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Nata [24]
3 years ago
7

Gibbons, small Asian apes, move by brachiation, swinging below a handhold to move forward to thenext handhold. A 9.3kggibbon has

an arm length (hand to shoulder) of 0.60m. We can model its motionas that of a point mass swinging at the end of a 0.60-m-long, massless rod. At the lowest point of its swing,the gibbon is moving at 3.3m/s. What upward force must a branch provide to support the swinging gibbon?
Physics
1 answer:
Mumz [18]3 years ago
5 0

Upward force provided by the branch is 260 N

<u>Explanation:</u>

Given -

Mass of Gibbon, m = 9.3 kg

Length of the branch, r = 0.6 m

Speed of the movement, v = 3.3 m/s

Upward force, T = ?

The tension force in the rod must be greater than the weight at the bottom of the swing in order to provide an upward centripetal acceleration.

Therefore,

F net = T - mg

F net = ma = mv²/r

Thus,

T = mv²/r + mg

T = m ( v²/r + g)

T = 9.3 [ ( 3.3)² / 0.6 + 9.8]

T = 259.9 N ≈ 260 N

Therefore, upward force provided by the branch is 260 N

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A positive charge of 0.00047 C is 15 m from a negative charge of 0.00089 C. What is the force of one of the charges due to the o
Lady_Fox [76]

Answer:

16.732 N

Explanation:

Given:

q1 = 0.00047 C = 4.7 x 10^-4 C

q2 = 0.00089 C = 8.9 x 10^-4 C

d = 15 m

k = 9 x 10^9 N m^2 / C^2

To Find:

F = ?

Solution:

F = k x q1 x q2/d^2

F = 9 x 10^9 x 4.7 x 10^-4 x 8.9 x 10^-4 / 15 x 15

F = 9 x 4.7 x 8.9 x 10^9 x 10^-4 x 10^-4 / 225

F = 9 x 4.7 x 8.9 x 10^9 x 10^-8 / 225

F = 9 x 4.7 x 8.9 x 10 / 225

F = 418.3/25

F = 1673.2/100

Therefore, F = 16.732 N

PLZ MARK ME AS BRAINLIEST!!!

8 0
3 years ago
Which data set has the largest standard deviation
AfilCa [17]

Answer:

<em>The data set marked as B has the largest standard deviation</em>

Explanation:

<u>Standard Deviation</u>

It's a number used to show how a set of measurements is spread out from the average value. A low standard deviation means that most of the values are close to the average. A high standard deviation means that the numbers are more spread out.

The formula for the standard deviation is

\displaystyle \sigma=\sqrt{\frac{\sum (x_i-\mu)^2}{n}}

Where x_i is the value of each measurement, n is the number of elements in the set, and \mu is the average or media of the values, defined as

\displaystyle \mu=\frac{\sum x_i}{n}

Let's analyze each set of data:

A.3,4,3,4,3,4,3

The average is

\displaystyle \mu=\frac{3+4+3+4+3+4+3}{7}=3.43

Computing the stardard deviation:

\sigma=\sqrt{\frac{(3-3.43)^2+(4-3.43)^2+(3-3.43)^2+(4-3.43)^2+(3-3.43)^2+(4-3.43)^2+(3-3.43)^2}{7}}

\sigma=0.5

B.1,6,3,15,4,12,8

The average is

\displaystyle \mu=\frac{1+6+3+15+4+12+8}{7}=7

Computing the stardard deviation:

\sigma=\sqrt{\frac{(1-7)^2+(6-7)^2+(3-7)^2+(15-7)^2+(4-7)^2+(12-7)^2+(8-7)^2}{7}}

\sigma=4.7

C. 20, 21,23,19,19,20,20

The average is

\displaystyle \mu=\frac{20+21+23+19+19+20+20}{7}=20.29

Computing the stardard deviation:

\sigma=\sqrt{\frac{(20-20.29)^2+(21-20.29)^2+(23-20.29)^2+(19-20.29)^2+(19-20.29)^2+(20-20.29)^2+(20-20.29)^2}{7}}

\sigma=1.3

D.12,14,13,14,12,13,12

The average is

\displaystyle \mu=\frac{12+14+13+14+12+13+12}{7}=12.86

Computing the stardard deviation:

\sigma=\sqrt{\frac{(12-12.86)^2+(14-12.86)^2+(13-12.86)^2+(14-12.86)^2+(12-12.86)^2+(13-12.86)^2+(12-12.86)^2}{7}}

\sigma=0.8

We can see the data set marked as B has the largest standard deviation

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