Answer:
v = 15.8 m/s
Explanation:
Let's analyze the situation a little, we have a compressed spring so it has an elastic energy that will become part kinetic energy and a potential part for the man to get out of the barrel, in addition there is a friction force that they perform work against the movement. So the variation of mechanical energy is equal to the work of the fictional force
= ΔEm =
-Em₀
Let's write the mechanical energy at each point
Initial
Em₀ = Ke = ½ k x²
Final
= K + U = ½ m v² + mg y
Let's use Hooke's law to find compression
F = - k x
x = -F / k
x = 4400/1100
x = - 4 m
Let's write the energy equation
fr d = ½ m v² + mgy - ½ k x²
Let's clear the speed
v² = (fr d + ½ kx² - mg y) 2 / m
v² = (40 4.00 + ½ 1100 4² - 60.0 9.8 2.50) 2/60.0
v² = (160 + 8800 - 1470) / 30
v = √ (229.66)
v = 15.8 m/s
Answer:
According to Coulomb's Law, the potential energy of two charged particles is directly proportional to the product of the two charges and inversely proportional to the distance between the charges
Explanation:
According to Coulomb's Law, the potential energy of two charged particles is directly proportional to the product of the two charges and inversely proportional to the distance between the charges. Since the potential energy of two charged particles is directly proportional to the product of the two charges, its magnitude increases as the charges of the particles increases. For like charges, the potential energy is positive(the product of the two alike charges must be positive) and since potential energy is inversely proportional to the distance between the charges therefore it decreases as the particles get farther apart . For opposite charges, the potential energy is negative(the product of the two opposite charges must be negative) and since potential energy is inversely proportional to the distance between the two charges, it becomes more negative as the particles get closer together.
Answer:
below
Explanation:
Net accelerating force becomes 12-8 = 4 N
F = ma
4 = 2 * a
a = 2 m/s^2
To solve this problem we will apply the concepts related to volume, as a function of length and area, as of mass and density. Later we will take the same concept of resistance and resistivity, equal to the length per unit area. Once obtained from the known constants it will be possible to obtain the area by matching the two equations:
Mass of copper wire
Density
Resistively of copper 
Resistance (R) = 0.390\Omega
Volume is defined as,

(1)
We know that,


(2)
Multiplying equation we have




Therefore the length of the wire is 1.68m