Answer:58.28 N
Explanation:
Given data
Now velocity of jet at height of 2m
equating them
W=
W=
W=58.28 N
Answer: 150m
Explanation:
The following can be depicted from the question:
Dimensions of outer walls = 9.7m × 14.7m.
Thickness of the wall = 0.30 m
Therefore, the plinth area of the building will be:
= (9.7 + 0.30/2 + 0.30/2) × (14.7 × 0.30/2 + 0.30/2)
= 10 × 15
= 150m
Answer:
Explanation:
In the diagram there three gears in which gear 1 is input gear ,gear 2 is idle gear and gear 3 is out put gear.
Lets take
All external matting gears will rotates in opposite direction with respect to each other.
So the speed of gear third can be given as follows
Answer:
Tmax= 46.0 lb-in
Explanation:
Given:
- The diameter of the steel rod BC d1 = 0.25 in
- The diameter of the copper rod AB and CD d2 = 1 in
- Allowable shear stress of steel τ_s = 15ksi
- Allowable shear stress of copper τ_c = 12ksi
Find:
Find the torque T_max
Solution:
- The relation of allowable shear stress is given by:
τ = 16*T / pi*d^3
T = τ*pi*d^3 / 16
- Design Torque T for Copper rod:
T_c = τ_c*pi*d_c^3 / 16
T_c = 12*1000*pi*1^3 / 16
T_c = 2356.2 lb.in
- Design Torque T for Steel rod:
T_s = τ_s*pi*d_s^3 / 16
T_s = 15*1000*pi*0.25^3 / 16
T_s = 46.02 lb.in
- The design torque must conform to the allowable shear stress for both copper and steel. The maximum allowable would be:
T = min ( 2356.2 , 46.02 )
T = 46.02 lb-in
