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igomit [66]
3 years ago
6

Ian walks 2 km to his best friend’s house, then walks 0.5 km to the library. He then makes a 2.5 km walk home. The entire walk t

ook him 3 hours. What is his average speed in km/h?
SHOW YOUR WORK!
Physics
1 answer:
Sonbull [250]3 years ago
4 0

Average speed = (total distance covered) / (time to cover the distance)

Ian's total distance = (2km + 0.5km + 2.5km) = 5 km

Ian's total time = 3 hours

Average speed = (5km) / (3 hours)

Average speed = 1.66 km/hr

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g You shine orange laser light that has a wavelength of 600 nm through a narrow slit. The slit forms a diffraction pattern on a
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3 0
3 years ago
Suppose that a person gets hit by a bus moving at 30 mi/h with a 58,000 lbs of force in the direction of motion. If the mass of
alexandr402 [8]

The impulse of a force is due to the change in the motion of an object

A. The persons speed after impact is approximately 59.38 mi/h

B. The expected speed is <u>29.89 mi/h</u> which is less than the findings

Reason:

Known parameters are;

The speed of the bus, v = 30 mi/h

The force with which the person was hit, F = 58,000 lbs

Mass of the bus, M = 40,000 lbs

Mass of the person, m = 150 lbs

Duration of the impact, Δt = 0.007 seconds

A. The speed of the person at the end of the impact, <em>v</em>, is given as follows;

The impulse of the force = F × Δt = m × Δv

For the person, we get;

58,000 lbf ≈ 1866094.816 lb·ft./s²

58,000 lbf × 0.007 s = 150 lbs × Δv

1,866,094.816 lb·ft./s²

\Delta v = \dfrac{1,866,094.816\ lbs \times 0.007 \, s}{150 \, lbs} \approx  87.084  \ ft./s

Δv = v₂ - v₁

The initial speed of the person at the instant, can be as v₁ = 0

The final speed, v₂ = Δv - v₁

∴ v₂ ≈  87.084 ft./s - 0 = 87.084 ft./s

≈ <u>87.084 ft./s</u>

<u />v_2 \approx \dfrac{87.084 \ ft./s}{y} \times\dfrac{1 \ mi}{5280 \ ft.} \times \dfrac{3,600 \ s}{1 \, hour} \approx 59.38 \ mi/h<u />

The speed of the person at the end of the impact, v₂ ≈ <u>59.38 mi/h</u>

B. Where the momentum is conserved, we have;

m₁·v₁ + m₂v₂ = (m₁ + m₂)·v

v = \dfrac{m_1 \cdot v_1 + m_2 \cdot v_2}{m_2 + m_1}

v = \dfrac{40,000 \times 30  + 150 \times 0}{40,000 + 150} \approx 29.89

The expected speed of the person at the end of the impact is 29.89 mi/h, and therefore, <u>the findings does not agree with the expectation</u>

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