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mamaluj [8]
2 years ago
4

The following data was taken during a student's experiment with an object moving at a relatively constant velocity. Use the data

to create a position-time graph (on the accompanying graph paper). Be sure to include a best-fit line. After the graph is completed, use your best-fit line to calculate the average speed of the object. Show speed calculations below.​
Physics
1 answer:
katrin2010 [14]2 years ago
4 0

Based on the best-fit line, the average speed of the object is equal to 6.15 m/s.

<h3>What is a scatter plot?</h3>

A scatter plot is also referred to as scatter chart, scatter diagram or scattergram and it can be defined as a type of graph which is used for the graphical representation of the values of two (2) variables, with the resulting points showing any association (correlation) between the data set.

<h3>What is a position vs time graph?</h3>

A position vs time graph can be defined as a type of graph that is used to graphically represent the distance traveled (covered) by an object from its starting position with respect to the time when it is started moving.

By critically observing the graph (see attachment) which models the data in the given table, we can infer and logically deduce that the linear function from the best-fit line is given by:

y = 6.35x + 0.86

<h3>What is a slope?</h3>

In Mathematics, the slope of a straight line on a position vs time graph simply refers to the ratio of displacement to time interval and it represents the average speed.

For the average speed, we have:

Average speed, ΔV = Δd/Δt

Average speed, ΔV = (40 - 8)/(6.2 - 1.0)

Average speed, ΔV = 32/5.2

Average speed, ΔV = 6.15 m/s.

Read more on scatterplot here: brainly.com/question/6592115

#SPJ1

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poizon [28]

Answer:

Gravity, normal force, friction, and air resistance.

Explanation:

Assuming the ball is on earth, there is always gravity acted upon the object.

Since the problem said that the golf ball is moving throughout the air, we can tell that there will be air resistance and friction(air resistance is a type of friction). There is also normal force pushing on the ball as it bumps into the air, as the air is pushing back with an equal and opposite force.

I don't know what you mean by "applied", so I don't think there's that type of force exerted on the ball.

Lastly, tension only occurs when you pull a string/rope. There is no string/rope in this case, so there is no tension exerted on the ball.

I hope this helped you.

8 0
3 years ago
Suppose you design an apparatus in which a uniformly charged disk of radius R is to produce an electric field. The field magnitu
ANEK [815]

Answer:

The electric field will be decreased by 29%

Explanation:

The distance between point P from the distance z = 2.0 R

Inner radius = R/2

Outer raidus = R

Thus;

The electrical field due to disk is:

\hat {K_a} = \dfrac{\sigma}{2 \varepsilon _o} \Big( 1 - \dfrac{z}{\sqrt{z^2+R_i^2}} \Big))

\implies \dfrac{\sigma}{2 \vaepsilon _o} \Big ( 1 - \dfrac{2.0 \ R}{\sqrt{ (2.0\ R)^2+(R)^2}} \Big)

Similarly;

\hat {K_b} = \hat {k_a} - \dfrac{\sigma}{2 \varepsilon_o} \Big( 1 - \dfrac{2.0 \ R}{\sqrt{(2.0 \ r)^2 + (\dfrac{R}{2}^2)}}\Big)

However; the relative difference is: \dfrac{\hat {k_a} - \hat {k_b}}{\hat {k_a} }= \dfrac{E_a -E_a + \dfrac{\sigma}{2 \varepsilon_o  \Big[1 - \dfrac{2.0 \ R}{\sqrt{(2.0 \ R)^2 + (\dfrac{R}{2})^2}} \Big] } } { \dfrac{\sigma}{2 \varepsilon_o \Big [ 1 - \dfrac{2.0 \ R}{\sqrt{ (2.0 \ R)^2 + (R)^2}} \Big] }}

\dfrac{\hat {k_a} - \hat {k_b}}{\hat {k_a} }= \dfrac{1 - \dfrac{2.0}{\sqrt{(2.0)^2 + \dfrac{1}{4}}} }{1 - \dfrac{2.0 }{\sqrt{(2.0)^2 + 1}}}

= 0.2828 \\ \\ \mathbf{\simeq  29\%}

3 0
3 years ago
If a star's radiation peaks in the ultraviolet region of the spectrum, you would not be able to see it. true or false.
slega [8]

the answer would be false

hope I helped and goodluck

:)

6 0
3 years ago
Read 2 more answers
If a car travels 30 kilometers in 2 hours, it’s average speed is?
7nadin3 [17]

Answer:

15km/h

Explanation:

Time taken to complete 30km/h=2 hours=15km/h hence the average speed is 15km/h.

5 0
3 years ago
A 67 kg kg driver gets into an empty taptap to start the day's work. The springs compress 2.3×10−2 m m . What is the effective s
amid [387]

Point of correction, spring constant is 2.3×10−2 m not 2.3×10−2 m m

Answer:

28577 N/m

Explanation:

From Hooke's law, F=kx where F is force applied, k is spring constant and x is compression

F=mg=67*9.81

k=\frac {mg}{x}

k=\frac {67*9.81}{2.3\times 10^{-2}}=28576.95652 N/m

Approximately, 28577 N/m

8 0
3 years ago
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