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andrew11 [14]
3 years ago
8

What is the relation between Physical states of solution and solvent

Physics
1 answer:
posledela3 years ago
4 0

Explanation:

The solute does not have to be in the same physical state as the solvent, but the physical state of the solvent usually determines the state of the solution. As long as the solute and solvent combine to give a homogeneous solution, the solute is said to be soluble in the solvent.

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You are trying to hear your friend give directions to new store in town. But from your distance (1 point) of 15 m you only hear
Marat540 [252]

Answer:

option D

Explanation:

given,

Intensity of sound = 20 dB

distance = 15 m

intensity of sound is increased to = 50 dB

distance between the sound level = ?

Using relation

L_2 = L_1 - |20(log \dfrac{r_2}{r_1})|

L₁ = 20 dB        L₂ = 50 dB         r₁ = 15 m      r₂ = ?

log (\dfrac{r_2}{r_1}) = \dfrac{L_1 -L_2}{20}

\dfrac{r_2}{r_1}= 10^{\dfrac{|L_1 -L_2|}{20}}

r_2 =r_1 10^{\dfrac{|L_1 -L_2|}{20}}

r_2 =15 \times 10^{\dfrac{|20-50|}{20}}

r_2 =15 \times 10^{-1.5}

r₂ = 0.47 m

r₂ = 47 cm

hence, the correct answer is option D

7 0
4 years ago
5/6 When switched on, the grinding machine accelerates from rest to its operating speed of 3450 rev/min in 6 seconds. When switc
ludmilkaskok [199]

Answer:

Δθ₁ =  172.5 rev

Δθ₁h =  43.1 rev

Δθ₂ =   920 rev

Δθ₂h = 690 rev

Explanation:

  • Assuming uniform angular acceleration, we can use the following kinematic equation in order to find the total angle rotated during the acceleration process, from rest to its operating speed:

       \Delta \theta = \frac{1}{2} *\alpha *(\Delta t)^{2}  (1)  

  • Now, we need first to find the value of  the angular acceleration, that we can get from the following expression:

       \omega_{f1}  = \omega_{o} + \alpha * \Delta t  (2)

  • Since the machine starts from rest, ω₀ = 0.
  • We know the value of ωf₁ (the operating speed) in rev/min.
  • Due to the time is expressed in seconds, it is suitable to convert rev/min to rev/sec, as follows:

       3450 \frac{rev}{min} * \frac{1 min}{60s} = 57.5 rev/sec (3)

  • Replacing by the givens in (2):

       57.5 rev/sec = 0 + \alpha * 6 s  (4)

  • Solving for α:

       \alpha = \frac{\omega_{f1}}{\Delta t} = \frac{57.5 rev/sec}{6 sec} = 9.6 rev/sec2 (5)

  • Replacing (5) and Δt in (1), we get:

       \Delta \theta_{1} = \frac{1}{2} *\alpha *(\Delta t)^{2} = \frac{1}{2} * 6.9 rev/sec2* 36 sec2 = 172.5 rev  (6)

  • in order to get the number of revolutions during the first half of this period, we need just to replace Δt in (6) by Δt/2, as follows:

       \Delta \theta_{1h} = \frac{1}{2} *\alpha *(\Delta t/2)^{2} = \frac{1}{2} * 6.9 rev/sec2* 9 sec2 = 43.2 rev  (7)

  • In order to get the number of revolutions rotated during the deceleration period, assuming constant deceleration, we can use the following kinematic equation:

       \Delta \theta = \omega_{o} * \Delta t + \frac{1}{2} *\alpha *(\Delta t)^{2}  (8)

  • First of all, we need to find the value of the angular acceleration during the second period.
  • We can use again (2) replacing by the givens:
  • ωf =0 (the machine finally comes to an stop)
  • ω₀ = ωf₁ = 57.5 rev/sec
  • Δt = 32 s

       0 = 57.5 rev/sec + \alpha * 32 s  (9)

  • Solving for α in (9), we get:

       \alpha_{2}  =- \frac{\omega_{f1}}{\Delta t} = \frac{-57.5 rev/sec}{32 sec} = -1.8 rev/sec2 (10)

  • Now, we can replace the values of ω₀, Δt and α₂ in (8), as follows:

        \Delta \theta_{2}  = (57.5 rev/sec*32) s -\frac{1}{2} * 1.8 rev/sec2\alpha *(32s)^{2} = 920 rev (11)

  • In order to get finally the number of revolutions rotated during the first half of the second period, we need just to replace 32 s by 16 s, as follows:
  • \Delta \theta_{2h}  = (57.5 rev/sec*16 s) -\frac{1}{2} * 1.8 rev/sec2\alpha *(16s)^{2} = 690 rev (12)
7 0
3 years ago
A planned high-speed train between Houston and Dallas will travel a distance of 386 kilometers in 5.40 × 10^3 seconds. What is t
Mazyrski [523]

¡Hellow!

For this problem, first, lets convert the seconds in hours:

5,4x10³\rightarrow 5400

h = sec / 3600

h = 5400 s / 3600

h = 1,5

Let's recabe information:

d (Distance) = 386 km

t (Time) = 1,5 h

v (Velocity) = ?

For calculate velocity, let's applicate formula:

                                                    \boxed{\boxed{\textbf{d = v * t} } }

Reeplace according we information:

386 km = v * 1,5 h

v = 386 km / 1,5 h

v = 257,33 km/h

The velocity of the train is of <u>257,33 kilometers for hour.</u>

<u></u>

Extra:

For convert km/h to m/s, we divide the velocity of km/h for 3,6:

m/s = km/h / 3,6

Let's reeplace:

m/s = 257,33 km/h / 3,6

m/s = 71,48

¿Good Luck?

7 0
4 years ago
What is the pressure at a depth of 15 cm brine of density 1.2/cm³? ​
Anna [14]

P = density × gravity acceleration × height

P = 1200 × 9.81 × 15/100

P = 1765.8

6 0
3 years ago
Read 2 more answers
(I will give you 10 pts) What elements are formed after a star runs out of hydrogen?
babunello [35]
<span>The core finally cools into a white dwarf, then a black dwarf. This is what happens when a normal-sized star dies. If a really huge star dies, it has so much mass that after the helium is used up, it still has enough carbon to fuse it into heavy elements like iron. When the core turns to iron, it no longer burns.

please give me </span>Brainliest answer?


4 0
3 years ago
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