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AveGali [126]
3 years ago
14

The object represented by this graph is moving

Physics
2 answers:
natali 33 [55]3 years ago
6 0

Answer: the third one, toward the origin at a constant velocity.

Explanation:

Vinil7 [7]3 years ago
4 0

The object represented by this graph is moving toward the origin at constant velocity.

Option 3.

<u>Explanation:</u>

In the figure, x-axis is representing increase in the time and y-axis is presenting increase in the distance from bottom to up. But the line in the graph which is plotted is decreasing from high distance to small distance with increase in time. So this indicates that as the time is increasing, the distance is decreasing.

And the object is moving toward the origin as the distance of the object motion is found to decrease with increase of time as per the graph. But the slope of the graph is found to be almost constant, this indicates that the velocity of the object is constant. Thus, the object represented by this graph is moving toward the origin at constant velocity.

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You have two contentious friends, Chris and Pat, and you’ve quickly discovered that they need you to resolve arguments they have
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Answer:

v_average = (d₂-d₁) / Δt

this average velocity is not necessarily the velocity of the extreme points,

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To resolve the debate, it must be shown that the two have part of the reason, the space or distance between the two points divided by time is the average speed between the points.

             v_average = (d₂-d₁) / Δt

this average velocity is not necessarily the velocity of the extreme points, in the only case that it is so is when there is no acceleration.

Therefore neither of them is right.

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A spaceship hovering over the surface of Venus drops an object from a height of 17 m. How much longer does it take to reach the
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1.96s and 1.86s. The time it takes to a spaceship hovering the surface of Venus to drop an object from a height of 17m is 1.96s, and the time it takes to the same spaceship hovering the surface of the Earth to drop and object from the same height is 1.86s.

In order to solve this problem, we are going to use the motion equation to calculate the time of flight of an object on Venus surface and the Earth. There is an equation of motion  that relates the height as follow:

h=v_{0} t+\frac{gt^{2}}{2}

The initial velocity of the object before the dropping is 0, so we can reduce the equation to:

h=\frac{gt^{2}}{2}

We know the height h of the spaceship hovering, and the gravity of Venus is g=8.87\frac{m}{s^{2}}. Substituting this values in the equation h=\frac{gt^{2}}{2}:

17m=\frac{8.87\frac{m}{s^{2} } t^{2}}{2}

To calculate the time it takes to an object to reach the surface of Venus dropped by a spaceship hovering from a height of 17m, we have to clear t from the equation above, resulting:

t=\sqrt{\frac{2(17m)}{8.87\frac{m}{s^{2} } }} =\sqrt{\frac{34m}{8.87\frac{m}{s^{2} } } }=1.96s

Similarly, to calculate the time it takes to an object to reach the surface of the Earth dropped by a spaceship hovering from a height of 17m, and the gravity of the Earth g=9.81\frac{m}{s^{2}}.

t=\sqrt{\frac{2(17m)}{9.81\frac{m}{s^{2} } }} =\sqrt{\frac{34m}{9.81\frac{m}{s^{2} } } }=1.86s

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