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3 years ago

A beetle crawls from x=1.09m to x=2.55m in 12.8s. what was the average velocity?

Physics

1 answer:

Ivan3 years ago

6 0

The average velocity of the beetle is equal to the ratio between the displacement and the time:

$v=\frac{S}{t}$

The displacement of the beetle is

$S=2.55 m-1.09 m=1.46 m$

The time is t=12.8 s

Therefore, the average velocity is

$v=\frac{S}{t}=\frac{1.46 m}{12.8 s}=0.11 m/s$

in the positive x-direction.

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A car is moving with the velocity of 90km/h. If the car come to rest after 10 seconds. Calculate the final velocity and distance

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3 years ago

I need help in this pls i really need a answer

IrinaK [193]

Answer:

In terms of distance, average speed is 20 km/h

In terms of displacement, average speed is 0 km/h.

Explanation:

Total distance:

$= (40.0 - 10.0) + (20.0 - 10.0) + (40.0 - 20.0) \\ = 30.0 + 10.0 + 20.0 \\ = 60.0 \: km$

Total time is 3.0 hours

but:

$average \: speed = \frac{total \: distance}{total \: time} \\$

In terms of distance.

substitute:

$average \: speed = \frac{60.0}{3.0} \\ \\ = 20 \: {kmh}^{ - 1}$

$displacement = ( - 30.0) + 10 .0+ 20.0 \\ = 0$

In terms of displacement:

$speed = \frac{0}{3} \\ \\ = 0 \: {kmh}^{ - 1}$

3 0

3 years ago

Kalea throws a baseball directly upward at time t = 0 at an initial speed of 13.7 m/s. How high h does the ball rise above its r

Trava [24]

Answer:

h = 9.57 seconds

Explanation:

It is given that,

Initial speed of Kalea, u = 13.7 m/s

At maximum height, v = 0

Let t is the time taken by the ball to reach its maximum point. It cane be calculated as :

$v=u-gt$

$u=gt$

$t=\dfrac{u}{g}$

$t=\dfrac{13.7}{9.8}$

t = 1.39 s

Let h is the height reached by the ball above its release point. It can be calculated using second equation of motion as :

$h=ut+\dfrac{1}{2}at^2$

Here, a = -g

$h=ut-\dfrac{1}{2}gt^2$

$h=13.7\times 1.39-\dfrac{1}{2}\times 9.8\times (1.39)^2$

h = 9.57 meters

So, the height attained by the ball above its release point is 9.57 meters. Hence, this is the required solution.

7 0

3 years ago

A car traveling at 15 m/s starts to decelerate steadily. It comes to a complete stop in 10 seconds. What is it’s acceleration

tresset_1 [31]

So the initial velocity is 15 m/s, the final velocity is 0 since it's at a complete stop and time is 10 seconds. Therefore:

$Acceleration=\frac{v_{final}-v_{initial}}{t_{final}-t_{initial}} =\frac{0m/s-15m/s}{10s-0s} =-1.5\frac{m}{s^2}$

Therefore, the acceleration is -1.5 m/s^2. The reason it's negative is due to the fact that the vector is going against it's original movement since it's decelerating.

7 0

2 years ago

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