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Dafna1 [17]
3 years ago
5

Part B For this activity, you'll analyze the motion of the marked spot on the skateboarder's leg. For each dropdown labeled “cm”

or “model cm,” change the option to “Legs.” In each frame containing a plot or table, you can use the buttons at the top of the frame to choose whether it shows a plot (graph) or a table. If it is showing a plot, you can select the y-axis label to choose which measurement to plot. See how to change the graph and table options. Now look at the graphs for the horizontal and vertical displacement of the legs against time. How do the two graphs, x vs. t and y vs. t appear? What can you say about the displacement of the snowboarder?
Physics
1 answer:
Anika [276]3 years ago
7 0

Answer:

Explanation:

c. By using the Select Data button and the Select Data Source optionExplanation:A scatter plot is a plot which is used to plot the points of the data on the horizontal and the vertical axis also it depicts how one variable is affected by the another. After preparing the scatter plot to enter the data in the scatter plot we need to use the data button and then data source option so that the data could be entered in the scatter plothence, option c is correct

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You have to design a loop-the-loop for a new amusement park so that when each car passes the top of the loop inverted (upside-do
elena-14-01-66 [18.8K]

first of all go to slader .com to serach up this answer your welcome they have the best answers

6 0
3 years ago
A two slit pattern is viewed on a screen 1.00m from the slits if the two third-order minima are 22.0 cm apart what is the width
Bingel [31]

Answer:

4.4 cm

Explanation:

Given:

Distance of the screen from the slit, D = 1 m

Distance between two third order interference minimas, x = 22 cm

Let's say, minima occurs at:

x_n = (n + \frac{1}{2}) \frac{wL}{d}

We have:

2x_2 = 2(2 + \frac{1}{2}) * \frac{w*22}{d}

Calculating further for the width of the central bright fringe, we have:

\frac{w}{d} = \frac{22}{5}

= 4.4 cm

Note: w in representswavelength

8 0
3 years ago
Suppose we have a laser emitting a diffraction-limited beam (=632 nm) with a 2-mm diameter. How big a light spot from this lase
expeople1 [14]

Answer:

Explanation:

λ = wave length = 632 x 10⁻⁹

slit width a = 2 x 10⁻³ m

angular separation of central maxima

= 2 x λ /a

= 2 x 632 x 10⁻⁹ / 2 x 10⁻³

= 632 x 10⁻⁶ rad

width in m of light spot.

= 632 x 10⁻⁶  x 376000 km

= 237.632 km

5 0
3 years ago
A damped mass/spring system takes 14.0 s for its amplitude of the oscillator to decrease by a factor of 9. By what factor does t
fiasKO [112]

Answer:

The correct answer is "0.246".

Explanation:

Given that the amplitude is decreased by a factor of 9, then

A \rightarrow (A-\frac{A}{9} )

A \rightarrow \frac{8A}{9}

As we know,

Energy will be:

⇒  E_{1}=\frac{1}{2}KA^2

and,

⇒  E_{2}=\frac{1}{2}K(\frac{8A}{9} )^2

          =\frac{64KA^2}{162}

⇒  \Delta E=E_1-E_2

On putting the estimated values, we get

           =\frac{1}{2}KA^2-\frac{64KA^2}{162}

⇒  \frac{\Delta E}{E}=\frac{\frac{20}{162}KA^2}{\frac{1}{2}KA^2}

          =\frac{40}{162}

          =0.246

3 0
3 years ago
A particle on a spring moves in simple harmonic motion along the x axis between turning points at x1 = 95 cm and x2 = 135 cm. (i
uranmaximum [27]

Answer:

(i) x = 115\,cm, (ii) x = 95\,cm, (iii) x = 95\,cm

Explanation:

(i) x_{1} and x_{2} represent the points where particle has a velocity of zero and spring reach maximum deformation, Given the absence of non-conservative force and by the Principle of Energy Conservation, the position where particle is at maximum speed is average of both extreme positions:

x = 115\,cm

(ii) Maximum accelerations is reached at x_{1} and x_{2}.

x = 95\,cm

(iii) Greatest net forces exerted on the particle are reached at  x_{1} and x_{2}.

x = 95\,cm

8 0
3 years ago
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