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3 years ago

SOUND IS A FORM OF WAVE.LIST AT LEAST 3 REASONS TO SUPPORT IDEA THAT SOUND IS A WAVE.

1 answer:

6 0

1 The question asks for a certain quantity of examples in a list (Name 6 factors that contributed to the start of World War I, What 3 subatomic particles constitute an atom? etc).
2 The question is academically precise and, therefore, indecisive in the wording (What are the 2 kinds of loading most professional engineers and academics in the field of engineering today generally consider to be relevant in most cases when considering typical types of structure usually made of common materials using well-understand methods?)
3 The question challenges the answerer to defend a position as opposed to merely rattling off a list based on knowledge alone, thereby invoking higher levels of Bloom's Taxonomy. (What are 4 arguments that could be used to defend arguments made by the physicists of the day that electromagnetic waves must move through an illusive substance called 'the ether?)

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A car traveling on a straight road at 10 meters per second accelerates uniformly to a speed of 21 meters per second in 12 second

labwork [276]

21+10=31 because you can see that 21 and 10 are in metres while 12 is in seconds so 21+10=31 is the answer.

6 0

3 years ago

A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1.4 ft/s,

Art [367]

Answer:

$\dfrac{d\theta}{dt} =-0.233\ rad/s$

Explanation:

given,

length of ladder = 10 ft

let x be the distance of the bottom and y be the distance of the top of ladder.

x² + y² = 100

differentiating with respect to time we get

$2 x\dfrac{dx}{dt}+2y\dfrac{dy}{dt} = 0$ ..............(1)

when x = 8 and y = 6 and when \dfrac{dx}{dt} = 1.4ft/s

from equation (1)

now,

$16\times 1.4 + 12\dfrac{dy}{dt} = 0$

$\dfrac{dy}{dt} = -\dfrac{5.6}{3}$

let the angle between the ladders be θ

$tan\theta = \dfrac{y}{x}$

y = xtan θ

$\dfrac{dy}{dt} =\dfrac{dy}{dt} tan\theta + x sec^2\theta\dfrac{d\theta}{dt}$

$-\dfrac{5.6}{3} =1.4\times \dfrac{6}{8} + 8 (1+\dfrac{9}{16})\dfrac{d\theta}{dt}$

$\dfrac{25}{2} \dfrac{d\theta}{dt} =\dfrac{-17.5}{6}$

$\dfrac{d\theta}{dt} =-0.233\ rad/s$

6 0

3 years ago

Which best explains why magnets can push on or pull other magnets without touching them?

V125BC [204]

Magnetism is a property of materials that respond at an atomic or subatomic level to an applied magnetic field. For example, the most well known form of magnetism is ferromagnetism such that some ferromagnetic materials produce their own persistent magnetic field. However, all materials are influenced to greater or lesser degree by the presence of a magnetic field. Some are attracted to a magnetic field (paramagnetism); others are repulsed by a magnetic field (diamagnetism); others have a much more complex relationship with an applied magnetic field. Substances that are negligibly affected by magnetic fields are known as non-magnetic substances. They include copper, aluminium, gases, and plastic.

The magnetic state (or phase) of a material depends on temperature (and other variables such as pressure and applied magnetic field) so that a material may exhibit more than one form of magnetism depending on its temperature, etc.
Or
If it's a multiple choice question this is the best answer: 
A magnetic field surrounds each magnet, which affects other objects with magnetic fields
hope this helpsss.
and can you help me as well with two questions if you dont mind

5 0

3 years ago

A baseball player leads off the game and hits a long home run. The ball leaves the bat at an angle of 70.0 from the horizontal w

professor190 [17]

Answer: 211.059 m

Explanation:

We have the following data:

$\theta=70\°$ The angle at which the ball leaves the bat

$V_{o}=55 m/s$ The initial velocity of the ball

$g=-9.8 m/s^{2}$ The acceleration due gravity

We need to find how far (horizontally) the ball travels in the air: $x$

Firstly we need to know this velocity has two components:

Horizontally:

$V_{ox}=V_{o}cos \theta$ (1)

$V_{ox}=55 m/s cos(70\°)=18.811 m/s$ (2)

Vertically:

$V_{oy}=V_{o}sin \theta$ (3)

$V_{oy}=55 m/s sin(70\°)=51.683 m/s$ (4)

On the other hand, when we talk about parabolic movement (as in this situation) the ball reaches its maximum height just in the middle of this parabola, when $V=0$ and the time $t$ is half the time it takes the complete parabolic path.

So, if we use the following equation, we will find $t$ :

$V=V_{o}+gt=0$ (5)

Isolating $t$ :

$t=\frac{-V_{o}}{g}$ (6)

$t=\frac{-55 m/s}{-9.8 m/s^{2}}$ (7)

$t=5.61 s$ (8)

Now that we have the time it takes to the ball to travel half of is path, we can find the total time $T$ it takes the complete parabolic path, which is twice $t$ :