Answer:
a) KA = 1.2 J
b) vB = 5.00 m/s
c) W = 6.30 J
Explanation:
m = 0.600 kg
vA = 2.00 m/s
KB = 7.50 J
a) KA = ?
b) vB = ?
c) W = ?
We can apply the folowing equations
K = 0.5*m*v²
and
W = ΔK = KB- KA
then
a) KA = 0.5*m*vA² = 0.5*(0.600 kg)*(2.00 m/s)²
⇒ KA = 1.2 J
b) KB = 0.5*m*vB² ⇒ vB = √(2*KB / m)
⇒ vB = √(2*7.50 J / 0.600 kg)
⇒ vB = 5.00 m/s
c) W = ΔK = KB- KA = (7.50 J) - (1.2 J)
⇒ W = 6.30 J
Answer:
The change in the mechanical energy of the projectile is 43,750 J
Explanation:
Given;
mass of the projectile, m = 5 kg
initial velocity of the projectile, u = 200 m/s
final velocity of the projectile, v = 150 m/s
The change in mechanical energy is calculated from the principle of conservation of energy;
ΔP.E = ΔK.E
The change in potential energy is zero (0)
0 = ΔK.E
ΔK.E = K.E₁ - K.E₂
ΔK.E = ¹/₂mu² - ¹/₂mv²
ΔK.E = ¹/₂m(u² - v²)
ΔK.E = ¹/₂ x 5(200² - 150²)
ΔK.E = 43,750 J
Therefore, the change in the mechanical energy of the projectile is 43,750 J
Answer:
it depends on wether the + and - are facing eachother
or away from eachother
Explanation:
