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Oksanka [162]
2 years ago
5

Tahir travel twice as far as ahmed, but onley one third as fast. Ahmed starts travel on tuesday at noon at point x to point z 30

0km, by 9:00pm. Tahir starts travel same day after 3hr of ahmed. In which time tahir reach at point z?
Engineering
1 answer:
shepuryov [24]2 years ago
7 0

Answer:

6:00 pm the next day

Explanation:

Given that

Tahir traveled twice as far as Ahmed. We say,

Ahmed traveled a distance, D

Tahir would travel a distan, 2D

Tahir traveled 1/3 as fast as Ahmed, so we say

Ahmed traveled at a speed, S

Tahir would travel at a speed, S/3

If Ahmed starts travel on tuesday at noon at point x to point z 300km, by 9:00pm,

Time taken by Ahmed to travel is

9:00 pm - 12:00 pm = 9 hours

Ahmed, traveled 300 km in 9 hours, meaning he traveled at 33.3 km in an hour.

Speed, S that Ahmed traveled with is 33.3 km/h

Remember, we stated that Tahir travels at a speed of S/3, that is, The speed of Tahir is

33.3/3 = 11.1 km/h.

300 km would then be traveled in 300 km/11.1 km/h = 27 hours.

Tahir started traveling, 3 hours after Ahmed, that is 12:00 pm + 3:00 hrs = 3:00 pm, and if he's to spend 27 hours on the journey he would reach destination z at 6:00 pm the next day

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Answer:

Sound barrier.

Explanation:

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Describe what V1-V4 is
photoshop1234 [79]

Answer:

it is wave inversion

Explanation:

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3 years ago
A cooling system load is 96,000 BTUh sensible. How much chilled air is required to satisfy the load if the system is designed fo
Natalija [7]

Answer:

For 20^{\circ} - 5.556 lb/s

For 15^{\circ} - 7.4047 lb/s

Solution:

As per the question:

System Load = 96000 Btuh

Temperature, T = 20^{\circ}

Temperature rise, T' = 15^{\circ}

Now,

The system load is taken to be at constant pressure, then:

Specific heat of air, C_{p} = 0.24 btu/lb ^{\circ}F

Now, for a rise of 20^{\circ} in temeprature:

\dot{m}C_{p}\Delta T = 96000

\dot{m} = \frac{96000}{C_{p}\Delta T} = \frac{96000}{0.24\times 20} = 20000 lb/h = \frac{20000}{3600} = 5.556 lb/s

Now, for 15^{\circ}:

\dot{m}C_{p}\Delta T = 96000

\dot{m} = \frac{96000}{C_{p}\Delta T} = \frac{96000}{0.24\times 15} = 26666.667 lb/h = \frac{26666.667}{3600} = 7.4074 lb/s

4 0
2 years ago
Which basic principle influences how all HVACR systems work?
bezimeni [28]

Answer:

B) An increase in pressure can lower the boiling point of a liquid and change the temperature at which it turns to a gas.

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6 0
3 years ago
A number 12 copper wire has a diameter of 2.053 mm. Calculate the resistance of a 37.0 m long piece of such wire.
Alinara [238K]

Answer:

R=1923Ω

Explanation:

Resistivity(R) of copper wire at 20 degrees Celsius is 1.72x10^-8Ωm.

Coil length(L) of the wire=37.0m

Cross-sectional area of the conductor or wire (A) = πr^2

A= π * (2.053/1000)/2=3.31*10^-6

To calculate for the resistance (R):

R=ρ*L/A

R=(1.72*10^8)*(37.0)/(3.31*10^-6)

R=1922.65Ω

Approximately, R=1923Ω

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2 years ago
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