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irinina [24]
3 years ago
12

Two​ vehicles, a car and a​ truck, leave an intersection at the same time. the car heads east at an average speed of 50 miles pe

r​ hour, while the truck heads south at an average speed of 60 miles per hour. find an expression for their distance apart d​ (in miles) at the end of t hours.
Physics
1 answer:
olya-2409 [2.1K]3 years ago
5 0

The car heads east at an average speed of 50 miles per​ hour from the intersection point towards East. The truck heads east at an average speed of 60 miles per​ hour from the intersection point towards South.

The distance of car from the intersection point after t hours is 50t.

The distance of truck from the intersection point after t hours is 60t.

Since these distances are perpendicular to each other, distance apart d​ (in miles) at the end of t hours is

d=\sqrt{(50t)^2+(60t)^2} \\ d=10\sqrt{61} t\\ d=78.1t

Thus the distance apart is d=78.1t \;miles

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Which of the following is a unit of acceleration? i
WINSTONCH [101]

Answer:

The SI unit of acceleration is metres/second2 (m/s2).

Explanation:

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6 0
3 years ago
A bead slides without friction around a loop the-loop. The bead is released from a height of 17.6 m from the bottom of the loop-
solong [7]

Answer:

 N₁ = 393.96 N   and  N = 197.96 N

Explanation:

In This exercise we must use Newton's second law to find the normal force. Let's use two points the lowest and the highest of the loop

Lowest point, we write Newton's second law n for the y-axis

          N -W = m a

where the acceleration is ccentripeta

          a = v² / r

           

          N = W + m v² / r

          N = mg + mv² / r

         

we can use energy to find the speed at the bottom of the circle

starting point. Highest point where the ball is released

           Em₀ = U = m g h

lowest point. Stop curl down

           Em_{f} = K = ½ m v²

           Emo = Em_{f}

           m g h = ½ m v²

           v² = 2 gh

we substitute

             N = m (g + 2gh / r)

            N = mg (1 + 2h / r)

let's calculate

          N₁ = 5 9.8 (1 + 2 17.6 / 5)

          N₁ = 393.96 N

headed up

we repeat the calculation in the longest part of the loop

          -N -W = - m v₂² / r

            N = m v₂² / r - W

             N = m (v₂²/r  - g)

we seek speed with the conservation of energy

           Em₀ = U = m g h

final point. Top of circle with height 2r

             Em_{f} = K + U = ½ m v₂² + mg (2r)

              Em₀ =   Em_{f}

            mgh = ½ m v₂² + 2mgr

             v₂² = 2 g (h-2r)

we substitute

            N = m (2g (h-2r) / r - g)

            N = mg (2 (h-r) / r 1) = mg (2h/r  -2 -1)

             N = mg (2h/r  - 3)

            N = 5 9.8 (2 17.6 / 5 -3)

            N = 197.96 N

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3 0
3 years ago
  Consider a bathtub filled with water, with a plug in the drain. How long would this be considered a closed system?  A. a few s
MatroZZZ [7]
Is this a trick question?

<u><em> . . a closed system doesn't exchange any matter with its surroundings and isn't subject to any external force, it is essentially an isolated system.</em></u>

In the case of a bathtub, <u><em>it is </em></u><em><u>open immediately </u></em><u><em>to evaporation</em></u>, so the truth is this system is <u><em>never </em></u>closed

<u><em>the correct answer is E.) none of the above</em></u>



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