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sergiy2304 [10]
3 years ago
11

The truck travels in a circular path having a radius of 50 m at a speed of v = 4 m>s. For a short distance from s = 0, its sp

eed is increased by v # = (0.05s) m>s 2, where s is in meters. Determine its speed and the magnitude of its acceleration when it has moved s = 10 m.
Engineering
1 answer:
miss Akunina [59]3 years ago
6 0

Answer:

Magnitude of velocity when s is 10 m= 4.58m/s

Magnitude of acceleration when s is 10 m = 0.653 m/s²

Explanation:

Radius of circular path = 50m

Speed = 4m/s

increase in acceleration = (0.05s)m/s²

speed and acceleration when s is 10 m = ?

Magnitude of Velocity:

From third equation of motion

v_{f}^{2}-v_{i}^{2}=2aS\\\\v_{f}=\sqrt{2aS+v_{i}^{2}} \\\\v_{f}=\sqrt{2(0.05)(10)+(4)^{2}} \\\\v_{f}=4.58\,m/s

Magnitude of Acceleration:

a=\sqrt{a_{t}^{2}+a_{n}^{2}}--(1)

The tangential component of acceleration is

a_{t}=(0.05)(10)\\a_{t}=0.5\,m/s^{2}

The normal component of acceleration is

a_{n}=\frac{v^{2}}{R}\\\\a_{n}=\frac{16}{50}\\\\a_{n}=0.42\,m/s^{2}

Substituting tangential and normal component in (1)

a=\sqrt{(0.5)^{2} +(0.42)^{2} }

a=0.653 m/s²

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A composite wall is composed of 20 cm of concrete block with k = 0.5 W/m-K and 5 cm of foam insulation with k = 0.03 W/m-K. The
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Answer:

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Explanation:

The rate of heat transfer through the wall is given by:

q=\frac{Ak}{L}dT

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Assumptions:

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{k_{fi}= 0.03 W/m.K

{L_{fi}= 5 cm = 0.05 m

{T_{fi}= 25 \°C

{k_{cb} = 0.5 W/m.K

{L_{cb}= 20 cm = 0.20 m

{T_{cb}= 0 \°C

{T_{m}= ? \°C = temperature at the interface

Solving for {T_{m} will give the temperature at the interface:

\frac{q}{A}=\frac{k_{fi} }{L_{fi} }(T_{fi} -T_{m})=\frac{k_{cb} }{L_{cb} }(T_{m} -T_{cb})

\frac{0.03}{0.05 }(25 -T_{m})=\frac{0.5}{0.2}(T_{m} -0})

15 -0.6T_{m}=2.5T_{m}

3.1T_{m}=15

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Two well-known NP-complete problems are 3-SAT and TSP, the traveling salesman problem. The 2-SAT problem is a SAT variant in whi
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both the statements are true.

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