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Katyanochek1 [597]
3 years ago
14

10Ω resistor and a 5 Ω are connected in parallel. This pair is then connected in series in with another 4 Ω resistor. What is th

e equivalent resistance of the circuit?
Physics
1 answer:
Dmitrij [34]3 years ago
5 0

Answer:

7.33 Ω

Explanation:

10Ω resistor and a 5 Ω are connected in parallel

so,

R₁=(10X5)/(10+5)

   =(50/15) Ω

   =3.33 Ω

This pair is then connected in series in with another 4 Ω resistor

R₂=R₁+4

    =(3.33+4) Ω

    =7.33 Ω

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We measure the loudness of sound in decibels.<br> a. True<br> b. False
Tatiana [17]

Answer: The correct answer is True.

Explanation:

Loudness of sound is referred to how soft or loud a sound is for the listener.

This term is measured in a unit known as decibels referred to as dB.

This unit is used to measure the relative intensity of sounds on a scale from zero to 100 dB.

More the value of decibels, it will be uncomfortable for a person to hear that sound.

So Yes, the loudness of sound is measured in decibels.

7 0
3 years ago
Two concentric current loops lie in the same plane. The smaller loop has a radius of 3.4 cmcm and a current of 12 AA. The bigger
Free_Kalibri [48]

Answer:

Explanation:

Given that,

Current in loops are

i1 = 12A

i2 = 20A

The loops are 3.4cm apart

The magnetic field at the center is found to be zero, so when want to find the radius of bigger loop

Magnetic Field is given as

B= μoi/2πr

Where,

μo is a constant = 4π×10^-7 Tm/A

r is the distance between the two wires

i is the current in the wires

B is the magnetic field

NOTE

Field due to large loop should be equal to the smaller loop.

B1 = B2

μo•i1 / 2π•r1 = μo•i2 / 2π•r2

Then, μo, 2π cancels out, so we have

i1 / r1 = i2 / r2

Make r2 subject of formula

i1•r2 = i2•r1

r2 = i2•r1 / i2

r2 = 20×3.4/12

r2 = 5.67cm

The radius of the bigger loop is 5.67cm.

4 0
3 years ago
Read this /https://www.carbonbrief.org/polar-bears-and-climate-change-what-does-the-science-say
DerKrebs [107]

Melting ice would damage this polar bears habitat meaning the polar bear may decrease


5 0
3 years ago
A spherical shell contains three charged objects. The first and second objects have a charge of − 14.0 nC and 31.0 nC , respecti
allsm [11]

Answer:

The charge on the third object is − 21.7nC

Explanation:

From Gauss's Law

Φ = Q/ε₀

where;

Φ is the total electric flux through the shell = − 533 N⋅m²/C

Q is the total charge Q in the shell = ?

ε₀ is the permittivity of free space = 8.85 x 10⁻¹²

From this equation; Φ = Q/ε₀

Q = Φ * ε₀ = − 533 * 8.85 x 10⁻¹²

Q =  −4.7 X 10⁻⁹ C = -4.7nC

Q = q₁ + q₂ + q₃

− 4.7nC = − 14.0 nC + 31.0 nC + q₃

− 4.7nC − 17nC = q₃

− 21.7nC = q₃

Therefore, the charge on the third object is − 21.7nC

8 0
3 years ago
half life worksheet answer key 3. What percent of a sample As-81 remains un-decayed after 43.2 seconds
Ivahew [28]

The final mass after decay can be obtained by using under given relation:

half life period of As-81 = 33 seconds

mf = mi x (1/2^n)

= 100 x ( 1/2^(43.2/33))

= 40.4 %


3 0
3 years ago
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