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Rama09 [41]
4 years ago
12

A car is traveling at 36 km/h on an acceleration lane to a freeway. What acceleration is required to obtain a speed of 72 km/h i

n a distance of 100m? What time is required to travel this distance?
Engineering
1 answer:
dmitriy555 [2]4 years ago
6 0

First, write down the information given and the change units if necessary (we must have similar units to operate on).

Initial speed, u = 36 km/h = 10 m/s

Final speed, v = 72 km/h = 20 m/s

Distance, s = 100 m

We know that

{v}^{2}  -  {u}^{2}  = 2as \\  {20}^{2}  -  {10}^{2}  = 2 \times a \times 100 \\ 400 - 100 = 200 \times a \\ a =  \frac{300}{200 }  =  \frac{3}{2}  \: m {s}^{ - 2}

Now, we substitute v, u, and a in the formula

v = u + at \\ 20 = 10 +  \frac{3}{2} t \\  \frac{3}{2} t = 10 \\ 3t = 20 \\ t =  \frac{20}{3}  = 6.67 \: seconds

Please mark Brainliest if this helps!

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A rigid, well-insulated tank of volume 0.9 m is initially evacuated. At time t = 0, air from the surroundings at 1 bar, 27°C beg
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\dot{w}= -0.303 KW

Explanation:

This is the case of unsteady flow process because properties are changing with time.

From first law of thermodynamics for unsteady flow process

\dfrac{dU}{dt}=\dot{m_i}h_i+\dot{Q}-\dot{m_e}h_i+\dot{w}

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\dot{m_e}=0

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m_2u_2-m_1u_1=(m_2-m_1)h_i- \dot{w}\Delta t

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m_1,m_2 is the initial and final mass in the system respectively.

Initially tank is evacuated so m_1=0

We know that for air u=C_vT ,h=C_p T,P_2v_2=m_2RT_2

m_2=0.42 kg

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