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3 years ago

for an object that is speeding up in the positive direction, what does the displacement vs. time graph look like?

Physics

2 answers:

Irina18 [472]3 years ago

8 0

Answer:

displacement vs time graph will look like a parabolic shaped graph

Explanation:

As we know that the displacement of an object is given by kinematics equation when object is moving with uniform acceleration

So we know that

$d = v t + \frac{1}{2}at^2$

now here since velocity is increasing in positive X direction as well as the acceleration is also in same direction then the displacement vs time graph must be a straight line graph

so we will have

slava [35]3 years ago

7 0

Answer:

The speeding up is steady, it is a parabola (a=V*t+(at^2)/2), and give it's an equation in connection to time, at that point, it is conceivable to discover the separation recipe by utilizing more substantial amount mathematics(integrals).

Explanation:

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A 783kg elevator rises straight up 164 meters. What is the potential energy of the elevator?

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Answer:

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•<br> What happens to similar charged<br> particles and differently charged<br> particles

mixas84 [53]

Answer:

Like charges repel

Different charges attract

Explanation:

When particles of similar charges are brought together, they repel each other and increase the distance of separation. Repulsion occurs because both two electrons have negative electrical charge forcing their lines of force to repel. However, when particles of opposite charges are brought nearer to each other, they attract each other and reduce the distance of separation.

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3 years ago

In magazine car tests an important indicator of performance is the zero to 60 mph (0 to 96.6 km/h) acceleration time. A time bel

slavikrds [6]

Answer:

2.73414 seconds

467622.66798 J

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

g = Acceleration due to gravity = 9.81 m/s² = a

$v=60\times \dfrac{1609.34}{3600}=26.822\ m/s$

$mgh=\dfrac{1}{2}mv^2\\\Rightarrow h=\dfrac{v^2}{2g}\\\Rightarrow h=\dfrac{26.822^2}{2\times 9.81}\\\Rightarrow h=36.66766\ m$

$s=ut+\frac{1}{2}at^2\\\Rightarrow 36.66766=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{36.66766\times 2}{9.81}}\\\Rightarrow t=2.73414\ s$

$v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{26.822-0}{9.81}\\\Rightarrow t=2.73414\ s$

The time taken is 2.73414 seconds

The potential energy is given by

$U=mgh\\\Rightarrow U=1300\times 9.81\times 36.66766\\\Rightarrow U=467622.66798\ J$

The change in potential energy is 467622.66798 J

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3 years ago