The forces acting on a falling leaf are.

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 negative 4 to positive 5, 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 negative -4 and point E is at positive 5, 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 5:6 ratio, 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