An object is thrown straight up into the air at 10 m/s. how fast is the object traveling after 2 seconds?
1 answer:
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The concept required to solve this problem is the optical relationship that exists between the apparent depth and actual or actual depth. This is mathematically expressed under the equations.
Where,
Depth of glass
Refraction index of water
Refraction index of glass
Refraction index of air
Depth of water
I enclose a diagram for a better understanding of the problem, in this way we can determine that the apparent depth in the water of the logo would be subject to
Therefore the distance below the upper surface of the water that appears to be the logo is 4.041cm
Answer:
E = 0.437 N/C
Explanation:
Given that,
Charge,
Electric force,
Let the strength of the electric field is E. We know that, the electric force is given by :
F = qE
Where
E is the electric field strength
So, the strength of the electric field is equal to 0.437 N/C.
To solve this exercise we will use the concept related to heat loss which is mathematically given as
Where,
m = mass
= Specific Heat
Change in temperature
Replacing with our values we have that
Specific heat of mercury
Replacing
Therefore the heat lost by mercury is 0.09J