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NemiM [27]
4 years ago
10

A car traveling at a rate of 30 ft/sec is approaching an intersection. When the car is 120 ft from the intersection, a truck tra

veling at a rate of 40 ft/sec crosses the intersection. If the roads are at right angles to each other, how fast are the car and the truck separating 2 seconds after the truck crosses the intersection?
Physics
1 answer:
kozerog [31]4 years ago
5 0

Answer:

how fast are the car and the truck separating 2 seconds after the truck crosses the intersection: = 140ft

Explanation:

A car traveling at a rate of 30 ft/sec is approaching an intersection.

When the car is 120 ft from the intersection,

a truck traveling at a rate of 40 ft/sec crosses the intersection.

If the roads are at right angles to each other,

how fast are the car and the truck separating 2 seconds after the truck

for the car,

speed = distance/time

30 = 120/time

when the car is at 120ft to the intersection, he will used  t = 4sec to the intersection

for every 1sec, the car move 30ft

for the truck

speed = distance/time

40 = d/2sec

the distance of the truck at 2sec = 80ft

how fast are the car and the truck separating 2 seconds after the truck crosses the intersection: = 140ft

the car will be 60ft to the intersection and the truck will be 80ft after the intersection

therefore, the distance between the car and truck = 60+80 = 140ft

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5. A stream of air at 12 bar and 900 K is mixed with another stream of air at 2 bar and 400 K with 2 times the mass flowrate. If
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The anserrs to the question are

(a)  The temperature would be 566.67  K and

(b) The pressure of the resulting air stream is 14 bar

Explanation:

Here the two streams of gases meet ad form a single stream

The steady-flow energy equation can be implemented at the mixing point of the to streams as follows

mₐhₐ₁ + mₙ hₙ₁ + Q° + W° = mₐhₐ₂ + mₙhₙ₂

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Q = 0 and  W = 0  hence

mₐhₐ₁ + mₙ hₙ₁ = mₐhₐ₂ + mₙhₙ₂ where h = cp×T we have

mₐcpₐ×Tₐ + mₙcpₙ×Tₙ  = mₐcpₐ×T + mₐcpₙ×T

Therefore the final temperature T is given by

T = \frac{m_ac_{pa}T_a + m_nc_{pn}T_n}{m_ac_{pa} + m_nc_{pn}}  for the same kind of gas cpₐ =cpₙ therefore

T = \frac{m_aT_a + m_nT_n}{m_a + m_n} where mₐ and mₙ are mass flow rate

Therefore we have where Tₐ = = 900 K and Tₙ = 400 K and mₙ = 2mₐ gives

T = \frac{m_a*900 + 2*m_a*400}{m_a + 2*m_a} = T = \frac{900 + 800}{1 + 2} = 1700/3 = 566.67 K  

From Dalton's law, the total pressure of a mixture of gases is equal to the sum of the partial pressures of the components of the mixture

Therefore the total pressure of the combined stream = pₐ + pₙ = p

= 12 bar + 2 bar = 14 bar

Stream pressure = 14 bar

4 0
3 years ago
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Answer:

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Using the law of conservation of momentum to solve the problem. According to the law, the sum of momentum of the bodies before collision is equal to the sum of the bodies after collision. The bodies move with the same velocity after collision.

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