Which options lists a form of kinetic energy followed by a form of potential

Which of the following results when a crest and trough meet?

kifflom [539]

D. Destructive interference. An easy way to think about it is the waves are opposite each other, so they essentially cancel each other out, or make an effort to.

6 0

3 years ago

Science: Could a dysfunction or disease of the nervous system affect other body systems, such as the muscular or digestive syste

SashulF [63]

Answer:

You may experience the sudden onset of one or more symptoms, such as: Numbness, tingling, weakness, or inability to move a part or all of one side of the body (paralysis). Dimness, blurring, double vision, or loss of vision in one or both eyes. Loss of speech, trouble talking, or trouble understanding speech.

Request: Please mark me as the brainliest.

3 0

2 years ago

a stone with a mass of 2.40 kg is moving with velocity (6.60î − 2.40ĵ) m/s. find the net work (in j) on the stone if its velocit

ch4aika [34]

By the work energy theorem, the total work done on the stone is given by its change in kinetic energy,

$W = \Delta K = \dfrac m2 ({v_2}^2 - {v_1}^2)$

We have

$\vec v_1 = (6.60\,\vec\imath - 2.40\,\vec\jmath)\dfrac{\rm m}{\rm s} \implies {v_1}^2 = \|\vec v_1\|^2 = 49.32 \dfrac{\rm m^2}{\rm s^2}$

$\vec v_2 = (8.00\,\vec\imath + 4.00\,\vec\jmath) \dfrac{\rm m}{\rm s} \implies {v_2}^2 = \|\vec v_2\|^2 = 80.0\dfrac{\mathrm m^2}{\mathrm s^2}$

Then the total work is

$W = \dfrac{2.40\,\rm kg}2 \left(80.0\dfrac{\rm m^2}{\rm s^2} - 49.32\dfrac{\rm m^2}{\rm s^2}\right) \approx \boxed{36.8\,\rm J}$

5 0

2 years ago

If he were able to maintain this constant rate of acceleration, what would have been his speed as he finished the race?

Law Incorporation [45]

a) 4.8 m/s² his average acceleration during the first 2.5 s

b) 15 m distance he cover during the first 2.5 seconds.

c) 31 m/s his speed as he finished the race.

<h3>(a) How to calculate the Average Acceleration ?</h3>

Average acceleration is calculated by using Newtons first law of motion

v = u + at

where

u is initial velocity

v is final velocity

a is constant acceleration

t is time

Given u = 0m/s , v = 12 m/s and t = 2.5 sec

Therefore average acceleration is given by

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

$a=\frac{12}{2.5}$

a = 4.8 m/sec²

b) Here we will use newtons second law of motion

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

On substituting value we get

s = 15m

c) Here we will use newtons third law of motion

v² = u² + 2as

Here u = 0 , a = 4.8 m/s² and s = 100 m

Therefore

v² = 960

v = 31 m/s

Disclaimer: the question was given incomplete in the portal. Here is the complete question.

Question: Usain Bolt's 100m sprint Runner set the world record for the100 meter sprint during the 2009World Championships in Berlin with a time of 9.58 s, reaching a top speed of 12 m/s in about 2.5 s.

a. What was his average acceleration during the first 2.5 s?

b. What distance did he cover during the first 2.5 seconds, assuming his acceleration was constant?

c. If he were able to maintain this constant rate of acceleration, what would have been his speed as he finished the race?

Learn more about Newtons law of motion here:

brainly.com/question/12525794

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5 0

2 years ago

Drag each tool to the correct category. Identify the common use of each tool used by space scientists and earth scientists

inysia [295]

<h3>Answer</h3>

Tools that magnify objects Tools that don't magnify objects

magnifying glass rule

telescope binoculars

microscope

<h3>Explanation</h3>

radio dish: is use to receive radio waves from the air.

binoculars: provide a magnified stereoscopic view of distant objects.

streak plate: is a technique which is used to grow bacteria on a growth media surface so that individual bacterial colonies are isolated and sampled.

wind vane: is used to determine wind direction.

ruler: is used for measuring a line.

microscope: is used to magnify the objects which are not visible by naked eyes.

magnifying glass: is used to magnifying tiny objects which are difficult to see yet they are visible.

telescope: use by space scientist to examine astronomical event that may take place in the sky which is too far.

8 0

2 years ago

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