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3 years ago

Why aren’t descriptive investigations repeatable

1 answer:

8 0

It involves questions but it doesn't involve hypothesis. So we are still trying to find info by asking the question, and it's also a waste of time asking the question over and over, because the hypothesis wasn't formulated yet so that's why they are not repeatable.

Hope this helps!

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The velocity of a wave with a wavelength of 4.700 m and frequency of 54.00 Hz is

Feliz [49]

The answer is 253.8 m/s

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3 years ago

What kind of symmetry do you have

Keith_Richards [23]

During the daytime, I have mostly line symmetry.

During the night, I often have almost spherical symmetry.

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3 years ago

Read 2 more answers

Calculate the electric field at the center of a square

pantera1 [17]

Answer:

$E_y=1175510.2\ N.C^{-1}$

The Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

Explanation:

Given:

side of a square, $a=52.5\ cm$

charge on one corner of the square, $q_1=+45\times 10^{-6}\ C$

charge on the remaining 3 corners of the square, $q_2=q_3=q_4=-27\times 10^{-6}\ C$

Distance of the center from each corners $=\frac{1}{2} \times diagonals$

$diagonal=\sqrt{52.5^2+52.5^2}$

$diagonal=74.25\cm=0.7425\ m$

∴Distance of center from corners, $b=0.3712\ m$

Now, electric field due to charges is given as:

$E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}$

For charge $q_1$ we have the field lines emerging out of the charge since it is positively charged:

$E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}$

$E_1=2938775.5\ N.C^{-1}$

Force by each of the charges at the remaining corners:

$E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}$

$E_2=E_3=E_4=1763265.3\ N.C^{-1}$

Now, net electric field in the vertical direction:

$E_y=E_1-E_4$

$E_y=1175510.2\ N.C^{-1}$

Now, net electric field in the horizontal direction:

$E_y=E_2-E_3$

$E_y=0\ N.C^{-1}$

So the Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

8 0

3 years ago

Because of the curvature of the earth, the maximum distance D that you can see from the top of a tall building of height h is es

mojhsa [17]

Answer:

D = 9.9 10⁶ mi

Explanation:

In the exercise they give the expression for maximum viewing distance

D = 2 r h + h²

Ask for this distance for a height of 1100 feet

Let's calculate

D = 2 3960 1100 + 1100²

D = 8.712 10⁶ + 1.21 10⁶

D = 9.92 10⁶ mi

D = 9.9 10⁶ mi

8 0

3 years ago

Red shift data shows that galaxies are O expanding O shrinking O moving away O moving closer

Anon25 [30]

Answer:

Should be moving away

Explanation:

Red is a longer wavelength therefore further away. Wavelength is stretched out more and on the red end. I hope this is right. I decided to research and answer since you didn’t have other answers. Are you taking this on edg? I hope I helped!

7 0

3 years ago