Answer:
Vmax=11.53 m/s
Explanation:
from conservation of energy
Spring potential energy =potential energy due to elevation
0.5*k*x²= mg
=mgh
0.5*k*2.3²= 430*9.81*6
k=9568.92 N/m
For safety reason
k"=1.13 *k= 1.13*9568.92
k"=10812.88 N/m
agsin from conservation of energy
spring potential energy=change in kinetic energy
0.5*k"*x²=0.5*m*
10812.88 *2.3²=430*
=11.53 m/s
Answer:
Average acceleration on first part of the chunk is given as
Average acceleration on second part of the chunk is given as
Explanation:
By momentum conservation along x direction we will have
so we have
also by energy conservation
by solving above equation we will have
Average acceleration on first part of the chunk is given as
Average acceleration on second part of the chunk is given as
Thermal conductions
K= QL/ART
Aluminium T₁ = 10 + 273.15
T₂ = 283.15k
205 = 2.0 × 0.30/4× 10⁻⁴ × (T₂ - 283.15)
Copper
385 = Q × 0.70/4×10⁻⁴ ×(433.15 - T₂)
Where T₃ = 160 + 273.15
T₃ = 433.15K
From 2 to 3
205/385 = 0.30/0.70 × 433.15 - T₂/T₂ - 283.15
= 0.53T₂ -150.06 = 181.92 - 0.42 T₂
→ 0.95T₂ = 331.98 ⇒ T₂ = ₂349.45k
T₂ = 76.3°c
=77°c.
