Physics, 04.10.2019 22:20 natalieoppelt

What is the average acceleration of the object represented in the graph above over the 10 seconds? i know the answer isn’t c

What is the average acceleration of the object represented in the graph above over the 10 seconds?

Answers: 1

Show answers

Answers

Answer from: falabit

The average acceleration of the object represented in the graph over the 10 seconds is 2 m/s²

From the question,

We are to determine the average acceleration of the object represented in the graph

Average acceleration is the rate at which the velocity changes.

Average acceleration can be calculated using the formula

$Average\ acceleration = \frac{Change\ in\ velocity}{Change\ in\ time}$

$Average\ acceleration=\frac{v_{f}-v_{0} }{t_{f}-t_{0} }$

Where $v_{f}$ is the final velocity

$v_{0}$ is the initial velocity

$t_{f}$ is the final time

and $t_{0}$ is the initial time

From the graph

$v_{f} = 20 \ m/s$

$v_{0} = 0 \ m/s$

$t_{f}= 10 \ secs$

$t_{0}= 0 \ secs$

Putting the parameters into the formula, we get

$Average\ acceleration = \frac{20-0}{10-0}$

$Average\ acceleration = \frac{20}{10}$

$Average\ acceleration = 2 \ m/s^{2}$

Hence, the average acceleration of the object represented in the graph over the 10 seconds is 2 m/s²

Learn more here: link

Answer from: jacamron

It's always a good idea to start with the definition of the thing you're trying to find.

This problem is just trying to find out whether you KNOW the definition of acceleration. You may know it, but you haven't used it yet.

Average acceleration = (change in velocity) divided by (total time).

Change in velocity = (end value) minus (start value)

Change in velocity = (20m/s) - 0

Change in velocity = 20 m/s

Time = 10 s

Average acceleration = (20m/s)/(10s)

Average acceleration = 2 m/s^2

Answer from: Quest

most voltage drop calculators want the distance between the source and destination. then internally, they double it so both the source and return wires are calculated. but you need to know which distance the calculator uses so you're not half of what you should be or double what you should be.

most circuit breakers will not take a #6 directly, so the electrician will have to splice on a smaller pigtail. you also have to upsize the grounding wire when you upsized the ungrounded (hot) wire (see note in 250.122). this will preclude using most cable assemblies, as a cable with #6 copper will most likely only have a #10 grounding conductor. you need a #6 grounding conductor if the circuit is 20a. a #8 cable is no better, as it will have a #10 grounding conductor too. but most 20a breakers will take a #8 directly.

in your related post, you didn't say what type of load this is. if it is a light or heater, voltage drop won't be a big issue. if this is a motor, the current needs are huge at startup and the voltage drop can pose a problem. if this is a motor, knowing if the motor is under much load during startup will answer the voltage drop effect (a saw is easier to start than a compressor).

Answer from: Quest

Ineed the problem dude