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WINSTONCH [101]
3 years ago
15

How would a small bar magnet be oriented when placed at position X​

Physics
2 answers:
kotykmax [81]3 years ago
6 0

Answer:

D

Explanation:

Well opposites attract, don't they?

Alex Ar [27]3 years ago
4 0

The answer is right here

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What is the average velocity of the particle from rest to 9 seconds?
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Answer: v = 2[m/s]Explanation:This avarage velocity can be found with the ... B. 2 meters/ second. C. 3 meters/second. D. 4 meters/second. 1.
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What Is Rotational Motion?​
mars1129 [50]

Answer:

the motion of an object in a circular path , in a fixed orbit

8 0
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A bowling ball is pushed with a force of 22.0N and accerlates at 5.5 m/s square. What is the mass of the bowling ball
stealth61 [152]
Mass of bowling ball is 4.0kg (A)
7 0
3 years ago
A girl and a boy pull in opposite directions on strings attached to each end of a spring balance. Each child exerts a force of 2
BabaBlast [244]

Answer:

yo they deleted my answer. The answer is 0N

Explanation:

so when two forces pull on an object from opposite sides with the same force (in this case its 20N), then the object is in equilibrium at 0N.

So its clear that there is one person on the the opposite side.

SOOO  generally<u>: (left or down) would be considered </u><u>negative</u><u> in an equation. And the other person  (right or up) would be considered </u><u>positive</u><u>.</u> So if both forces are the same numbers on opposite sides then the answer is 0 (if you add both of them).

<em>0 is the number of equilibrium.</em>

OK, so the equation would be -20N + 20N and then badda bing badda boom viola, the answer: 0N

thanks for coming to my TED talk. I hope they don't delete this answer.

5 0
3 years ago
An air filter can remove dust particles from air but will reach capacity (saturation) at 50.0 mg. Of air containing 225 µg dust
zubka84 [21]

Answer:

Time to Reach Saturation = 0.0146 day

Explanation:

In order to solve this problem, we first need to calculate the dust filtered by the filter per cubic meter of air:

Filtered Dust per m³ = Dust Particles Entering per m³ - Dust Particles Leaving per m³

Filtered Dust per m³ = 225 μg/m³ - 15 μg/m³

Filtered Dust per m³ = 210 μg/m³ = 210 x 10⁻³ mg/m³

Now, we find volume flow rate of air through filter:

Volume Flow Rate of Air = (400 ft³/min)(0.3048 m/1 ft)³

Volume Flow Rate of Air = 11.33 m³/min

Now, we calculate rate of dust filtered:

Rate of Dust Filtered = (Filtered Dust per m³)(Volume Flow Rate of Dust)

Rate of Dust Filtered = (210 x 10⁻³ mg/m³)(11.33 m³/min)

Rate of Dust Filtered = 2.38 mg/min

Now, for the time required to reach saturation:

Time to Reach Saturation = (Saturation Capacity)/(Rate of Dust Filtered)

Time to Reach Saturation = (50 mg)/(2.38 mg/min)

Time to Reach Saturation = (21.02 min)(1 day/24 h)(1 h/60 min)

<u>Time to Reach Saturation = 0.0146 day</u>

3 0
4 years ago
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