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Reptile [31]
3 years ago
12

As an object moves along the x axis, many measurements are made of its position, enough to generate a smooth, accurate graph of

x versus t. Which of the following quantities for the object cannot be obtained from this graph alone? (Select all that apply.)
the displacement during some time interval
the speed at any instant
the velocity at any instant
the average velocity during some time interval
the acceleration at any instant
Physics
2 answers:
Bas_tet [7]3 years ago
6 0
The best and most correct answers among the choices provided by your question are he second and third choices. 
<span>The velocity at any instant the average velocity during some time interval cannot be obtained from the graph alone.</span>


I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
vladimir1956 [14]3 years ago
3 0

Answer:

the acceleration at any instant

Explanation:

1). the displacement during some time interval

We can determine the displacement of the object from the graph because we know that displacement is defined as the change in position of the object and from graph change in position can be obtained easily

2). the speed at any instant

3). the velocity at any instant

speed at any instant of time and velocity at any instant of time must be same in the magnitude and we can determine it by the slope of the graph

As we know that instantaneous velocity is defined as

v = \frac{dx}{dt}

so slope of the graph at any instant will give an idea about instantaneous speed of velocity

4). the average velocity during some time interval

To find average velocity we need the displacement of object at a given interval of time and ratio of this displacement and time interval is known as average velocity

So we can find it from graph

5). the acceleration at any instant

To find the instantaneous acceleration we need to find

a = \frac{dv}{dt}

this is not possible to find from x-t graph because in order to find this slope we need to find velocity time graph.

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Explanation:

The gut microbiota has recently emerged as an important, and previously unappreciated, player in host physiology (1). In particular, the gut microbiota contributes to a variety of physiological and pathophysiological processes in the host including immune disorders (2–4), atherosclerosis (5), irritable bowel syndrome (6, 7), blood pressure regulation (8), and chronic kidney disease (9, 10). Bacteria residing in the human gut are an important component of human physiology: the total wet weight of gut microbes in the human has been estimated to be 175 g–1.5 kg (11, 12), and the cells of the microbiota outnumber human cells by 10:1 (1). These bacteria interact with the immune system of the host (13), and secrete a variety of metabolites, which enter host circulation and can affect a variety of physiological parameters (8, 14), reviewed in Ref. (15). In fact, metabolites produced by the gut microbiota have been found to play key roles in renal disease (16), blood pressure regulation (8), and immune disorders (2–4). Therefore, just as we consider the genetic background of an animal or an individual to be an important contributing factor to their physiology, so too must we consider the genetic background of the microbiota associated with that animal.

Gut microbiota vary greatly amongst laboratory animals, and these differences result in notable differences in experimental results. Mice of the same strain from different vendors have different microbiota profiles (17), and similarly, the same mice housed at different institutions have different microbiota profiles (18, 19). Conversely, inoculating two different inbred mouse strains with the same gut bacteria leads to differences in host gene expression between the two mouse strains (20). Clearly, there is a complex interplay between the genetics of the microbiota and that of the host organism, which has only recently begun to be appreciated.

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A ball of mass m is thrown into the air in a 45° direction of the horizon, after 3 seconds the ball is seen in a direction 30° f
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Velocity (magnitude) is 98.37 m/s

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We use the vertical component of the initial velocity, which is:

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Using kinematics expression of vertical velocity (in y direction) for an accelerated motion (constant acceleration, which is gravity):

v_{y}=v_{0y}+a*t=\frac{\sqrt{2} }{2}v_0-9.8t

Now we need to find v_y as a function of v_0. We use the horizontal velocity, which is always the same as follow:

v_x=v_0cos(45\º)=\frac{\sqrt{2} }{2}v_0=v_{t=3}*cos(30\º) \\

We know the angle at 3 seconds:

v_y(t=3)=v_{t=3}*sin(30\º)\\v_{t=3}=\frac{v_y}{sin(30\º)}

Substitute  v_{t=3} in  v_x and then solve for  v_y

\frac{\sqrt{2} }{2}v_0=\frac{v_y*cos(30\º) }{sin(30\º)} \\v_y=\frac{\sqrt{6} }{6}v_0

With this expression we go back to the kinematic equation and solve it for initial speed

\frac{\sqrt{6} }{6} v_0 =\frac{\sqrt{2} }{2}v_0-29.4\\v_0(\frac{\sqrt{6}-3\sqrt{2}}{6} )=-29.4\\v_0=98.37 m/s

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