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3 years ago

Number of electrons in Na

Physics

2 answers:

sukhopar [10]3 years ago

7 0

There are 11 electrons

Naddik [55]3 years ago

3 0

Answer:

Explanation:

Because 11 is its atomic number

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If an organism didn't have DNA, it would be missing which sign of life ? a, b , c

sergey [27]

Answer:

Explanation:

4 0

3 years ago

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A dime is placed in front of a concave mirror that has a radius of curvature R = 0.40 m. The image of the dime is inverted and t

andrew11 [14]

Answer:

distance between the dime and the mirror, u = 0.30 m

Given:

Radius of curvature, r = 0.40 m

magnification, m = - 2 (since,inverted image)

Solution:

Focal length is half the radius of curvature, f = $\frac{r}{2}$

f = $\frac{0.40}{2} = 0.20 m$

Now,

m = - $\frac{v}{u}$

- 2 = - $\frac{v}{u}$

$\frac{v}{u}$ = 2 (2)

Now, by lens maker formula:

$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$

$\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$

v = $\frac{uf}{u - f}$ (3)

From eqn (2):

v = 2u

put v = 2u in eqn (3):

2u = $\frac{uf}{u - f}$

2 = $\frac{f}{u - f}$

2(u - 0.20) = 0.20

u = 0.30 m

6 0

3 years ago

As a pole of a 2nd-order discrete-time system moves away from the origin in the z-plane, while its phase remains constant, the d

Rudik [331]

Answer:

False

Explanation:

When the location of the poles changes in the z-plane, the natural or resonant frequency (ω₀) changes which in turn changes the damped frequency (ωd) of the system.

As the poles of a 2nd-order discrete-time system moves away from the origin then natural frequency (ω₀) increases, which in turn increases damped oscillation frequency (ωd) of the system.

ωd = ω₀√(1 - ζ)

Where ζ is called damping ratio.

For small value of ζ

ωd ≈ ω₀

4 0

3 years ago

An object has a height of 0.066 m and is held 0.210 m in front of a converging lens with a focal length of 0.140 m. (Include the

Anvisha [2.4K]

Explanation:

Given that,

Size of object, h = 0.066 m

Object distance from the lens, u = 0.210 m (negative)

Focal length of the converging lens, f = 0.14 m

If v is the image distance from the lens, we can find it using lens formula as follows :

$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{0.14 }+\dfrac{1}{(-0.21)}\\\\v=0.42\ m$

(a) Magnification,

$m=\dfrac{v}{u}\\\\m=\dfrac{0.42}{(-0.21)}\\\\m=-2$

(b) Magnification, $m=\dfrac{h'}{h}$

h' is image height

$-2=\dfrac{h'}{0.066}\\\\h'=-2\times 0.066\\\\h'=-0.132\ m$

Hence, this is the required solution.

4 0

3 years ago

A uniform string of length 0.50 m is fixed at both ends. Find the

kozerog [31]

Answer:

configuration of string:

Node - Antinode - Node or N-A-N

This is 1/2 wavelength since a full wavelength is N-A-N-A-N

f (fundamental) = V / wavelength

F0 = 300 m/s / 1 m = 100 / sec

F1 = 300 m/s / .5 m = 600 / sec

Each increase is a multiple of the fundamental since the wavelength

increases by 1/2 wavelength to keep nodes at both ends of the string

4 0

3 years ago