Answer:
3.75 m/s south
Explanation:
Momentum before collision = momentum after collision
m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂
Since the car and truck stick together, v₁ = v₂.
m₁ u₁ + m₂ u₂ = (m₁ + m₂) v
Given m₁ = 1500 kg, u₁ = -15 m/s, m₂ = 4500 kg, and u₂ = 0 m/s:
(1500 kg) (-15 m/s) + (4500 kg) (0 m/s) = (1500 kg + 4500 kg) v
-22500 kg m/s = 6000 kg v
v = -3.75 m/s
The final velocity is 3.75 m/s to the south.
Answer:
Explanation:
Two identical bodies are sliding toward each other on a frictionless surface.
Initial speed of body 1, m₁ = 1 m/s
Initial speed of body 2, m₂ = 2 m/s
They collide and stick.
We need to find the speed of the combined mass. Let V is the speed of the combined mass.
Using the conservation of momentum.
We have, m₁ = m₂ = m
So, the speed of the combined mass is 
.
Hey again!
Ok..
Now... The melting Point of this solid is 90°C.
Meaning That as soon as it gets to this temp... It STARTS Melting.
So at that temp... It still has some solid parts in it.
You can say its a Solid Liquid Mixture.
Additional Heat being applied at that point is not raising the temperature;rather its used in breaking the bonds in the solid. This is the Fusion stage.
After Fusion...It'd then Be a Pure Liquid with no solids in it.
So
Q'=MC∆0----- This is the heat needed to take the solid's temp from 30°c - 90°c
Q"=ml ----- This is the heat used in breaking the bonds holding the solids in the solid-liquid phase.
So
Q= Q' + Q"
Q= mc∆0 + ml
∆0 = 90°c - 30°c = 60°c
Q= 2.5(390)(60) + (2.5)(4000)
Q=6.9 x 10⁴Joules
