K, P, K, K, P, K, K, P, K, P. If it is moving, it is kinetic, if it isn't, it's potential. the sugar one is a little tricky using that method though, because we generally consider this in terms of spacial movement, but sugar holds energy which is later released by your body to allow you to move.the chemical bonds have potential energy because they release energy when broken.
Answer:
The awnser is A.
Explanation:
I got it right on edgenuity. If im wrong sorry ;-;
<h3>
Answer:</h3>
19.3 g/cm³
<h3>
Explanation:</h3>
Density of a substance refers to the mass of the substance per unit volume.
Therefore, Density = Mass ÷ Volume
In this case, we are given;
Mass of the gold bar = 193.0 g
Dimensions of the Gold bar = 5.00 mm by 10.0 cm by 2.0 cm
We are required to get the density of the gold bar
Step 1: Volume of the gold bar
Volume is given by, Length × width × height
Volume = 0.50 cm × 10.0 cm × 2.0 cm
= 10 cm³
Step 2: Density of the gold bar
Density = Mass ÷ volume
Density of the gold bar = 193.0 g ÷ 10 cm³
= 19.3 g/cm³
Thus, the density of the gold bar is 19.3 g/cm³
Answer : The exit temperature of the product is, 
Explanation :
Total heat = Heat lost by liquid + Latent heat of fusion + Heat lost by frozen

where,
Q = Total heat = 6000 kJ
m = mass of product = 15 kg
= specific heat of liquid = 
= latent heat of fusion = 
= specific heat of frozen = 
= initial temperature of liquid = 
= final temperature of liquid = 
= initial temperature of frozen = ?
= final temperature of frozen = 
Now put all the given value in the above expression, we get:
![6000kJ=[15kg\times 4kJ/kg^oC\times (10-2)^oC]+[15kg\times 275kJ/kg]+[15kg\times 2.5kJ/kg^oC\times (2-T_3)^oC]](https://tex.z-dn.net/?f=6000kJ%3D%5B15kg%5Ctimes%204kJ%2Fkg%5EoC%5Ctimes%20%2810-2%29%5EoC%5D%2B%5B15kg%5Ctimes%20275kJ%2Fkg%5D%2B%5B15kg%5Ctimes%202.5kJ%2Fkg%5EoC%5Ctimes%20%282-T_3%29%5EoC%5D)

Thus, the exit temperature of the product is, 