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3 years ago

The reactivity of an element is based on its __________.

Physics

1 answer:

Goryan [66]3 years ago

7 0

Answer:

The number of electrons in the outermost shell of an atom determines its reactivity. Noble gases have low reactivity because they have full electron shells. Halogens are highly reactive because they readily gain an electron to fill their outermost shell.

Explanation:

i think

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Which medium is the greatest threat to the printed newspaper?

icang [17]

<span>Internet because the news travels faster than in a newspaper.</span>

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3 years ago

How does a mutation in a gene affect a protein?

Olenka [21]

Answer:

a variant can cause a protein to malfunction or to not be produced at all.

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2 years ago

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Soap bubbles can display impressive colors, which are the result of the enhanced reflection of light of particular wavelengths f

Bogdan [553]

Answer:

the minimum wall thickness that will enhance the reflection of light is 146.9 nm

Explanation:

Given the data in the question;

At the first interface, a phase shift occurs as the incident light is in air that has less refractive index compare to the thin film of soap bubble.

At the second interface, no shift occurs,

condition for constructive interference;

t = ( m + 1/2) × λ/2n

where m = 0, 1, 2, 3 . . . . . .

now, the condition for the constructive interference;

t = mλ/2n

where t is the thickness of the soap bubble, λ is the wavelength of light and n is the refractive index of soap bubble.

so the minimum thickness of the film which will enhance reflection of light will be;

t $_{min$ = ( m + 1/2) × λ/2n

we substitute

t $_{min$ = ( 0 + 1/2) × 711 /2(1.21)

t $_{min$ = 0.5 × 711/2.42

t $_{min$ = 0.5 × 293.80165

t $_{min$ = 146.9 nm

Therefore, the minimum wall thickness that will enhance the reflection of light is 146.9 nm

8 0

3 years ago

A 2.1 kg block is dropped from rest from a height of 5.5 m above the top of the spring. When the block is momentarily at rest, t

ella [17]

Answer:

The speed of the block is 8.2 m/s

Explanation:

Given;

mass of block, m = 2.1 kg

height above the top of the spring, h = 5.5 m

First, we determine the spring constant based on the principle of conservation of potential energy

¹/₂Kx² = mg(h +x)

¹/₂K(0.25)² = 2.1 x 9.8(5.5 +0.25)

0.03125K = 118.335

K = 118.335 / 0.03125

K = 3786.72 N/m

Total energy stored in the block at rest is only potential energy given as:

E = U = mgh

U = 2.1 x 9.8 x 5.5 = 113.19 J

Work done in compressing the spring to 15.0 cm:

W = ¹/₂Kx² = ¹/₂ (3786.72)(0.15)² = 42.6 J

This is equal to elastic potential energy stored in the spring,

Then, kinetic energy of the spring is given as:

K.E = E - W

K.E = 113.19 J - 42.6 J

K.E = 70.59 J

To determine the speed of the block due to this energy:

KE = ¹/₂mv²

70.59 = ¹/₂ x 2.1 x v²

70.59 = 1.05v²

v² = 70.59 / 1.05

v² = 67.229

v = √67.229

v = 8.2 m/s

8 0

3 years ago

Read 2 more answers

A uniformly charged ring of radius 10.0 cm has a total charge of 75.0 mC. Find the electric field on the axis of the ring at (a)

wlad13 [49]

Answer:

(a) 6650246.305 N/C

(b) 24150268.34 N/C

(c) 6408227.848 N/C

(d) 665024.6305 N/C

Explanation:

Given:

Radius of the ring (r) = 10.0 cm = 0.10 m [1 cm = 0.01 m]

Total charge of the ring (Q) = 75.0 μC = $75\times 10^{-6}\ \mu C$ [1 μC = 10⁻⁶ C]

Electric field on the axis of the ring of radius 'r' at a distance of 'x' from the center of the ring is given as:

$E_x=\dfrac{kQx}{(x^2+r^2)^\frac{3}{2}}$

Plug in the given values for each point and solve.

(a)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=1.00\ cm=0.01\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.01)}{((0.01)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{6750}{1.015\times 10^{-3}}\\\\E_x=6650246. 305\ N/C$

(b)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=5.00\ cm=0.05\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.05)}{((0.05)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{33750}{1.3975\times 10^{-3}}\\\\E_x=24150268.34\ N/C$

(c)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=30.0\ cm=0.30\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.30)}{((0.30)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{202500}{0.0316}\\\\E_x=6408227.848\ N/C$

(d)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=100\ cm=1\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(1)}{((1)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{675000}{1.015}\\\\E_x=665024.6305\ N/C$

7 0

3 years ago