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Paladinen [302]
2 years ago
6

Assuming that the density of air is 1.3 g/l, how many grams of carbon monoxide are in 1.0 l of air at the maximum allowable conc

entration?
Physics
1 answer:
Monica [59]2 years ago
4 0
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
<span>35x1.3/1000000 = 0.0000455g = 0.0455 mg = 45.5 microgram/L</span>


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2 years ago
What are common features of both the MMPI-2 and CPI? Select all that apply.
Amiraneli [1.4K]

Answer:     D

Both MMPI-2 and CPI check for serious personality disorders.

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3 years ago
A sound wave travels through water. What best describes the direction of the water particles?
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7 0
3 years ago
A food packet is dropped from a helicopter during a flood-relief operation from a height of 750 meters. Assuming no drag (air fr
Leviafan [203]

1. Velocity at which the packet reaches the ground: 121.2 m/s

The motion of the packet is a uniformly accelerated motion, with constant acceleration a=g=9.8 m/s^2 directed downward, initial vertical position d=750 m, and initial vertical velocity v_0 = 0. We can use the following SUVAT equation to find the final velocity of the packet after travelling for d=750 m:

v_f^2 -v_i^2 =2ad

substituting, we find

v_f^2 = 2ad\\v_f = \sqrt{2ad}=\sqrt{2(9.8 m/s^2)(750 m)}=121.2 m/s

2. height at which the packet has half this velocity: 562.6 m

We need to find the heigth at which the packet has a velocity of

v_f=\frac{121.2 m/s}{2}=60.6 m/s

In order to do that, we use again the same SUVAT equation substituting v_f with this value, so that we find the new distance d that the packet travelled from the helicopter to reach this velocity:

v_f^2-v_i^2=2ad\\d=\frac{v_f^2}{2a}=\frac{(60.6 m/s)^2}{2(9.8 m/s^2)}=187.4 m

Which means that the heigth of the packet was

h=750 m-187.4 m=562.6 m

3 0
2 years ago
HELP ME PLEASE *50 points*
aev [14]
Their is no document for us to look at , can you add it so i can help you
8 0
2 years ago
Read 2 more answers
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