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svet-max [94.6K]
3 years ago
13

The _________ is the difference between two times

Physics
1 answer:
Sergeu [11.5K]3 years ago
7 0
<h3>Answer:</h3>

The <em><u>time </u></em><em><u>duration </u></em>is the difference between two times.

<h3>Hope it helps..</h3>

ray4918 here to help..

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A 24-gauge copper wire has a diameter of 0.51 mm and is used to connect a speaker to an amplifier. The speaker is located 8.5 m
Sergeu [11.5K]

Given Information:

Diameter of the wire = d = 0.51 mm

Length of the wire = L = 8.5 m

Resistance of the speaker = Rsp = 8 Ω

Required Information:

Part A: Resistance of the connecting wire = R = ?

Part B: Ratio of the resistance of connecting wire and speaker = ?

Answer:

Part A: R = 1.43 Ω

Part B: R is 17.8% of Rsp

Explanation:

R = ρL/A

where ρ is the resistivity of the copper wire 1.72x10⁻⁸ Ω/m (at 20 ˚C)

A is the area of the copper wire A = πr²

r = d/2 = 0.51/2 = 0.255 = 0.0255 cm 0.000255 m

Part A: What is the minimum resistance of the connecting speaker wires at 20 ˚C?

R = ρL/πr

R = 1.72x10⁻⁸(8.5)/π*(0.000255)²

R =  0.715 Ω

Since two connecting wires are needed so total resistance becomes

R = 2*715 = 1.43 Ω

Part B: Compare the resistance of the wire to the resistance of the speaker (RSP = 8 Ω)

Resistance of connecting wires = 1.43 Ω

Resistance of the speaker = 8 Ω

1.43/8 = 0.178 = 17.8 %

Therefore, resistance of the connecting wires is 17.8% of the speaker resistance.

4 0
4 years ago
The atomic mass of gold is 0.197 kg/mole. how many moles are in 0.566 kg of gold.
iogann1982 [59]

Explanation:

0.566kg *(1mol/0.197 kg)= 2.87 mol gold

note how the units cancel out, if the units do not cancel out (kg/kg=1) then u did something wrong

7 0
3 years ago
Read 2 more answers
(a) two protons in a molecule are 4.50 10-10m apart. find the magnitude of the electric force exerted by one proton on the other
Anuta_ua [19.1K]

Answer:

a)  1. 1365 × 10⁻⁹N

b) 9.1862 × 10⁻⁴⁶N

c) 8.61 × 10⁻¹¹C/Kg

Explanation:

Ke = 8.99× 109 N.m2 / C2 ,      G = 6.67 × 10-11 N. m2 / kg2

a)  F = Ke . q₁.q₂/r²  

=  (8.99× 109 N.m2 / C2 ) ×(1.60×10⁻¹⁹C)²/(4.50×10⁻¹⁰C)²

= 1. 1365 × 10⁻⁹N

b)  

F = Gm₁m₂/r²

= (6.67× 10⁻¹¹)×(1.67×10⁻²⁷)²/(4.50×10⁻¹⁰)²

= 9.1862 × 10⁻⁴⁶N

The electric force is larger by 8.0497 ×10³⁷ times

c)

if Keq₁q₂/r² Gm₁m₂/r²,

with q₁=q₂ = q,  and m₁ =m₂ = m

Then q/m =\sqrt{G/M} = \sqrt{(6.67 X 10^{-11} } /8.89 X 10^{9}

= 8.61 × 10⁻¹¹C/Kg

8 0
3 years ago
How much heat energy is produced by 0.5 Wh of electrical energy ?
babymother [125]
1.7 Btu

1 watt = 3.41214 Btu/h

1watt * 1h = 3.41214 Btu/h * h

1 = 3.41214 Btu/ (watt*h)/

0.5 watt * h = 0.5 watt*h * 3.41214 Btu/(watt*h) = 1.706 Btu
6 0
3 years ago
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What is the net force acting on the object above?
Maru [420]
5 N to the right

There’s a force of 6 N pulling to the right (the positive direction) and 1 N pulling left (the negative direction).
Net force= 6 N - 1 N = 5 N to the right
8 0
3 years ago
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