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ollegr [7]
3 years ago
9

You can make a solute dissolve more quickly in a solvent by

Physics
1 answer:
Gekata [30.6K]3 years ago
3 0

Answer:Stirring.

Explanation:Stirring a solute into a solvent speeds up the rate of dissolving because it helps distribute the solute particles throughout the solvent. For example, when you add sugar to iced tea and then stir the tea, the sugar will dissolve faster.

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Most people will experience hyposmia at some point in their life.<br> OA True<br> OB. False
suter [353]

Answer:

False

Explanation:

it is very rare to get hyposmia

5 0
2 years ago
Read 2 more answers
Please help me with question B.
Step2247 [10]
It accelerates in the y component (bc of gravity) AND the x-component (b/c of the friction force).
5 0
3 years ago
Debido al desorden en el laboratorio un científico tiene 2 termómetros diferentes pero no sabe en qué escalas están por lo que d
just olya [345]

Answer:

La escala del termómetro ''A'' es grados Celsius.

La escala del termómetro ''B'' es grados Fahrenheit.

Explanation:

Para hallar en qué escalas están los termómetros partimos de que la mezcla a la cuál se midió su temperatura mantuvo su temperatura constante.

Esto quiere decir que los termómetros están expresando la misma temperatura pero en una escala distinta.

Sabemos que dada una temperatura en grados Celsius ''C'' si la queremos convertir a grados Fahrenheit ''F'' debemos utilizar la siguiente ecuación :

F=(\frac{9}{5})C+32 (I)

Ahora, si reemplazamos y asumimos que la temperatura de 18° es en grados Celsius, entonces si reemplazamos C=18 en la ecuación (I) deberíamos obtener F=64.4 ⇒

F=(\frac{9}{5}).(18)+32=32.4+32=64.4

Efectivamente obtenemos el valor esperado. Finalmente, corroboramos que la temperatura del termómetro ''A'' está medida en grados Celsius y la temperatura del termómetro ''B'' en grados Fahrenheit.

6 0
3 years ago
A 10 kg ball is held above a building with a height of 30 m. What is the
Darina [25.2K]

Answer: 2940 J

Explanation: solution attached:

PE= mgh

Substitute the values:

PE= 10kg x 9.8 m/s² x 30 m

= 2940 J

6 0
3 years ago
a 5.0 kg ball is dropped from a 2.5 m high window. what is the velocity of the ball just before it hits the ground?
julsineya [31]

Answer:

Approximately 7.0\; \rm m \cdot s^{-1}.

Assumption: the ball dropped with no initial velocity, and that the air resistance on this ball is negligible.

Explanation:

Assume the air resistance on the ball is negligible. Because of gravity, the ball should accelerate downwards at a constant g = 9.81 \; \rm m \cdot s^{-2} near the surface of the earth.

For an object that is accelerating constantly,

v^2 - u^2 = 2\, a \, x,

where

  • u is the initial velocity of the object,
  • v is the final velocity of the object.
  • a is its acceleration, and
  • x is its displacement.

In this case, x is the same as the change in the ball's height: x = 2.5\; \rm m. By assumption, this ball was dropped with no initial velocity. As a result, u = 0. Since the ball is accelerating due to gravity, a = 9.81\; \rm m \cdot s^{-2}.

v^2 - u^2 = 2\, g \cdot h.

In this case, v would be the velocity of the ball just before it hits the ground. Solve for

v^2 = 2\, a\, x + u^2.

\begin{aligned}v &= \sqrt{2\, a\, x + u^2} \\ &= \sqrt{2\times 9.81 \times 2.5 + 0} \\ &\approx 7.0\; \rm m\cdot s^{-1}\end{aligned}.

3 0
3 years ago
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