In the following reaction, which substance is the precipitate? (NH4)2SO4(aq) + Ba(NO3)2(aq) BaSO4(s) + 2NH4NO3(aq) (NH4)2SO4 Ba(
2 answers:
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This problem uses the relationships among current I, current density J, and drift speed vd. We are given the total of electrons that pass through the wire in t = 3s and the area A, so we use the following equation to to find vd, from J and the known electron density n, so:
<span>The current I is any motion of charge from one region to another, so this is given by:
</span>
The magnitude of the current density is:
Being:
<span>
Finally, for the drift velocity magnitude vd, we find:
</span>
Notice: The current I is very high for this wire. The given values of the variables are a little bit odd
<span>The current I is any motion of charge from one region to another, so this is given by:
</span>
The magnitude of the current density is:
Being:
<span>
Finally, for the drift velocity magnitude vd, we find:
</span>
Notice: The current I is very high for this wire. The given values of the variables are a little bit odd
Answer:
Both objects will undergo the same change in velocity
Explanation:
m = Mass of the Earth = 5.972 × 10²⁴ kg
G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²
r = Radius of Earth = 6371000 m
m = Mass of object
Any object which is falling has only the acceleration due to gravity.
The acceleration due to gravity on Earth is 9.81364 m/s²
So, the speeds of the objects will change at an equal rate of 9.81364 m/s² but the change will be negative when an object is thrown up.
Hence, both objects will undergo the same change in velocity.