Answer:
Airbags will deploy in almost any angle.
Explanation:
Cars have sensors around them, so when the car gets hit, the sensors detect a crash and deploy the airbags to keep you safe.
Answer:
The differential equation and the boundary conditions are;
A) -kdT(r1)/dr = h[T∞ - T(r1)]
B) -kdT(r2)/dr = q'_s = 734.56 W/m²
Explanation:
We are given;
T∞ = 70°C.
Inner radii pipe; r1 = 6cm = 0.06 m
Outer radii of pipe;r2 = 6.5cm=0.065 m
Electrical heat power; Q'_s = 300 W
Since power is 300 W per metre length, then; L = 1 m
Now, to the heat flux at the surface of the wire is given by the formula;
q'_s = Q'_s/A
Where A is area = 2πrL
We'll use r2 = 0.065 m
A = 2π(0.065) × 1 = 0.13π
Thus;
q'_s = 300/0.13π
q'_s = 734.56 W/m²
The differential equation and the boundary conditions are;
A) -kdT(r1)/dr = h[T∞ - T(r1)]
B) -kdT(r2)/dr = q'_s = 734.56 W/m²
A chemical engineer can clearly see from this kind of test if a substance stays in a system and builds up or if it just passes through.
<h3>What is a chemical engineer?</h3>
- Processes for manufacturing chemicals are created and designed by chemical engineers.
- To solve issues involving the manufacture or usage of chemicals, fuel, medications, food, and many other goods, chemical engineers use the concepts of chemistry, biology, physics, and math.
- A wide range of sectors, including petrochemicals and energy in general, polymers, sophisticated materials, microelectronics, pharmaceuticals, biotechnology, foods, paper, dyes, and fertilizers, have a significant demand for chemical engineers.
- Chemical engineering is undoubtedly difficult because it requires a lot of physics and math, as well as a significant number of exams at the degree level.
To learn more about chemical engineer, refer to:
brainly.com/question/23542721
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Answer:
Velocity of ball B after impact is 
and ball A is 
Explanation:
= Initial velocity of ball A
= Initial velocity of ball B = 0
= Final velocity of ball A
= Final velocity of ball B
= Coefficient of restitution = 0.8
From the conservation of momentum along the normal we have
Coefficient of restitution is given by
Adding the above two equations we get
From the conservation of momentum along the plane of contact we have
Velocity of ball B after impact is and ball A is .
