer:
The amount of kilograms of ice at -20.0°C that must be dropped into the water to make the final temperature of the system 40.0°C = 0.0674 kg
Explanation:
Heat gained by ice in taking the total temperature to 40°C = Heat lost by the water
Total Heat gained by ice = Heat used by ice to move from -20°C to 0°C + Heat used to melt at 0°C + Heat used to reach 40°C from 0°C
To do this, we require the specific heat capacity of ice, latent heat of ice and the specific heat capacity of water. All will be obtained from literature.
Specific heat capacity of ice = Cᵢ = 2108 J/kg.°C
Latent heat of ice = L = 334000 J/kg
Specific heat capacity of water = C = 4186 J/kg.°C
Heat gained by ice in taking the total temperature to 40°C = mCᵢ ΔT + mL + mC ΔT = m(2108)(0 - (-20)) + m(334000) + m(4186)(40 - 0) = 42160m + 334000m + 167440m = 543600 m
Heat lost by water = mC ΔT = 0.25 (4186)(75 - 40) = 36627.5 J
543600 m = 36627.5
m = 0.0674 kg = 67.4 g of ice.
