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3 years ago

9

Before taking off a plane travels at a speed of 1/4 km per second. The runaway is 5 km. How many seconds does it take the plane

to get to the end of the runaway?

1 answer:

Vikentia [17]3 years ago

8 0

Answer:

1 5segundos

Explanation:

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According to the decreasing order of toughness. list the following materials (note: the steels are assumed to have no cold work

fiasKO [112]

Answer:

1090 Steel >1040 Steel > Pure aluminium >Diamond.

Explanation:

Toughness:

Toughness can be define as the are of load -deflection diagram up to fracture point.

Modulus of toughness can be defines as the area of stress-strain diagram up to fracture point.Modulus of toughness is the property of material.

So the decreasing order of toughness can be given as follows

1090 Steel >1040 Steel > Pure aluminium >Diamond.

8 0

3 years ago

Entropy change is evaluated using Eq. 6.2a based on an internally reversible process. Can the entropy change between two states

Vadim26 [7]

Answer:

YES

Explanation:

Entropy is an extensive property of the system entropy change that value of entropy change can be determined for any process between the states whether reversible or not. i have attached the formula to calculate entropy change which is independent of whether the system is reversible or not and can be determined for any process.

4 0

3 years ago

Four race cars are traveling on a 2.5-mile tri-oval track. The four cars are traveling at constant speeds of 195 mi/h, 190 mi/h,

Snezhnost [94]

Answer:

Explanation:

1) The number of times, the car with the speed of 195 mph will cross the given point is equal to 30 minutes divided by the time taken by car to cross the 2.5 miles.

0 .5*195/2.5 = 39

Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times

and car with the speed of 180 mph crosses it 36 times

here, the time-mean speed, vt is given below,

vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

= 186.433 mph

and space mean speed is given by,

= (39+38+37+36)/(39/195+38/190+37/1850+36/180)

1) The number of times, the car with the speed of 195 mph will cross the given point is equal to 30 minutes divided by the time taken by car to cross the 2.5 miles.

0 .5*195/2.5 = 39

Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times

and car with the speed of 180 mph crosses it 36 times

here, the time-mean speed, vt is given below,

vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

= 186.433 mph

and space mean speed is given by,

= (39+38+37+36)/(39/195+38/190+37/1850+36/180)

=187.5 mph

2) There would be only four number of observations when the aerial photo is given, therefore time mean speed, vt in that condition will be calculated as

Vt = 195+190+185+180/4

= 187.5

Vs= 4/(1/195+1/190+1/185+1/180)

= 188.36 mph

2) There would be only four number of observations when the aerial photo is given, therefore time mean speed, vt, in that condition will be calculated as

Vt = 195+190+185+180/4

= 187.5

Vs= 4/(1/195+1/190+1/185+1/180)

= 188.36 mph

4 0

3 years ago

Advocard [28]

Answer:

I would say C

Explanation:

let me know if im right

3 0

1 year ago

Read 2 more answers

Travel Time Problem: Compute the time of concentration using the Velocity, Sheet Flow Method for Non-Mountainous Orange County a

Vaselesa [24]

Answer:

Total time taken = 0.769 hour

Explanation:

using the velocity method

for sheet flow ;

Tt = $\frac{0.007(nl)^{0.8} }{(Pl)^{5}s^{0.4} }$

Tt = travel time

n = manning CaH

Pl = 25years

L = how length ( ft )

s = slope

For Location ( 1 )

s = 0.045

L = 1000 ft

n = 0.06 ( from manning's coefficient table )

Tt1 = 0.128 hour

For Location ( 2 )

s = 2.5 %

L= 750

n = 0.13

Tt2 = 0.239 hour

For Location ( 3 )

s = 1.5%

L = 500 ft

n = 0.15

Tt3 = 0.237 hour

For Location (4)

s = 0.5 %

L = 250 ft

n = 0.011

Tt4 = 0.165 hour

hence the Total time taken = Tt1 + Tt2 + Tt3 + Tt4

= 0.128 + 0.239 + 0.237 + 0.165 = 0.769 hour

5 0

3 years ago