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3 years ago

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An object is traveling at a constant velocity of 15 m/s when it experiences a constant acceleration of 3.5 m/s2 for a time of 11

s. What will its velocity be after that acceleration?

1 answer:

Luda [366]3 years ago

7 0

Vf=Vi+at=15m/s+(3.5m/s^2)(11s)=53.5m/s

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Because the pressure falls, water boils at a lower temperature with increasing altitude. Consequently, cake mixes and boiled egg

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Answer:

1) The boiling point of water reduces by 3.28°C at 1,000 m above sea-level

2) The boiling point of water reduces by 6.56°C at 2,000 m above sea-level

Explanation:

The variation of the boiling point of water with elevation is given as follows

The boiling point reduces by 0.5°C for every 152.4 meter increase in elevation

At sea-level, the boiling point temperature of water = 100°C

1) At 1,000 m elevation, the boiling point temperature, T = 100 - (1,000/152.4) × 0.5 ≈ 96.72 °C

Therefore, the boiling point of water reduces by 100° - 96.72° = 3.28°C at 1,000 m above sea-level

2) At 2,000 m elevation, the boiling point temperature, T = 100 - (2,000/152.4) × 0.5 ≈ 93.44°C

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Answer:

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Explanation:

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4 years ago

A piston is filled with gas. When the pressure is increasing, what is happening to the volume? Assume that all other properties

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3 years ago

a hammer drops from a height of 8 meters. calculate the speed with which it hits the ground. show work

ioda

Answer:

12.5 m/s

Explanation:

The motion of the hammer is a free fall motion, so a uniformly accelerated motion, therefore we can use the following suvat equation:

$v^2-u^2=2as$

Where, taking downward as positive direction, we have:

s = 8 m is the displacement of the hammer

u = 0 is the initial velocity (it is dropped from rest)

v is the final velocity

$a=g=9.8 m/s^2$ is the acceleration of gravity

Solving the equation for v, we find the final velocity:

$v=\sqrt{u^2+2as}=\sqrt{0+2(9.8)(8)}=12.5 m/s$

So, the final speed is 12.5 m/s.

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3 years ago