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tekilochka [14]
3 years ago
14

The human ear is most sensitive to the sounds in what range?

Physics
1 answer:
jeyben [28]3 years ago
3 0

Answer:

The human ear is most sensitive to sounds in the range of 2000 to 3000 Hz

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A step-down transformer providing electricity for a residential neighbor-hood has exactly 2680 turns in its primary. when the po
Fynjy0 [20]
The turns ratio is equal to the voltage ratio. Let n1 and n2 be the primary and secondary turns. Then
5850V/120V=n1/n2
48.75=2680/n2
n2=2680/48.75
n2=55
6 0
3 years ago
A man has a mass of 110 kg. What is his weight?<br> A. 110N<br> B. 1325 N<br> C. 559 N<br> D. 1078 N
Slav-nsk [51]

Answer:

Option D

Explanation:

<u><em>Given:</em></u>

Mass = m = 110 kg

Acceleration due to gravity = g = 9.8 m/s

<u><em>Required:</em></u>

Weight = W = ?

<u><em>Formula</em></u>

W = mg

<u><em>Solution:</em></u>

W = (110)(9.8)

W = 1078 N

7 0
3 years ago
What is the Binary compound for KP?
pychu [463]

Answer:

A binary compound is a chemical compound that is composed of two elements. Learn more about the definition of a binary compound, and find out examples of binary compounds.

Explanation:

3 0
2 years ago
A 0.5 kg block of aluminum (caluminum=900j/kg⋅∘c) is heated to 200∘c. the block is then quickly placed in an insulated tub of co
Alex_Xolod [135]

To solve this problem, we should recall the law of conservation of energy. That is, the heat lost by the aluminium must be equal to the heat gained by the cold water. This is expressed in change in enthalpies therefore:

- ΔH aluminium = ΔH water

where ΔH = m Cp (T2 – T1)

The negative sign simply means heat is lost. Therefore we calculate for the mass of water (m):

- 0.5 (900) (20 – 200) = m (4186) (20 – 0)

m = 0.9675 kg

 

Using same mass of water and initial temperature, the final temperature T of a 1.0 kg aluminium block is:

- 1 (900) (T – 200) = 0.9675 (4186) (T – 0)

- 900 T + 180,000 = 4050 T

4950 T = 180,000

T = 36.36°C

 

The final temperature of the water and block is 36.36°C

4 0
4 years ago
Read 2 more answers
A two slit pattern is viewed on a screen 1.00m from the slits if the two third-order minima are 22.0 cm apart what is the width
Bingel [31]

Answer:

4.4 cm

Explanation:

Given:

Distance of the screen from the slit, D = 1 m

Distance between two third order interference minimas, x = 22 cm

Let's say, minima occurs at:

x_n = (n + \frac{1}{2}) \frac{wL}{d}

We have:

2x_2 = 2(2 + \frac{1}{2}) * \frac{w*22}{d}

Calculating further for the width of the central bright fringe, we have:

\frac{w}{d} = \frac{22}{5}

= 4.4 cm

Note: w in representswavelength

8 0
3 years ago
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