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Elina [12.6K]
3 years ago
7

30 POINTS! (Question is located in picture under graph)

Physics
2 answers:
harkovskaia [24]3 years ago
8 0
Hi i think its E because the the numbers on both sides are to spread out and are not even one side is 4 and the other is 3
a_sh-v [17]3 years ago
5 0

Answer:

B) The unit of the variable on the y-axis is incorrect.

Explanation:

The question to this exercise is in the picture. It is '' What makes the above graph incorrect? ''

Let's analize each option :

A) The variables are plotted along incorrect axes. This is wrong because actually the graph is well-plotted. In a displacement versus time graph we put the time variable in the x-axis and the displacement variable in the y-axis.

B) The unit of the variable on the y-axis is incorrect. This is true. In the graph we can see that ''Displacement(\frac{m}{s})'' when the units of displacement are units of lengths such as meters or miles for example. Therefore, the B) is a correct option

C) The scale for each axis is not appropriate for the data. This is wrong. Because of the scale for each axis we are able to see the five  data points which are (1,2) ; (2,4) ; (3,6) ; (4,8) and (5,10)

D) The data points have not been plotted clearly. This is wrong. All data points are well-plotted signposted with a point at the intersection of the dashed coordinated lines. For example, the point (1,2) is at the intersection of two lines : one dashed line perpedicular to the x-axis at 1 and one dashed line perpendicular to the y-axis at 2

E) The data points have not been plotted accurately. This is wrong. The data points have been plotted accurately at the intersection of the coordinated dashed lines.

Finally, the only correct option is B)

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Answer:

Angle: 48.19^o

Explanation:

<u>Two-Dimension Motion</u>

When the object is moving in one plane, the velocity, acceleration, and displacement are vectors. Apart from the magnitudes, we also need to find the direction, often expressed as an angle respect to some reference.

Our boy can swim at 3 m/s from west to east in still water and the river he's attempting to cross interacts with him at 2 m/s southwards. The boy will move east and south and will reach the other shore at a certain distance to the south from where he started. It happens because there is a vertical component of his velocity that is not compensated.

To compensate for the vertical component of the boy's speed, he only has to swim at a certain angle east of the north (respect to the shoreline). The goal is to make the boy's y component of his velocity equal to the velocity of the river. The vertical component of the boy's velocity is

v_b\ cos\alpha

where v_b is the speed of the boy in still water and \alpha is the angle respect to the shoreline. If the river flows at speed v_s, we now set

v_b\ cos\alpha=v_s

\displaystyle cos\alpha=\frac{v_s}{v_b}=\frac{2}{3}

\alpha=48.19^o

8 0
3 years ago
Explain what happens to the energy in a group in a system if one object loses energy according to the Law of Conservation of Ene
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Explanation:

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880nm = 6.6E-34/√ 2.9.1E-31 x me

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I Think its ABC

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