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3 years ago

For an object to appear red, what must happen?

Physics

1 answer:

Lapatulllka [165]3 years ago

3 0

Answer:

<h2><em><u>C. It must reflect red light wavel</u></em></h2>

Explanation:

<em>The color of the object is reflected by the object other colors are absorbed by the object.</em>

Here <em>object appears in red color so red color is reflected by the object and other colors are absorbed.</em>

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How do i do a Wavelength and frequency problem

KengaRu [80]

Answer:

wavelength = v/f or wavelength equals to velocity over frequency

frequency= v/w or velocity over wavelength

frequency= 1/p or one over period or time

7 0

4 years ago

Question 1 of 10

Debora [2.8K]

Answer:

V = λ f (wavelength * frequency)

λ = V / f = 343 m/s / 262 / s = 1.3 m

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2 years ago

A car drives 23m/s east for 35 seconds. What is the displacement?

Lina20 [59]

Answer:

805m

Explanation:

Speed = displacement/time

Speed = 23m/s

Time = 35s

Displacement = speed × time

= 23 × 35

= 805m

5 0

3 years ago

After flying for 15 min in a wind blowing 42 km/h at an angle of 19° south of east, an airplane pilot is over a town that is 48

masha68 [24]

Answer:

The speed of the airplane relative to the air is 209.47km/hr

Explanation:

Whenever we are solving a physics problem, it's really useful to start by drawing a diagram of the problem (See picture attached). It will help us visualize the problem better.

Now, we know that the plane flew for an amount of time of 15 minutes. For our dimensions to be the same, we need to turn those 15min to hours, like this:

$15min*\frac{1hr}{60min}=0.25hr$

Once our time is rewritten as hours, we can now calculate the velocity towards north of the plane.

$V=\frac{distance}{time}$

the plane traveled a distance to the north of 48km so the velocity is:

$V=\frac{48km}{0.25hr}$

V=192km/hr j

Now, we can calculate the x and y-components of the velocity of the wind. The problem states that the wind is blowing at 42km/hr at an angle of 19° south of east, so the x and y-components of the velocity of the wind are:

$V_{x}=42km/hr*cos(-19^{o} )=39.71 i$

and

$V_{y}=42km/hr*sin(-19^{o} )=-13.67 j$

So the velocity of the wind can be expressed as a vector as:

$V_{wind}=(39.71i - 13.67j)km/hr$

Once we know this, we can find the velocity of the plane with respect of the wind on x and on y:

$V_{plane x}=V_{plane/wind x}+V_{wind x}$

$V_{plane/wind x}=V_{plane x}-V_{wind x}$

$V_{plane/wind x}=(0-39.71 i)km/hr$

$V_{plane/wind x}= -39.71 i km/hr$

and

$V_{plane y}=V_{plane/wind y}+V_{wind y}$

$V_{plane/wind y}=V_{plane y}-V_{wind y}$

$V_{plane/wind y}=192km/hr j - (- 13.67j)km/hr$

$V_{plane/wind x}= 205.67 j km/hr$

So the velocity of the plane with respect to the wind can be rewritten as:

$V_{plane/wind x}= (-39.71i + 205.67 j) km/hr$

Since the problem asks us to find the speed of the plane with respect to the wind, this means that we need to find the magnitude of the velocity, since the speed is a scalar defined to be the magnitude of the velocity.

so:

$speed=\sqrt{(-39.71)^{2}+(205.67)^{2} }$