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2 years ago

A position versus time graph is shown:

Physics

1 answer:

Stella [2.4K]2 years ago

8 0

Answer: Answer: The object moves forward at 5 m/s, stops, and then changes velocity.

Explanation:

With the information given in the question we can graph the points (image attached).

As we can observe, in the first segment of the graph the velocity is increasing linearly (at a constant rate) and is 5 m/s, then in the second segment we can see the position of the object remains the same from second 2 to second 4, which means the object is stopped.

Finally, in the third and last segment, we can observe a change in velocity (at a negative constant rate, because is decreasing), which is decreasing until the object stops.

Explanation:

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Beginning at the NW corner of the intersection of Pine & 675, thence north 950 feet, thence west 380 feet, thence south 950

Serjik [45]

Answer:

this description is valid for mediadle displacement, bone is an acceptable description

Explanation:

The description of a person's position must be done with a position vector. These vectors must have magnitude, a given direction and a starting point.

In the description this has a starting point corner NO of pine and 675.

Each displacement occurs with respect to the previous one, indicating the magnitude of the displacement and its direction.

After analyzing this description is valid for mediadle displacement, bone is an acceptable description

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3 years ago

Why are atoms in a covalent bond usually a certain distance away from each other?

ANTONII [103]

I think it is because the electrons repel each other

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3 years ago

What are 2 things that effect work on an object

uysha [10]

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2 years ago

If a car has a kinetic energy of 40000 J and is moving at a velocity of 25 m/s, what is the mass of the car? KE=1/2mv

ohaa [14]

Answer:

3200 Kg

Explanation:

Just apply the formula:

40.000 = 1/2 . m . 25

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2 years ago

What percentage of the takeoff velocity did the plane gain when it reached the midpoint of the runway? a plane accelerates from

ElenaW [278]

When is at the end of the runway the velocity of the plane is given by the equation $vf^{2}=0+2*a*s$ where s=1800 m is the runway length. Thus
$vf^{2}=2*5*1800=18000 (m/s)^{2}$
$vf =134.164 (m/s)$

At half runway the velocity of the plane is
$v^{2}=2*5* \frac{1800}{2}=9000 ( \frac{m}{s} )^{2}
$
$v= \sqrt{9000}=94.87 ( \frac{m}{s})$

Therefore at midpoint of runway the percentage of takeoff velocity is
‰ $P= \frac{v}{vf}= \frac{94.87}{134.164}=0.707$

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2 years ago