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3 years ago

An example of a qualitative observation would be ___________.

Physics

2 answers:

Maslowich3 years ago

5 0

D all of these because it makes the most sense

bazaltina [42]3 years ago

4 0

All of those! A qualitative observation is any one that doesn't include numbers. <em>Quantitative </em>observations include numbers, but not usually any other kind of information.

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Using complete sentences and your own words describe some of the ways humans use water.

Lera25 [3.4K]

Answer:

Humans use water for many different things. We use water to stay hydrated. Our bodies need water to live. Back in the day people used water for transportation and trading. This was a way to become wealthy and exchange goods and ideas from one place to another. We also use water to clean ourselves off. If we don't we can become sick with illnesses that can harm our bodies.

Explanation:

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2 years ago

If these gas molecules were compressed further, how would the average kinetic energy of the molecules be affected? A. It would i

Advocard [28]

I believe the correct response would be B. It would decrease.

8 0

3 years ago

1.) I’m bored, so I decide to play catch by myself. I throw a 1.5kg ball in the air at 25 m/s. How long do I have to wait to cat

navik [9.2K]

Answer:

technically yes

Explanation:

with a gun depending on how fast it shoots so when you fire at something you shoot in front of it a little bit so you hit it but a plane that fast you shoot like 100 feet infront of it...

4 0

3 years ago

Read 2 more answers

Write an expression for a harmonic wave with an amplitude of 0.19 m, a wavelength of 2.6 m, and a period of 1.2 s. The wave is t

zlopas [31]

Answer:

$y = 0.19 sin(5.23 t - 2.42x + \frac{\pi}{2})$

Explanation:

As we know that the wave equation is given as

$y = A sin(\omega t - k x + \phi_0)$

now we have

$A = 0.19 m$

$\lambda = 2.6 m$

so we have

$k = \frac{2\pi}{\lambda}$

$k = \frac{2\pi}{2.6}$

$k = 2.42 per m$

also we have

$T = 1.2 s$

so we have

$\omega = \frac{2\pi}{T}$

$\omega = \frac{2\pi}{1.2}$

$\omega = 5.23 rad/s$

now we know that at t = 0 and x = 0 wave is at y = 0.19 m

so we have

$\phi_0 = \frac{\pi}{2}$

so we have

$y = 0.19 sin(5.23 t - 2.42x + \frac{\pi}{2})$

6 0

3 years ago

An electron that has a velocity with x component 2.6 × 106 m/s and y component 3.2 × 106 m/s moves through a uniform magnetic fi

elena55 [62]

Answer:

a) $\vec F_{B} = 8.766\times 10^{-14}\,T\,k$ , b) $\vec F_{B} = -8.766\times 10^{-14}\,T\,k$

Explanation:

a) The magnetic force experimented by a particle has the following vectorial form:

$\vec F_{B} = q\cdot \vec v \times \vec B$

The charge of the electron is equal to $-1.602\times 10^{-19}\,C$ . Then, cross product can be solved by using determinants:

$\vec F_{B} = \begin{vmatrix}i&j&k\\-4.165\times 10^{-13}\, C\cdot \frac{m}{s} &-5.126\times 10^{-13}\,C\cdot \frac{m}{s} &0\,C\cdot \frac{m}{s} \\0.041\,T&-0.16\,T&0\,T\end{vmatrix}$

The magnetic force is:

$\vec F_{B} = 8.766\times 10^{-14}\,T\,k$

b) The charge of the proton is equal to $1.602\times 10^{-19}\,C$ . Then, cross product has the following determinant:

$\vec F_{B} = \begin{vmatrix}i&j&k\\4.165\times 10^{-13}\, C\cdot \frac{m}{s} &5.126\times 10^{-13}\,C\cdot \frac{m}{s} &0\,C\cdot \frac{m}{s} \\0.041\,T&-0.16\,T&0\,T\end{vmatrix}$

The magnetic force is:

$\vec F_{B} = -8.766\times 10^{-14}\,T\,k$

8 0

2 years ago