Give an example of when your average speed would be higher than instantaneous speed

1 answer:

3 0

When I drive to the office, I drive through two school zones,
and four intersections that are controlled by traffic lights.

My average speed for the trip is higher than my instantaneous
speed is at any point in the school zones, or at any time when
I'm waiting for a red light to change.

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What will the resistance be for a lamp that draws 4.6 amps of current from a 120-volt outlet? A. 550 ohms B. 115 ohms C. 26 ohms

Paraphin [41]

R = V/I = 120 / 4.6 ≈ 26 ohms.

Option C.

3 0

3 years ago

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The following situation will be used for the next three problems: A rock is projected upward from the surface of the moon, at ti

chubhunter [2.5K]

Explanation:

Given that,

Initial speed of the rock, u = 30 m/s

The acceleration due to gravity at the surface of the moon is 1.62 m/s².

We need to find the time when the rock is ascending at a height of 180 m.

The rock is projected from the surface of the moon. The equation of motion in this case is given by :

$h=ut-\dfrac{1}{2}gt^2\\\\180=30t-\dfrac{1}{2}\times 1.62t^2$

It is a quadratic equation, after solving whose solution is given by:

t = 7.53 s

t = 8 seconds

(e)If it is decending, v = -20 m/s

Now t' is the time of descending. So,

$v=-u+gt\\\\t=\dfrac{v+u}{g}\\\\t=\dfrac{20+30}{1.62}\\\\t=30.86\ s$

Let h' is the height of the rock at this time. So,

$h'=ut-\dfrac{1}{2}gt^2\\\\h'=30\times 30.86-\dfrac{1}{2}\times 1.62\times 30.86^2\\\\h'=154.40\ m$

h' = 155 m

6 0

3 years ago

A soccer ball is kicked from the top of one building with a height of H1 = 30.2 m to another building with a height of H2 = 12.0

viktelen [127]

Hi there!

Initially, we have gravitational potential energy and kinetic energy. If we set the zero-line at H2 (12.0m), then the ball at the second building only has kinetic energy.

We also know there was work done on the ball by air resistance that decreased the ball's total energy.

Let's do a summation using the equations:
$KE = \frac{1}{2}mv^2 \\\\PE = mgh$