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3 years ago

Which solute will dissolve first in the illustration?

Physics

1 answer:

pentagon [3]3 years ago

4 0

B explanation : they are both filled to the same pint

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According to Coulomb's law, if the separation between two particles of the same charge increases four times, the potential energ

Lera25 [3.4K]

Answer:

c. V = k Q1 * Q2 / R1 potential energy of Q1 and Q2 separated by R

V2 / V1 = (R1 / R2) = 1/4

V2 = V1 / 4

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3 years ago

The longest wavelength within the visible spectrum is _______ light.

Goryan [66]

<span>The longest wavelength within the visible spectrum is the red light. The answer is letter C. It is called visible light because it is the only light that can be seen by the human eye. Red light is the longest wavelength around 620 to 750 nanometer. It is followed by orange which has a wavelength of 590 t 620 nanometer. And then blue which has a wavelength of 450 to 495 nanometer. And the shortest wavelength is violet which has a wavelength of 380 to 459 nanometer. </span>

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3 years ago

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Your favorite taco stand is 800 m away, and closes in 15 minutes. How fast do you have to run to make it on

Dimas [21]

You must run at least 53.3333333 meters a minute.

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3 years ago

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Two identical loudspeakers 2.30 m apart are emitting sound waves into a room where the speed of sound is 340 m/s. Abby is standi

Dimas [21]

Answer:

Abby is standing (4.5^2 + 2.3^2)^1/2 from the far speaker

D2 = 5.05 m from the far speaker

The difference in distances from the speakers is

5.05 - 4.5 = .55 m (Let y be wavelength, lambda)

n y = 4.5

(n + 1) y = 5.05 for the speakers to be in phase at smallest wavelength

y = .55 m subtracting equations

f = v / y = 340 / .55 = 618 / sec should be the smallest frequency

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3 years ago

You can use any coordinate system you like in order to solve a projectile motion problem. To demonstrate the truth of this state

posledela

Answer:

a) y₂ = 49.1 m , t = 1.02 s , b) y = 49.1 m , t= 1.02 s

Explanation:

a) We will solve this problem with the missile launch kinematic equations, to find the maximum height, at this point the vertical speed is zero

$v_{y}$ ² = $v_{oy}$ ² - 2 g (y –yo)

The origin of the coordinate system is on the floor and the ball is thrown from a height

y-yo = $v_{oy}² /2 g
 y- 0 = 10.0²/2 9.8
 y - 0 = 5.10 m
 
The height from the ground is the height that rises from the reference system plus the depth of the ground from the reference system
 y₂ = 5.1 + 44
 y₂ = 49.1 m
Let's use the other equation to find the time
 [tex]v_{y}$ = $v_{oy}$ - g t

t = $v_{oy}$ / g

t = 10 / 9.8

t = 1.02 s

b) the maximum height

y- 44.0 = $v_{y}$ ² / 2 g

y - 44.0 = 5.1

y = 5.1 +44.0

y = 49.1 m

The time is the same because it does not depend on the initial height

t = 1.02 s

7 0

3 years ago