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eimsori [14]
2 years ago
9

How many bonds can antimony form?

Physics
1 answer:
bija089 [108]2 years ago
4 0

Answer:

An antimony can form 3 bonds

Let me know if this helps

Thanks!

:)

Explanation:

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A positive point charge q is placed at the center of an uncharged metal sphere insulated from the ground. The outside of the sph
Olenka [21]

Answer:

B.The charge on A is -q; there is no charge on B.

Explanation:

We are given that

Charge=+q

We have to find the correct statement.

When positive charge is placed at center of uncharged metal sphere

insulated from the ground then negative charge(-q) induced on inner

surface  A  of sphere  and the outer  surface B  is grounded then the surface is neutral .

It means there is no charge on surface B.

Hence, option B is true .

B.The charge on A is -q; there is no charge on B.

4 0
3 years ago
iron β is a solid phase of iron still unknown to science. The only difference between it and ordinary iron is that Iron β forms
saw5 [17]

Answer:

8.60 g/cm³

Explanation:

In the lattice structure of iron, there are two atoms per unit cell. So:

\frac{2}{a^{3} }  = \frac{N_{A} }{V_{molar} } where V_{molar}  = \frac{A}{\rho } an and A is the atomic mass of iron.

Therefore:

\frac{2}{a^{3} } = \frac{N_{A} * p }{A}

This implies that:

A = (\frac{2A}{N_{A} * p)^{\frac{1}{3} }  }

  = \frac{4}{\sqrt{3} }r

Assuming that there is no phase change gives:

\rho = \frac{4A}{N_{A}(2\sqrt{2r})^{3}   }

  = 8.60 g/m³

3 0
3 years ago
PY85
aliya0001 [1]

Answer:

460 g

Explanation:

Heat lost by the warm water = heat gained by the cold water

-mCΔT = mCΔT

-m (4.184 J/g/K) (37°C − 85°C) = (1000 g) (4.184 J/g/K) (37°C − 15°C)

-m (37°C − 85°C) = (1000 g) (37°C − 15°C)

-m (-48°C) = (1000 g) (22°C)

m = 458 g

Rounded to two significant figures, you need a mass of 460 g of water.

3 0
3 years ago
The outer layer of cable on a cable reel is 16.2 cm from the center of the reel. The reel is initially stationary and can rotate
ahrayia [7]

Answer:

B. w=12.68rad/s

C. α=3.52rad/s^2

Explanation:

B)

We can solve this problem by taking into account that (as in the uniformly accelerated motion)

\theta=\omega_{0}t+\frac{1}{2}\alpha t^{2}\\\theta = \frac{s}{r}      ( 1 )

where w0 is the initial angular speed, α is the angular acceleration, s is the arc length and r is the radius.

In this case s=3.7m, r=16.2cm=0.162m, t=3.6s and w0=0. Hence, by using the equations (1) we have

\theta=\frac{3.7m}{0.162m}=22.83rad

22.83rad=\frac{1}{2}\alpha (3.6s)^2\\\\\alpha=2\frac{(22.83rad)}{3.6^2s}=3.52\frac{rad}{s^2}

to calculate the angular speed w we can use\alpha=\frac{\omega _{f}-\omega _{i}}{t _{f}-t _{i}}\\\\\omega_{f}=\alpha t_{f}=(3.52\frac{rad}{s^2})(3.6)=12.68\frac{rad}{s}

Thus, wf=12.68rad/s

C) We can use our result in B)

\alpha=3.52\frac{rad}{s^2}

I hope this is useful for you

regards

3 0
3 years ago
Read 2 more answers
A rod of length 0.82 m, rotating with an angular speed, 4.2 rad/s, about axes that pass perpendicularly through one end, has a m
mariarad [96]

Answer:

Explanation:

KE = ½Iω²

ΚΕ = ½(mL²/3)ω²

ΚΕ = ½(0.63(0.82²)/3)4.2²

ΚΕ = 1.24541928

KE = 1.2 J

5 0
2 years ago
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