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bekas [8.4K]
1 year ago
13

How many outer-orbital electrons are found in an atom of:a) Na?b) P?c) Br?d) I?e) Te?f) Sr?

Physics
1 answer:
lys-0071 [83]1 year ago
6 0

Answer:

  1. Na=1
  2. P=5
  3. Br=7
  4. I=7
  5. Te=6
  6. Sr=2

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Define average velocity.Immersive Reader
vagabundo [1.1K]

Answer:

The slope of a graph of position vs time

4 0
3 years ago
An inductor is connected to a 26.5 Hz power supply that produces a 41.2 V rms voltage. What minimum inductance is needed to keep
alexira [117]

Answer:

The minimum inductance needed is 2.78 H

Explanation:

Given;

frequency of the AC, f = 26.5 Hz

the root mean square voltage in the circuit, V_{rms} = 41.2 V

the maximum current in the circuit, I₀ = 126 mA

The root mean square current is given by;

I_{rms} = \frac{I_o}{\sqrt{2} } \\\\I_{rms}  = \frac{126*10^{-3}}{\sqrt{2} }\\\\I_{rms}  =0.0891 \ A

The inductive reactance is given by;

X_l = \frac{V_{rms}}{I_{rms}} \\\\X_l= \frac{41.2}{0.0891}\\\\X_l = 462.4 \ ohms

The minimum inductance needed is given by;

X_l = \omega L\\\\X_l = 2\pi  fL\\\\L = \frac{X_l}{2\pi f}\\\\L = \frac{462.4}{2\pi *26.5}\\\\L = 2.78 \ H

Therefore, the minimum inductance needed is 2.78 H

7 0
3 years ago
The charges are in a uniform electric field whose direction makes an angle 36.8 ∘ with the line connecting the charges. What is
astraxan [27]

Complete question:

Point charges q1=- 4.10nC and q2=+ 4.10nC are separated by a distance of 3.60mm , forming an electric dipole. The charges are in a uniform electric field whose direction makes an angle 36.8 ∘ with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude 7.30×10−9 N⋅m ? Express your answer in newtons per coulomb to three significant figures.

Answer:

The magnitude of this field is 826 N/C

Explanation:

Given;

The torque exerted on the dipole, T = 7.3 x 10⁻⁹ N.m

PEsinθ = T

where;

E is the magnitude of the electric field

P is the dipole moment

First, we determine the magnitude dipole moment;

Magnitude of dipole moment = q*r

P = 4.1 x 10⁻⁹ x 3.6 x 10⁻³ = 1.476 x 10⁻¹¹ C.m

Finally, we determine the magnitude of this field;

E = \frac{T}{P*sin(\theta)}=  \frac{7.3 X 10^{-9}}{1.476X10^{-11}*sin(36.8)}\\\\E = 825.6 N/C

E = 826 N/C (in three significant figures)

Therefore, the magnitude of this field is 826 N/C

6 0
3 years ago
23692U = 14456Ba + 89 36Kr + 3n  is a
alekssr [168]
I think The answer is c
5 0
3 years ago
A sled travels 15 meters down a slope inclined at 30 degrees with a horizontal. What is the horizontal displacement of the sled?
Arte-miy333 [17]

Length of the slope is given as

L = 15 m

also the inclination is given as

\theta = 30 degree

now the horizontal displacement is let say "x"

now from geometry we can say

\frac{x}{L} = cos30

x = L cos30

now substitute all values in it

x = 15 * cos30

x = 12.99 m

<em>so it will displace horizontally by 12.99 m</em>

6 0
3 years ago
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