Answer:
In the acid processes, deoxidation can take place in the furnaces, leaving a reasonable time for the inclusions to rise into the sla*g and so be removed before casting. Whereas in the basic furnaces, deoxidation is rarely carried out in the presence of the sla*g, otherwise phosphorus would return to the metal.
Answer:
a) v = +/- 0.323 m/s
b) x = -0.080134 m
c) v = +/- 1.004 m/s
Explanation:
Given:
a = - (0.1 + sin(x/b))
b = 0.8
v = 1 m/s @ x = 0
Find:
(a) the velocity of the particle when x = -1 m
(b) the position where the velocity is maximum
(c) the maximum velocity.
Solution:
- We will compute the velocity by integrating a by dt.
a = v*dv / dx = - (0.1 + sin(x/0.8))
- Separate variables:
v*dv = - (0.1 + sin(x/0.8)) . dx
-Integrate from v = 1 m/s @ x = 0:
0.5(v^2) = - (0.1x - 0.8cos(x/0.8)) - 0.8 + 0.5
0.5v^2 = 0.8cos(x/0.8) - 0.1x - 0.3
- Evaluate @ x = -1
0.5v^2 = 0.8 cos(-1/0.8) + 0.1 -0.3
v = sqrt (0.104516)
v = +/- 0.323 m/s
- v = v_max when a = 0:
-0.1 = sin(x/0.8)
x = -0.8*0.1002
x = -0.080134 m
- Hence,
v^2 = 1.6 cos(-0.080134/0.8) -0.6 -0.2*-0.080134
v = sqrt (0.504)
v = +/- 1.004 m/s
Answer:
The correct option is;
A. proper protection
Explanation:
Motorcycle riders ride the motorcycle while at some level of speed while having the entire body exposed to be a major part of any collision.
Injuries sustained from motorcycle accidents are several times more severe than injuries sustained by occupants of a car that is fully protected by the metallic panel in the same and even more serious accident scenarios
Hence, motorcycle riders require adequate protection by putting on available motorcyclist safety gear
Therefore, to reduce the risk of severe injury n a crash, motorcycle riders require proper protection.
Answer:
total amount of water after 2 min will be 84.4 kg/s
Explanation:
Given data:
one tank inflow = 0.1 kg/s
2nd tank inflow = 0.3 kg/s
3rd tank outflow = 0.03 kg/s
Total net inflow in tank is = 0.3 +0.1 =0.4 kg/s
From third point, outflow is 0.03 kg/s
Therefore, resultant in- flow = 0.4 - 0.03
Resultant inflow is = 0.37 kg/s
Tank has initially 40 kg water
In 2 min ( 2*60 sec), total inflow in tank is 0.37*60*2 = 44.4 kg
So, total amount of water after 2 min will be = 40+44.4 = 84.4 kg
