OK. So you're pushing on the small box, and on the other side of it, the small
box is pushing on the big box. So you're actually pushing both of them.
-- The total mass that you're pushing is (5.2 + 7.4) = 12.6 kg.
-- You're pushing it with 5.0N of force.
-- Acceleration of the whole thing = (force)/(mass) = 5/12.6 = <em>0.397 m/s²</em> (rounded)
-- Both boxes accelerate at the same rate. So the box farther away from you ...
the big one, with 7.4 kg of mass, accelerates at the same rate.
The force on it to make it accelerate is (mass) x (acceleration) =
(7.4 kg) x (5/12.6 m/s²) = <em>2.936 N.</em>
The only force on the big box comes from the small box, pushing it from behind.
So that same <em>2.936N</em> must be the contact force between the boxes.
Explanation:
Momentum is conserved.
a) In the first scenario, Olaf and the ball have the same final velocity.
mu = (M + m) v
(0.400 kg) (10.9 m/s) = (70.2 kg + 0.400 kg) v
v = 0.0618 m/s
b) In the second scenario, the ball has a final velocity of 8.10 m/s in the opposite direction.
mu = mv + MV
(0.400 kg) (10.9 m/s) = (0.400 kg) (-8.10 m/s) + (70.2 kg) v
v = 0.108 m/s
Answer:
11.515 Joule
Explanation:
Volume of aluminium = V = 4.89×10⁻³ m³
Coefficient of volume expansion for aluminum = α = 69×10⁻⁶ /°C
Initial temperature = 19.1°C
Final temperature = 357°C
Pressure of air = 1.01×10⁵ Pa
Change in temperature = ΔT= 357-19.1 = 337.9 °C
Change in volume
ΔV = αVΔT
⇒ΔV = 69×10⁻⁶×4.89×10⁻³×337.9
⇒ΔV = 114010.839×10⁻⁹ m³
Work done
W = PΔV
⇒W = 1.01×10⁵×114010.839×10⁻⁹
⇒W = 11.515 J
∴ Work is done by the expanding aluminum is 11.515 Joule
Answer:
PE = 3.92x10^16J
potential energy
Explanation:
PE = m*g*h
mass of water = 1000kg/m³
(4*10^10m³)*1000kg = 4*10^13kg
PE = (4*10^13kg)*(9.81m/s²)*(100m)
PE = 3.92x10^16J
