First example: book, m= 0.75 kg, h=1.5 m, g= 9.8 m/s², it has only potential energy Ep,
Ep=m*g*h=0.75*9.8*1.5=11.025 J
Second example: brick, m=2.5 kg, v=10 m/s, h=4 m, it has potential energy Ep and kinetic energy Ek,
E=Ep+Ek=m*g*h + (1/2)*m*v²=98 J + 125 J= 223 J
Third example: ball, m=0.25 kg, v= 10 m/s, it has only kinetic energy Ek
Ek=(1/2)*m*v²=12.5 J.
Fourth example: stone, m=0.7 kg, h=7 m, it has only potential energy Ep,
Ep=m*g*h=0.7*9.8*7=48.02 J
The order of examples starting with the lowest energy:
1. book, 2. ball, 3. stone, 4. brick
Answer:
Explanation:
Given that,
Mass of the heavier car m_1 = 1750 kg
Mass of the lighter car m_2 = 1350 kg
The speed of the lighter car just after collision can be represented as follows
b) the change in the combined kinetic energy of the two-car system during this collision
substitute the value in the equation above
Hence, the change in combine kinetic energy is -2534.78J
Answer:
Electrons are not little balls that can fall into the nucleus under electrostatic attraction
Explanation:
Answer:
35%
Explanation:
The car's engine gives off 65% thermal energy
So only 35 % is converted into mechanical energy .
input heat = Q₁ = 100
output heat = Q₂ = 65
Work output = Q₁ - Q₂ = W
W = 100 - 65 = 35
Efficiency = W / Q₁ X 100
= (35/ 100) X 100
= 35%.
