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garri49 [273]
3 years ago
11

Which term describes an area of low density in a longitudinal wave?

Physics
2 answers:
LiRa [457]3 years ago
6 0

Answer:

Explanation:

Low density is an area of rarefaction.

kicyunya [14]3 years ago
6 0

Answer: rarefaction

Explanation: correct answer on a p e x

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TiliK225 [7]

Answer: Answer is D

I took the test little while back.

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3 years ago
Explain why liquids solidify when they are cooled.
Setler [38]
Because the molecules that move freely begin to compact closer together, with less heat, means less molecular activity. 
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Factor 72x^2 - 8/9<br><br> If you can, please show the steps to solve!
ludmilkaskok [199]
Factor out 8 and then facotr and u get

8/9(9x+1)(9x-1
7 0
2 years ago
Ksp for agbr is 5x10-13. what is the maximum concentration of silver ion that you can have in a 0.1 m solution of nabr?
liberstina [14]

Answer : The maximum concentration of silver ion is 5\times 10^{-12}m

Solution : Given,

K_{sp} for AgBr = 5\times 10^{-13}

Concentration of NaBr solution = 0.1 m

The equilibrium reaction for NaBr solution is,

NaBr(aq)\rightleftharpoons Na^++Br^-

The concentration of NaBr solution is 0.1 m that means,

[Na^+]=[Br^-]=0.1m

The equilibrium reaction for AgBr is,

                          AgBr\rightleftharpoons Ag^++Br^-

At equilibrium                     s       s

The expression for solubility product constant for AgBr is,

K_{sp}=[Ag^+][Br^-]

The concentration of Ag^+ = s

The concentration of Br^- = 0.1 + s

Now put all the given values in K_{sp} expression, we get

5\times 10^{-13}=(s)(0.1+s)

By rearranging the terms, we get the value of 's'

s=5\times 10^{-12}m

Therefore, the maximum concentration of silver ion is 5\times 10^{-12}m.

4 0
3 years ago
Read 2 more answers
Having difficulty finding the PE and KE for these values no mass is given. Does anyone know to go solve these?
Alexandra [31]

11) 1.04\cdot 10^7 J

12) 1.04\cdot 10^7 J

13) 50.0 m/s

14) 41.6 m/s

Explanation:

11)

The potential energy of an object is the energy possessed by the object due to its position relative to the ground. It is given by

PE=mgh

where

m is the mass of the object

g is the acceleration due to gravity

h is the height relative to the ground

Here in this problem, when the train is at the top, we have:

m = 8325 kg (mass of the train + riders)

g=9.8 m/s^2 (acceleration due to gravity)

h = 127 m (height of the train at the top)

Substituting,

PE=(8325)(9.8)(127)=1.04\cdot 10^7 J

12)

According to the law of conservation of energy, the total mechanical energy of the train must be conserved (in absence of friction). So we can write:

KE_t + PE_t = KE_b + PE_b

where

KE_t is the kinetic energy at the top

PE_t is the potential energy at the top

KE_b is the kinetic energy at the bottom

PE_b is the potential energy at the bottom

The kinetic energy is the energy due to motion; since the train is at rest at the top, we have

KE_t=0

Also, at the bottom the height is zero, so the potential energy is zero

PE_b=0

Therefore, we find:

KE_b=PE_t=1.04\cdot 10^7 J

13)

The kinetic energy of an object is the energy of the object due to its motion. Mathematically, it is given by

KE=\frac{1}{2}mv^2

where

m is the mass of the object

v is the speed of the object

From question 12), we know that the kinetic energy of the train at the bottom is

KE=1.04\cdot 10^7 J

We also know that the mass is

m = 8325 kg

Therefore, we can calculate the speed of the train at the bottom:

v=\sqrt{\frac{2KE}{m}}=\sqrt{\frac{2(1.04\cdot 10^7)}{8325}}=50.0 m/s

14)

At the top of the second hill, the total mechanical energy of the train is still conserved.

Therefore, we can write again:

KE_1 + PE_1 = KE_2 + PE_2

where

KE_1 is the kinetic energy at the top of the 1st hill

PE_1 is the potential energy at the top of the 1st hill

KE_2 is the kinetic energy at the top of the 2nd hill

PE_2 is the potential energy at the top of the 2nd hill

From the previous questions, we know that

KE_1=0

and

PE_1=1.04\cdot 10^7 J

The height of the second hill is

h = 39 m

So we can also find the potential energy at the second hill:

PE_2=mgh=(8325)(9.8)(39)=3.2\cdot 10^6 J

So, the kinetic energy at the second hill is

KE_2=PE_1-PE_2=1.04\cdot 10^7 - 3.2\cdot 10^6 =7.2\cdot 10^6 J

And so, the speed is

v=\sqrt{\frac{2KE_2}{m}}=\sqrt{\frac{2(7.2\cdot 10^6)}{8325}}=41.6 m/s

4 0
2 years ago
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