Physics, 27.06.2019 20:30 310000982

An object with an initial velocity of 3.50 m/s moves east along a straight level path the object then undergoes a constant acceleration of 1.80 m/s east for a period of 5.00 s. how far does the object move while it is accelerating

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Answer from: paaoolav

The distance covered by the object while it is accelerating is 40 meters.

Given the following data:

Initial velocity = 3.50 m/s.Acceleration = 1.80 m/s^2

Time = 5.00 seconds.

To determine the distance covered by the object while it is accelerating, we would apply the second equation of motion:

The second equation of motion.

Mathematically, the second equation of motion is given by this formula:

$S=ut+\frac{1}{2} at^2$

Where:

S is the distance.u is the initial velocity.a is the acceleration.t is the time.

Substituting the given parameters into the formula, we have;

$S=3.5(5)+\frac{1}{2} \times 1.80 \times 5^2\\\\S=17.5+0.9 \times 25\\\\S=17.5+22.5$

S = 40 meters.

Read more on acceleration here: link

Answer from: Starzdaze78

distance traveled by the object is given as

$x = v_i t + \frac{1}{2} at^2$

given that

v_i = 3.50 m/s

a = 1.80 m/s^2

t = 5 s

now plug in all the values above

$x = (3.50)(5) + \frac{1}{2}(1.80)(5)^2$

x = 40 m

so here it will move total distance of 40 m

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

what are the options.

explanation:

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