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nevsk [136]
3 years ago
14

A straight railroad track is being built to connect two cities. the measured distance of the track between the two cities is 160

.5 miles. a mailstop is 28.5 miles from the first city. how far is the mailstop from the second city?
Physics
1 answer:
elena55 [62]3 years ago
7 0
Let point A be 0.0 miles (first city)
Let point B be 160.5 miles (first city to second city)
Let point C be 28.5 miles (first city to mail stop)

Take C – A = C [28.5 - 0.0 = 28.5] (This checks the distance between city 1 and Mail stop) 

Then Take B – C = Distance from the first city to the second city [160.5 - 28.5 = 132 Miles]


Answer: The Mail stop is 132 miles from the Second City.
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here we know that

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3 years ago
You're carrying a 3.0-m-long, 24 kg pole to a construction site when you decide to stop for a rest. You place one end of the pol
kozerog [31]

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F=133N

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From the question we are told that:

Length l=3.0m

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Look at the illustration , what concepts of heat are being shown in this pictures <br>​
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3 0
2 years ago
you are piloting a small plane and you want to reach an airport 450 km due south in 3.0 h a wind is blowing from the west 50.0 k
alex41 [277]

Answer:

You should choose airspeed 158.11 km/h at 18.4° west of south

Explanation:

The distance to the air port is 450 km due to south

You should to reach the airport in 3 hours

→ Velocity = distance ÷ time

→ Distance = 450 km , time = 3 hours

→ The velocity of your plane = 450 ÷ 3 = 150 km/h due to south

A wind is blowing from west 50 km/h

We need to know what heading and airspeed you should choose to

reach your destination

At first we must find the resultant velocity of your plane and the wind

The south and west are perpendicular, then the resultant velocity is

→ v_{R}=\sqrt{(v_{p})^{2}+(v_{w})^{2}}

→ v_{p}=150 km/h ,  v_{w}=50 km/h

→ v_{R}=\sqrt{(150)^{2}+(50)^{2}}=158.11 km/h

To cancel the velocity of the wind, the pilot should maintain the velocity

of the plane at 158.11 km/h

The direction of the velocity is the angle between the resultant velocity

and the vertical (south)

→ The direction of the velocity is tan^{-1}\frac{50}{150}=18.4°

The direction of the velocity is 18.4° west of south

<em>You should choose airspeed 158.11 km/h at 18.4° west of south</em>

8 0
3 years ago
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