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Ad libitum [116K]
2 years ago
13

This property of matter describes the relationship of mass to volume?

Physics
1 answer:
DedPeter [7]2 years ago
7 0

Density is a physical property that describes the relationship between mass and volume. Density is the amount of matter in a given space, or volume.

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Use the f=ma formula to determine your weight in newtons AND pounds on the Moon and Pluto you have a mass of 70.3 kilograms
Julli [10]

Answer:

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6 0
3 years ago
Calculate the electric field at the center of a square 46.4 cm on a side, if one corner is occupied by a +42.0 µc charge and the
liraira [26]

centre of square disrance to each corner found by Pythagoras' theorem.

coulombs law used to clculate field of each charge at centre

fields added vectorially for res

8 0
3 years ago
Why are isotopes not shown on the periodic table
Fofino [41]

Hey there!

The periodic table is arranged by number of protons/electrons which is the same for all isotopes of an element. <u>A different isotope will have a different number of neutrons only, which does not qualify it for a separate space on the periodic table.</u>

I hope this helps!

3 0
3 years ago
Please explain what is a Energy is
professor190 [17]

Answer:

<h2><u>The capacity or power to do work/ The ability to do work. </u></h2>

Explanation:

such as the capacity to move an object (of a given mass) by the application of force. Energy can exist in a variety of forms, such as electrical, mechanical, chemical, thermal, or nuclear, and can be transformed from one form to another

I hope this help:)

8 0
2 years ago
A plane wave with a wavelength of 500 nm is incident normally ona single slit with a width of 5.0 × 10–6 m.Consider waves that r
kaheart [24]

To solve this exercise it is necessary to use the concepts related to Difference in Phase.

The Difference in phase is given by

\Phi = \frac{2\pi \delta}{\lambda}

Where

\delta = Horizontal distance between two points

\lambda = Wavelength

From our values we have,

\lambda = 500nm = 5*10^{-6}m

\theta = 1\°

The horizontal distance between this two points would be given for

\delta = dsin\theta

Therefore using the equation we have

\Phi = \frac{2\pi \delta}{\lambda}

\Phi = \frac{2\pi(dsin\theta)}{\lambda}

\Phi = \frac{2\pi(5*!0^{-6}sin(1))}{500*10^{-9}}

\Phi= 1.096 rad \approx = 1.1 rad

Therefore the correct answer is C.

6 0
3 years ago
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