Answer:
Explanation:
KE = ½mv² = ½(6.8)8² = 217.6 J
round as appropriate because that result is way too much precision for the inputs provided. Arguably should be 200 J based on the single significant digit of the velocity.
Answer:
7 deg
Explanation:
= mass of the rod = 
= weight of the rod = 
= spring constant for left spring = 
= spring constant for right spring = 
= stretch in the left spring
= stretch in the right spring
= length of the rod = 0.75 m
= Angle the rod makes with the horizontal
Using equilibrium of force in vertical direction for left spring

Using equilibrium of force in vertical direction for right spring

Angle made with the horizontal is given as

Answer: Yep! I made a yt vid on them so look this up:
Video result for rick roll with a different link copy and paste0:10
Rick roll, but with different link
Rick roll, but with different link
Video result for rick roll with a different link copy and paste0:10
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NEVER GOING TO GIVE YOU UP
Video result for rick roll with a different link copy and paste0:10
Rick roll, but with different link
Answer:
The current through the tube is 73.39A.
Explanation:
The relationship between the resistivity
, the electric field
, and the current density
is given by

This equation can be solved for
to get:

Since the current is 

Now, for the tube of mercury
,
, and the area is
; therefore,


Hence, the current through the mercury tube is 73.39A.
Answer:
Energy released as heat will be -189.417 Kj
Explanation:
The oxidation is the type of reaction in which reaction occurs in the presence of oxygen. The exothermic reaction is the type of reaction in which heat is released. The enthalpy is the type of physical quantity that is used to measure the energy in the thermodynamic system.
As we know the molar mass of Copper II Oxide is 79.545 g
Also the oxidation of copper(II) oxide, CuO(s), is an exothermic process, 2Cu2O (s) + O2-->4CuO (s) delta H reaction= -292.0 kj/mol.
So for 51.60g of Copper II Oxide,
Heat released will be (-292.0 * 51.60/79.545) = -189.417 Kj