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mote1985 [20]
3 years ago
9

How do salt and water form a solution?

Physics
2 answers:
iogann1982 [59]3 years ago
8 0
It would be D. The solute, salt, must dissolve IN the solvent, water (:
Klio2033 [76]3 years ago
5 0

<u>Answer:</u> The solute, salt must dissolve in the solvent, water.

<u>Explanation:</u>

Solute is defined as the substance which is present in smaller proportion in a solution.

Solvent is defined as the substance which is present in larger proportion in a solution.

Solution is the combination of solute and solvent.

In a solution, a solute will always dissolve in solvent.

We are given:

Salt and water to form a solution. Here, salt is present in smaller proportion, so it is a solute and water is a solvent.

Hence, the solute, salt must dissolve in the solvent, water.

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3 years ago
Desde el balcón de un edificio se deja caer una manzana y llega a la planta baja en 5 s. ¿Desde qué piso se dejó caer, si cada p
miv72 [106K]

Answer: 42

Explanation:

I will answer this in English.

We know that the apple needs 5 seconds to reach the ground.

Each floor of the building has a height of 2.88m.

Now, when we drop something, the only force acting on the object is the gravitational one, so the acceleration of the apple is:

a = -g

for the velocity, we integrate the acceleration over time, and as the apple is dropped, we do not have any initial velocity, so we do not have a constant in the integration:

v = -g*t

for the position we integrate again, now we have an initial height H, so the position is:

p = (-g/2)*t^2 + H

now the apple hits the ground when p = 0, so we can solve this equation to find H.

i will use g = 9.8m/s^2

0 = (-4.9m/s^2)*(5s)^2 + H  

H = 122.5 m

now knowing H, we can divide it by the height of a floor in the building and get the number of the floor.

N = 122.5m/2.88m = 42.5

this means that the apple was dropped in the floor 42 (the 0.5 means that the apple was not right where the floor 42 starts, it was dropped around the middle of the floor 42)

5 0
3 years ago
Initial velocity=3m/s, final velocity=5m/s, displacement=2m. Find the acceleration.
shusha [124]

Answer:

Explanation:

To analyze the behavior of black bears during the night. To describe how black bears adapt to life near human. To evaluate the decline in the number of black bear.

4 0
3 years ago
Read 2 more answers
A child's top is held in place upright on a frictionless surface. The axle has a radius of ????=3.21 mm . Two strings are wrappe
tekilochka [14]

Answer:

Angular momentum, L=6.47\times 10^{-3}\ m

Explanation:

It is given that,

Radius of the axle, r=3.21\ mm=3.21\times 10^{-3}\ m

Tension acting on the top, T = 3.15 N

Time taken by the string to unwind, t = 0.32 s

We know that the rate of change of angular momentum is equal to the torque acting on the torque. The relation is given by :

\tau=\dfrac{dL}{dt}

Torque acting on the top is given by :

\tau=F\times r

Here, F is the tension acting on it. Torque acting on the top is given by :

\tau=2F\times r

2T\times r=\dfrac{L}{t}

L=2T\times r \times t

L=2\times 3.15\times 3.21\times 10^{-3}\times 0.32

L=6.47\times 10^{-3}\ m

So, the angular momentum acquired by the top is 6.47\times 10^{-3}\ m. Hence, this is the required solution.

7 0
3 years ago
Honeybees acquire a charge while flying due to friction with the air. A 100 mg bee with a charge of +23 pC experiences an electr
Sedbober [7]

Answer:

(A) ratio of electric force to weight will be  23.469\times 10^{-10}

(b) Electric field will be E=4.26\times 10^{10}N/C

Explanation:

We have given mass of bee = 100 mg  = m=100\times 10^{-3}=0.1kg

Charge on bee q=23pC=23\times 10^{-12}C

Electric field E = 100 N/C

Weight of the bee W=mg=0.1\times 9.8=0.98N

Electric force on the bee F=qE=23\times 10^{-12}\times 100=23\times 10^{-10}N

So the ratio of electric force on the bee and weight is =\frac{F}{W}=\frac{23\times 10^{-10}}{0.98}=23.469\times 10^{-10}

(B) To hold the bee in air electric force must be equal to weight of bee

So mg=qE

0.1\times 9.8=23\times 10^{-12}E

E=4.26\times 10^{10}N/C

4 0
3 years ago
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