Answer:
The capacite is C=5.32 uF using the equations of voltage and energy in capacitance
Explanation:
The energy holds is 5 J and the resistor dissipates 2J so the energy total is 3J
Using:

Voltage in this case is the energy dissipated so



Using the equation to find capacitance

F
C= 5.32 uF because u is the symbol for micro that is equal to 
Answer:
Explanation:
Givens
vi = 10 m/s
a = 1.5 m/s^2
d = 600 m
vf = ?
Formula
vf^2 = vi^2 + 2*a*d
Solution
vf^2 = 10^2 + 2*1.5 * 600
vf^2 = 100 + 1800
vf^2 = 1900
sqrt(vf^2) = sqrt(1900)
vf = 43.59 m/s
Answer:



Explanation:
Given:
- volume of liquid content in the can,

- mass of filled can,

- weight of empty can,

<u>So, mass of the empty can:</u>



<u>Hence the mass of liquid(soda):</u>



<u>Therefore the density of liquid soda:</u>
(as density is given as mass per unit volume of the substance)


<u>Specific weight of the liquid soda:</u>



Specific gravity is the density of the substance to the density of water:

where:
density of water


So based on your question where there is a block of mass m1= 8.8kg in the inclined plane with an angle of 41 with respect to the horizontal. To find the spring constant of the problem were their is a coefficients of friction of 0.39 and 0.429, you must use the formula K*x^2=m*a*sin(angle). By calculating the minimum spring constant is 220.66 N/m^2