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tamaranim1 [39]
3 years ago
5

At positions and_______, _______kinetic energy is maximum. At position

Physics
2 answers:
goblinko [34]3 years ago
8 0

Answer:

<h2><em>At positions A and G kinetic energy is maximum. At position D, potential energy is maximum.</em></h2>

Explanation:

Kinetic energy can be defined as the energy of a body due to its movement. This energy depends directly on the speed, the more velocity, more kinetic energy will be.

Potential energy is defined as the energy of a body due to its position where the velocity is zero, for example, the position at 12 meters above sea level.

Now, in this case, at points A and G the kinetic energy is maximum because at such moment, the speed of the movement is the highest.

The point D have the maximum potential energy, because there the ball has no velocity, at that moment the ball instantly stops.

So, the complete paragraph is

<em>At positions A and G kinetic energy is maximum. At position D, potential energy is maximum.</em>

worty [1.4K]3 years ago
7 0
Hi there!

I believe it looks like this.

At positions A, G kinetic energy is maximum.
At position D, potential energy is maximum.

Hope this helps! ☺♥
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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

5 0
4 years ago
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4 years ago
A cart moves along a track at a velocity of 3.5 cm/s. When a force is applied to the cart, its velocity increases to 8.2 cm/s. I
Lorico [155]

Answer:

3.13cm/s²

Explanation:

Given

Initial velocity u = 3.5cm/s

Final velocity v = 8.2cm/s

Time t = 1.5secs

Required

Acceleration of the cart a

To get that, we will use the equation of motion

v = u+at

Substitute the given parameters

8.2 = 3.5+1.5a

1.5a = 8.2-3.5

1.5a = 4.7

a = 4.7/1.5

a = 3.13cm/s²

Hence the acceleration to the cart is 3.13cm/s²

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