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kirill [66]
3 years ago
6

The forces acting on a falling leaf are.

Physics
1 answer:
mojhsa [17]3 years ago
3 0
B. Gravity and air resistance.
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Fifteen identical particles have various speeds: one has a speed of 2.00 m/s, two have speeds of 3.00 m/s, three have speeds of
Evgesh-ka [11]

Answer:

a)  V=7.5m/s

b) rms=8.4m/s

c) Generally the most probable speed is 8m/s as it the most posses by particles being the average

Explanation:

From the question we are told that:

Sample size N=15

  Speed 1 v_1=2m/s\\\\Speed 2 v_2=3m/s\\\\Speed 3 v_2=5m/s\\\\Speed 4 v_4=8m/s\\\\Speed 3 v_5=9m/s\\\\Speed 2 v_6=15m/s\\\\

Generally the equation for Average speed is mathematically given by

 V_{avg}=\frac{\sum(nv)}{N}

Therefore

 V_{avg}=\frac{(2+2(3)+3(5)+4(8)+3(9)+2(15))}{15}

 V=7.5m/s

b)

Generally the equation for RMS speed of the particle is mathematically given by

  rms=\sqrt{\frac{\sum(nv^2)}{N}}

  rms=\sqrt{\frac{2^2+2(3)^2+3(5)^2+4(8)^2+3(9)^2+2(15)^2}{15}}

  rms=\sqrt{69.73}

  rms=8.4m/s

c

Generally the most probable speed is 8m/s as it the most posses by particles being the average

4 0
3 years ago
A number line goes from negative 5 to positive 5. Point D is at negative 4 and point E is at positive 5. A line is drawn from po
saul85 [17]

The location of the point F that partitions a line segment from D to E (\overline{DE}), that goes from <u>negative 4</u> to <u>positive 5,</u> into a 5:6 ratio is fifteen halves (option 4).  

We need to calculate the segment of the line DE to find the location of point F.

Since point D is located at <u>negative -4</u> and point E is at <u>positive 5</u>, we have:

\overline{DE} = E - D = 5 - (-4) = 9

Hence, the segment of the line DE (\overline{DE}) is 9.

Knowing that point F partitions the line segment from D to E (\overline{DE}) into a <u>5:6 ratio</u>, its location would be:

F = \frac{5}{6}\overline{DE} = \frac{5}{6}9 = 5*\frac{3}{2} = \frac{15}{2}  

Therefore, the location of point F is fifteen halves (option 4).

Learn more about segments here:

  • brainly.com/question/24472171?referrer=searchResults
  • brainly.com/question/13270900?referrer=searchResults

       

I hope it helps you!

5 0
3 years ago
Read 2 more answers
A horizontal force of 750 N is needed to overcome the force of static friction between a level floor and a 250-kg crate. What is
loris [4]

Answer:

The acceleration of the crate is 1.82\ m/s^2.

Explanation:

Given that,

Force, F = 750 N

Mass of the crate, m = 250 kg

The coefficient of friction is 0.12.

We need to find the acceleration of the crate. The net force acting on the crate is given by :

F=ma\\\\F-f=ma

f is frictional force, f=\mu N=\mu mg

F-\mu mg=ma\\\\a=\dfrac{F-\mu mg}{m}\\\\a=\dfrac{750-0.12\times 250\times 9.8}{250}\\\\a=1.82\ m/s^2

So, the acceleration of the crate is 1.82\ m/s^2. Hence, this is the required solution.

4 0
3 years ago
Which one is the answer ?
ollegr [7]
It’s blurry the pic




But I think that is the second one
6 0
3 years ago
Pls help i give brainliest
san4es73 [151]

Answer:

a.

Explanation:

cuz whenever the ball was traveling up the staircase it was building up energy to use (potential energy) unlike b.

5 0
3 years ago
Read 2 more answers
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