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Mkey [24]
1 year ago
7

7. If a graph is showing a positive slope, and the graph itself is a straight line, what must be the x and y axis if the straigh

t
line is showing a constant acceleration?
Physics
1 answer:
guajiro [1.7K]1 year ago
3 0

   An object has positive acceleration if it is accelerating and travelling in the right direction.

<h3>What does it signify if a position time graph has a straight line with a positive slope?</h3>
  • A straight line with a positive slope will appear on the Position vs. Time graph of an item travelling forward at a constant speed. It shows that the object's velocity is constant and that it is travelling ahead.
  • The y-axis of this sort of graph denotes position in relation to the starting point, while the x-axis denotes time. An object's distance from its beginning position at any given moment since it began moving is displayed on a position-time graph.
  • In a distance-time graph, the object's speed is equal to the slope or gradient of the line. The line becomes steeper as the thing goes more quickly (and the greater the gradient).
  • The starting point of the item is indicated by the y-intercept. An object's y-intercept is zero if it begins at the measuring instrument. Its y-intercept is 3 meters if it begins at a distance of 3 meters. When two items are in the same position, you may tell by where two lines intersect.

To Learn more About Positive acceleration, Refer:

brainly.com/question/460763

#SPJ13

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Answer:

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Explanation:

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allochka39001 [22]

poste en français s’il vous plaît

4 0
3 years ago
A 18.5-cm-diameter loop of wire is initially oriented perpendicular to a 1.9-T magnetic field. The loop is rotated so that its p
Alchen [17]

Answer:

0.3405V

Explanation:

#Given a magnetic field of 1.9T, diameter= 18.5cm(r=9.25cm or 0.0925m), we find the magnetic flux of the loop as:

\phi=B.(\pi r^2)cos 0\textdegree\\=1.9\pi\times 0.0925^2 \times cos 0\textdegree\\=5.107\times10^-^2 \ Tm^2

we can now calculate the induced emf, \frac{\phi}{\bigtriangleup t}:

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8 0
3 years ago
Identical balls are dropped from the same initial height and bounce back to half the initial height. In Case 1, the ball bounces
Kitty [74]

Answer:

<em>1. c. Same in both</em>

<em>2. a. Case 1</em>

<em></em>

Explanation:

1. The balls are identical in all sense, which means that if they are dropped from the same height, they should posses the same kinetic energy just before they collide with either the concrete floor or the stretchy rubber. Also, since they reach the same height when they bounced of the concrete floor or the piece of stretchy rubber, it means that they posses the same amount of kinetic energy at this point. Since their kinetic energy at these two points are the same, and they have the same masses, then this means that their momenta at these two instances will also be equal. Since all these is true, then the change in the momentum of the balls between the instance just before hitting the concrete floor or the stretchy rubber material and the instant the ball just leave the floor or the stretchy material is the same for both.

2. The ball that falls on the concrete will experience the greatest force, since the time of impact is small, when compared to the time spent by the other ball in contact with the stretchy rubber material; which will stretch, thereby extending the time spent in contact between them.

7 0
4 years ago
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