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sweet [91]
4 years ago
8

if a big stone and a small stone is dropped simultaneously from the certain height. they fall together why??​

Physics
2 answers:
Fiesta28 [93]4 years ago
5 0

Answer:

Technically the big one will go down last because it's bigger but the small one will go down faster cause it's small but if not you can try it out

Alenkasestr [34]4 years ago
4 0

Answer:

they are both stones that is why

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Plz! How to solve this question? The answer is B.
LekaFEV [45]

dhvjmdsgyddsfjdbfydgbgdfvdfshfgsdbfhdf nfdfh htiu4ewr huirhfhwref

4 0
3 years ago
The block shown above has a mass of 105 g.<br> What is the density of the block? SC.8.P.8.3
zaharov [31]

The density of the block is 1.25 cm³

The correct answer to the question is Option B. 1.25 cm³

To solve this question, we'll begin by calculating the volume of the block. This can be obtained as follow:

Length = 7 cm

Height = 4 cm

Width = 3 cm

<h3>Volume =? </h3>

Volume = Length × Width × Height

Volume = 7 × 3 × 4

<h3>Volume = 84 cm³</h3>

Thus, the volume of the block is 84 cm³

Finally, we shall determine the density of the block. This can be obtained as follow:

Density is defined as mass per unit volume i.e

Density = \frac{mass}{volume} \\\\

Mass of block = 105 g

Volume of block = 84 cm³

<h3>Density of block =? </h3>

Density = \frac{mass}{volume}\\\\Density = \frac{105}{84}\\\\

<h3>Density of block = 1.25 cm³</h3>

Therefore, the density of the block is 1.25 cm³.

Hence, Option B. 1.25 cm³ gives the correct answer to the question.

Learn more: brainly.com/question/2040396?referrer=searchResults

4 0
2 years ago
element x has two isotopes: x-27 and x-29. x-27 has an atomic mass of 26.975 and a relative abundance of 82.33%, and x-29 has an
meriva

Answer:

27.336

Explanation:

The atomic mass is the weighted average of the atomic masses of the isotopes.  Simply multiply each isotopes's atomic mass by its relative abundance, then sum the results.

m = 0.8233 × 26.975 + 0.1767 × 29.018

m = 27.336

4 0
3 years ago
*please refer to photo* An electric field of magnitude 5.25 ✕ 10^5N/C points due south at a certain location. Find the magnitude
kvv77 [185]

Answer:

Approximately 3.86\; {\rm N} (given that the magnitude of this charge is -7.35\; {\rm \mu C}.)

Explanation:

If a charge of magnitude q is placed in an electric field of magnitude E, the magnitude of the electrostatic force on that charge would be F = E\, q.

The magnitude of this charge is q = 7.35\; {\rm \mu C}. Apply the unit conversion 1\; {\rm \mu C} = 10^{-6}\; {\rm C}:

\begin{aligned} q &= 7.35\; {\mu C} \times \frac{10^{-6}\; {\rm C}}{1\; {\mu C}} = 7.35\times 10^{-6}\; {\rm C}\end{aligned}.

An electric field of magnitude E = 5.25\times 10^{5}\; {\rm N \cdot C^{-1}} would exert on this charge a force with a magnitude of:

\begin{aligned}F &= E\, q \\ &= 5.25 \times 10^{5}\; {\rm N \cdot C^{-1}} \times (-7.35\times 10^{-6}\; {\rm C}) \\ &\approx 3.86\; {\rm N}\end{aligned}.

Note that the electric charge in this question is negative. Hence, electrostatic force on this charge would be opposite in direction to the the electric field. Since the electric field points due south, the electrostatic force on this charge would point due north.

4 0
2 years ago
The vectors of the magnetic field around a long, straight, current-carrying wire are:
pochemuha

Answer:

concentric with the wire

Explanation:

the magnetic field around a long straight, current carrying wire form concentric circles around the wire.

7 0
3 years ago
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