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tester [92]
3 years ago
8

Are There some sounds for which we are all deaf

Physics
1 answer:
lana66690 [7]3 years ago
6 0

Answer: Infrasound is the span of low-frequency sounds below 20 Hz that fall below the hearing range of humans. While these sounds escape our ears, scientific instruments can detect them—and tell us some interesting things about the planet.

Explanation:

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An object floats in a beaker as shown. when it was put into the beaker, it displaced an amount of water into the graduated cylin
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Answer:

answer is  C. 10 g

Explanation:

: When an object floats, it displaces an amount of water that has the same mass as itself. Therefore, the mass of the water in the graduated cylinder is equal to the mass of the object. We can see that there are 10 mL of water in the graduated cylinder. We also know that the density of water is 1 g/mL. Since each mL of water has a mass of 1 g, then 10 mL of water has a mass of 10 g. If the mass of the displaced water is 10 g, then the mass of the floating object is also 10 g.

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Over an interval of 4.0 s the average rate of change of the concentration of B was measured to be -0.0760 M/s. What is the final
Serhud [2]

Complete question:

Consider the hypothetical reaction 4A + 2B → C + 3D

Over an interval of 4.0 s the average rate of change of the concentration of B was measured to be -0.0760 M/s. What is the final concentration of A at the end of this same interval if its concentration was initially 1.600 M?

Answer:

the final concentration of A is 0.992 M.

Explanation:

Given;

time of reaction, t = 4.0 s

rate of change of the concentration of B =  -0.0760 M/s

initial concentration of A = 1.600 M

⇒Determine the rate of change of the concentration of A.

From the given reaction: 4A + 2B → C + 3D

2 moles of B ---------------> 4 moles of A

-0.0760 M/s of B -----------> x

x = \frac{4(-0.076)}{2} \\\\x = -0.152 \ M/s

⇒Determine the change in concentration of A after 4s;

ΔA = -0.152 M/s  x 4s

ΔA = -0.608 M

⇒ Determine the final concentration of A  after 4s

A = A₀ + ΔA

A = 1.6 M + (-0.608 M)

A = 1.6 M - 0.608 M

A = 0.992 M

Therefore, the final concentration of A is 0.992 M.

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2 years ago
CHEGG You stretch a spring with spring constant k = 1.2x104 N/m to extend 6.0 cm away from its equilibrium position. How much do
lubasha [3.4K]

Answer:

The elastic potential energy of the spring change during this process is 21.6 J.    

Explanation:

Given that,

Spring constant of the spring, k=1.2\times 10^4\ N/m

It extends 6 cm away from its equilibrium position.

We need to find the elastic potential energy of the spring change during this process. The elastic potential energy of the spring is given by the formula as follows :

E=\dfrac{1}{2}kx^2\\\\E=\dfrac{1}{2}\times 1.2\times 10^4\times (0.06)^2\\\\E=21.6\ J

So, the elastic potential energy of the spring change during this process is 21.6 J.

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