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o-na [289]
3 years ago
13

When a 109 dB sound wave comes through an open window of area 0.485 m2, how much acoustic energy passes through the window in a

10.9 min interval?
Physics
1 answer:
Ksju [112]3 years ago
7 0

Answer:

57dB

Explanation:

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How do I draw a four-cell battery with the cells connected in series with each other?
Goshia [24]

Answer:

Explanation:

Arrangements of cells, when two or more cells are connected together they form what is called a battery

There are two ways cells can be connected

1. Parallel

2. Series

Find attached a sketch of serial arrangement

7 0
2 years ago
the maximum displacement of an oscillatory motion is A=0.49m. determine the position x at which the kinetic energy of the partic
kipiarov [429]

Answer:0.4 m

Explanation:

Given

Maximum displacement A=0.49

The sum of kinetic and elastic potential energy is \frac{1}{2}kA^2

where k=spring constant

U+K.E.=\frac{1}{2}kA^2

when K.E.=U/2

K.E.=kinetic energy

U=Elastic potential Energy

\rightarrow \ U+\frac{U}{2}=\frac{1}{2}KA^2\\\rightarrow \ \frac{3U}{2}=\frac{1}{2}KA^2\\\rightarrow \ U=\frac{1}{3}KA^2\\\rightarrow \ \frac{Kx^2}{2}=\frac{1}{3}KA^2\\\\x=\sqrt{\frac{2}{3}}A\\x=0.4\ m

3 0
3 years ago
How much work is done when a 5 N force moves a block 4 m?
katen-ka-za [31]
20N•m or 20J. Work is equal to force•distance, and 5N•4m is 20N•m, or J
4 0
3 years ago
Which one of the following statements is true concerning the magnitude of the electric field at a point in space? It is a measur
IceJOKER [234]

Answer:

It is a measure of the electric force per unit charge on a test charge.

Explanation:

The magnitude of the electric field is defined as the force per charge on the test charge.

Since we define electric field as the force per charge, it will have the units of  force divided by the unit of charge. This implies that the SI unit of electric field is given as Newton/Coulomb (N/C).

5 0
2 years ago
uniform electric field exists between two parallel plates separated by 2.0 cm. The intensity of the field is 15 kN/C. What is th
adoni [48]

Answer:

potential difference V= 300 volts

Explanation:

Given:

d= 2.0 cm = 0.02m

E = 15 kN/C = 15 × 10³ N/C

For a uniform field between two plates, the Electric Filed Intensity (E) is proportional to the potential difference (V) and inversely  proportional to distance between the plates.

E= V/d

⇒ V= E×d = 15 × 10³ N/C × 0.02 m = 300 volts  (∴1 Nm/C = 1 J/C= 1 volts)

7 0
3 years ago
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