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dsp73
3 years ago
9

How many elements did mendeleev place in group 2​

Physics
1 answer:
Pepsi [2]3 years ago
8 0

Answer:5

Explanation:

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How much will be the net displacement if a car travels 6 km east and then travels 14 km west
ExtremeBDS [4]
8km is the displacement
5 0
3 years ago
Question 6
Margaret [11]

Limestone and dolomite are the rocks present in the locations which leads to the formation of caves.

<h2>Formation of caves</h2>

The type of rocks that once existed in these  locations are limestone and dolomite whereas the pH of the nearby groundwater is slightly acidic which is responsible for the formation of caves. Caves are formed by the dissolution of limestone due to acid rain.

<h3>Acid rain</h3>

Rainwater reacts with carbon dioxide from the air and percolates through the soil, which turns into a weak acid. This slowly dissolves out the limestone which become turn to form caves so we can conclude that Limestone and dolomite are the rocks present in the locations which leads to the formation of caves.

Learn more about caves here: brainly.com/question/7965722

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4 0
2 years ago
The tangential acceleration of a cart moving at a constant speed in a horizontal circle is:
qaws [65]

Answer:

a=0                                  ∵ \alpha=0\ rad.s^{-1}

Explanation:

The tangential acceleration of a cart moving at a constant speed in a circle is:

The angular velocity is constant when the circular speed is constant.

We know that the (instantaneous) tangential velocity of such object is given by:

v=r.\omega

Now for angular acceleration we have a constant angular speed:

\alpha=0\ rad.s^{-1}

And angular acceleration is related to tangential acceleration as:

a=r.\alpha

\Rightarrow a=0

7 0
3 years ago
Which time interval has the greatest speed?
shutvik [7]

Answer:

es la 2

Explanation:

epero que te curva

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
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