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marshall27 [118]
3 years ago
14

A metal ion (X) with a charge of 4+ is attracted to a nonmetal ion (Z) with a

Physics
1 answer:
Keith_Richards [23]3 years ago
8 0

Answer:

A) X3Z4

Explanation:

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Which question cannot be answered using scientific methods?
serg [7]

Answer:

Questions that cannot be answered through scientific investigation are those that relate to personal preference, moral values, the supernatural, or unmeasurable phenomena.

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Two Force one of 12 Newton and another 24 Newton acts at 90 degree with each other find the resultant of two force and its direc
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Answer:

Fr = 26.83 [N]

Explanation:

To solve this problem we must use the Pythagorean theorem, since the forces are vector quantities, that is, they have magnitude and density. Therefore the Pythagorean theorem is suitable for the solution of this problem.

F_{r}=\sqrt{(12)^{2}+(24)^{2}  } \\F_{r}=26.83[N]

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Which of the following is true about light?
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It's hard to imagine something that's a wave and also a bundle.
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If you design an experiment to detect waves, then it responds to light.

And if you design an experiment to detect 'bundles' or particles, then
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How would the mathematical model change if the direction that the object traveled was reversed
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3 years ago
An object is 16.0cm to the left of a lens. The lens forms an image 36.0cm to the right of the lens.
CaHeK987 [17]

A) 11.1 cm

We can find the focal length of the lens by using the lens equation:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where

f is the focal length

p = 16.0 cm is the distance of the object from the lens

q = 36.0 cm is the distance of the image from the lens (taken with positive sign since it is on the opposide side to the image, so it is a real image)

Solving the equation for f:

\frac{1}{f}=\frac{1}{16.0 cm}+\frac{1}{36.0 cm}=0.09 cm^{-1}\\f=\frac{1}{0.09 cm^{-1}}=11.1 cm

B) Converging

The focal length is:

- Positive for a converging lens

- Negative for a diverging lens

In this case, the focal length is positive, so it is a converging lens.

C) 18.0 mm

The magnification equation states that:

\frac{h_i}{h_o}=-\frac{q}{p}

where

h_i is the heigth of the image

h_o is the height of the object

q=36.0 cm

p=16.0 cm

Solving the formula for h_i, we find

h_i = -h_o \frac{q}{p}=-(8.00 mm)\frac{36.0 cm}{16.0 cm}=-18.0 mm

So the image is 18 mm high.

D) Inverted

From the magnification equation we have that:

- When the sign of h_i is positive, the image is erect

- When the sign of h_i is negative, the image is inverted

In this case, h_i is negative, so the image is inverted.

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