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Alona [7]
3 years ago
12

A train travels 90 kilometers in 4 hours, and then 82 kilometers in 1 hours. What is its average speed?

Physics
1 answer:
kvasek [131]3 years ago
5 0
Average speed = total distance / total time.

That's 172 km / 5 hr = 34.4 km//hr
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Open circuit

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In such a circuit, the wires are cut of and not connected properly.

The reverse is a closed circuit.

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What could it be.... (science)
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"where are the majority of transform faults located?"
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5 0
3 years ago
You are designing a ski jump ramp for the next Winter Olympics. You need to calculate the vertical height from the starting gate
natita [175]

Answer:

44.6 m

Explanation:

From the law of conservation of energy, the total energy at the top of the ramp, E equals the total energy at the bottom of the ramp.

E = E'

U₁ + K₁ + W₁ = U₂ + K₂ + W₂ where U₁ = potential energy at top of ramp = mgh where = height of ramp, K₁ = kinetic energy at top of ramp = 1/2mv₁² where v₁ = speed at top of ramp = 1.8 m/s, W₁ = work done by friction and air resistance at top of ramp = 0 J, U₂ = potential energy at bottom of ramp = 0 J(since the skier is at ground level h = 0), K₂ = kinetic energy at bottom of ramp = 1/2mv₂² where v₂ = speed at bottom of ramp = 28.0 m/s, W₁ = work done by friction and air resistance at bottom of ramp = 3500 J

Substituting the values of the variables into the equation, we have

U₁ + K₁ + W₁ = U₂ + K₂ + W₂

mgh + 1/2mv₁² + W₁ = U₂ + 1/2mv₂² + W₂

mgh + 1/2m(1.8 m/s)² + 0 J = 0 J + 1/2m(28 m/s)² + 3500 J

9.8 m/s² × 75 kg h + 1/2 × 75 kg (3.24 m²/s²) + 0 J = 0 J + 1/2 × 75 kg (784 m²/s²) + 3500 J

(735 kgm/s²)h + 75  kg(1.62 m²/s²) = 75 kg(392m²/s²) + 3500 J

(735 kgm/s²)h + 121.5  kgm²/s² = 29400 kgm²/s² + 3500 J

(735 kgm/s²)h + 121.5 J = 29400 J + 3500 J

(735 kgm/s²)h + 121.5 J = 32900 J

(735 kgm/s²)h = 32900 J - 121.5 J

(735 kgm/s²)h = 32778.5 J

h = 32778.5 J/735 kgm/s²

h = 44.6 m

So, the maximum height of the ramp for which the maximum safe speed will not be exceeded is 44.6 m.

4 0
3 years ago
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