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2 years ago

6

Addition of _____________ to pure water causes the least increase in conductivity. A) weak bases B) acetic acid C) ionic compoun

ds D) organic molecules

2 answers:

geniusboy [140]2 years ago

8 0

Answer:

D

Explanation:

Water in itself is a bad conductor of electricity. Any compound that dissociates in water into ions and charged molecules, will increase conductivity of water. In this case ionic compounds weak basis and acids will put in free ions into the water. The ions can pass electricity because they are attracted to respective poles of electricity depending on their charges. In water, these ions are free moving unlike when they are immobilized in their lattice in solid form.

Organic compounds are mainly made of covalent bonds (around carbon) hence do not dissociate in water.

Mariana [72]2 years ago

3 0

Answer:

the answer is c

Explanation:

i just took it on usatp and its c

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Parallel rays of monochromatic light with wavelength 571nm illuminate two identical slits and produce an interference pattern on

Natasha2012 [34]

Answer:

$8.8\times 10^{-6} W/m^2$

Explanation:

We are given that

Wavelength, $\lambda=571 nm=571\times 10^{-9} m$

$1 nm=10^{-9} m$

R=75 cm= $\frac{75}{100}=0.75 m$

1 m=100 cm

$d=0.640 mm=0.64\times 10^{-3} m$

$1 mm=10^{-3} m$

$a=0.434 mm=0.434\times 10^{-3} m$

$y=0.830 mm=0.830\times 10^{-3} m$

$I_0=5.00\times 10^{-4}W/m^2$

$tan\theta=\frac{y}{R}$

$\theta=\frac{0.830\times 10^{-3}}{0.75\times 10^{-2}}$

$\theta=1.1\times 10^{-3}rad$

$tan\theta\approx \theta$

Because $\theta$ is small.

$\phi=\frac{2\pi dsin\theta}{\lambda}$

$\sin\theta\approx \theta$ ,

Therefore

$\phi=\frac{2\times\pi\times 0.64\times 10^{-3}\times 1.1\times 10^{-3}}{571\times 10^{-9}}$

$\phi=7.74 rad$

$\beta=\frac{2\pi a\theta}{\lambda}$

$\beta=5.3 rad$

$I=I_0cos^2(\frac{\phi}{2})(\frac{sin\frac{\beta}{2}}{\frac{\beta}{2}})^2$

$I=5\times 10^{-4}cos^2(\frac{7.74}{2})(\frac{sin\frac{5.3}{2}}{\frac{5.3}{2}})^2$

$I=8.8\times 10^{-6} W/m^2$

6 0

2 years ago

How do the tension of the cord and the force of gravity affect a pendulum?

LuckyWell [14K]

Answer:

<em>Force of gravity may not affect a pendulum during its equilibrium state</em>. But the gravity can affect the pendulum when a force occurs in any direction of the bob connected to the cord that makes a swing sideways. The gravity of pendulum never stops, it always accelerates. So the gravity affects the pendulum acceleration and speed.

<em>Similarly the tension in the cord will not affect the pendulum</em><em> </em>but if change in the length of the pendulum while keeping other factors constant changes the length of the period of pendulum. longer pendulum swings with lower frequency than shorter pendulums.

6 0

2 years ago

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A hiker walks due East for a distance of 6 km

solniwko [45]

Resultant displacement between the base camp and the cave will be 7.211 km

Displacement is a vector quantity that refers to "how far out of place an object is"; it is the object's overall change in position.

Her resultant displacement can be calculated using Pythagoras theorem

$H^{2} = P^{2} + B^{2}$

= $4^{2} + 6^{2}$

= 52

H = $\sqrt{52}$ = 7.211 km

Resultant displacement between the base camp and the cave will be 7.211 km

To learn more about displacement here :

brainly.com/question/13271165

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3 0

7 months ago

Question 1 (13 marks) A charge Q is located on the top-left corner of a square. Charges of 2Q, 3Q and 4Q are located on the othe

Colt1911 [192]

Answer:

Explanation:

First of all we shall calculate electric field near charge 2Q .

electric field due to charge Q = K x Q / (5 x 10⁻² )²

E₁ = KQ / 25 x 10⁻⁴ = KQ x 10⁴ / 25 . It is acting along positive x axis

E₁ = KQ x 10⁴ i / 25

Similarly electric field due to charge 3Q near 2Q

= 3KQ x 10⁴ i / 25 . It is acting along y-axis

E₂ = 3KQ x 10⁴ j / 25

Similarly electric field due to charge 4Q near 2Q

= 4KQ x 10⁴ j / (25 x 2 )

= 2 KQ x 10⁴ / 25 . It is acting acting along north east direction

unit vector in north east direction = ( i + j )/ √2

So E₃ can be represented by

E₃ = 2 KQ x 10⁴ ( i + j ) / 25 x √2

Total field = KQ x 10⁴ i / 25 + 3KQ x 10⁴ j / 25 + 2 KQ x 10⁴ ( i + j ) / ( 25 x √2 )

= KQ x 10⁴ [ i + 3 j + √2 i + √2 j ) / 25

= 400 KQ ( 2.414 i + 4.414 j ) N / C

Force on 2Q = Field x charge = 400 KQ ( 2.414 i + 4.414 j ) x 2Q N

= 800 KQ² ( 2.414 i + 4.414 j ) N

= 800 x 9 x 10⁹ x ( 2.5 x 10⁻⁶ )² x 2.414 x ( i + 2 j ) N

= 108.63 ( i + 2 j ) N .

Magnitude of this force

= 108.63 x √5

= 243 N approx .

4 0

2 years ago

State four law of photoelectric effect

Bogdan [553]

Answer:

LAW 1 : For a given metal and frequency, the number of photoelectrons emitted is directly proportional to the intensity of the incident radiation.

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LAW 2: For a given metal, there exists a certain frequency below which the photoelectric emission does not take place. This frequency is called threshold frequency.

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LAW 3: For a frequency greater than the threshold frequency, the kinetic energy of photoelectrons is dependent upon frequency or wavelength but not on the intensity of light.

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LAW 4: Photoelectric emission is an instantaneous process. The time lag between incidence of radiations and emission of electron is 10^-9 seconds.

Explanation:

7 0

1 year ago

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