9 × 10²¹ electrons flow through a cross section of the wire in one hour.
<h3>What is the relation between current and charge?</h3>
- Mathematically, current = charge / time
- In S.I. unit, Charge is written in Coulomb and time in second.
<h3>What is the amount of charge flown through a wire for one hour if it carries 0.4 A current?</h3>
- Charge= current × time
- Current= 0.4 A, time = 1 hour= 3600 s
- Charge= 0.4× 3600
= 1440 C
<h3>How many numbers of electrons present in 1440C of charge?</h3>
- One electron= 1.6 × 10^(-19) C
- So, 1440 C = 1440/1.6 × 10^(-19)
= 9 × 10²¹ electrons
Thus, we can conclude that the 9 × 10²¹ electrons flow through a cross section of the wire in one hour.
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Answer:
(a) A = 0.0800 m, λ = 20.9 m, f = 11.9 Hz
(b) 250 m/s
(c) 1250 N
(d) Positive x-direction
(e) 6.00 m/s
(f) 0.0365 m
Explanation:
(a) The standard form of the wave is:
y = A cos ((2πf) t ± (2π/λ) x)
where A is the amplitude, f is the frequency, and λ is the wavelength.
If the x term has a positive coefficient, the wave moves to the left.
If the x term has a negative coefficient, the wave moves to the right.
Therefore:
A = 0.0800 m
2π/λ = 0.300 m⁻¹
λ = 20.9 m
2πf = 75.0 rad/s
f = 11.9 Hz
(b) Velocity is wavelength times frequency.
v = λf
v = (20.9 m) (11.9 Hz)
v = 250 m/s
(c) The tension is:
T = v²ρ
where ρ is the mass per unit length.
T = (250 m/s)² (0.0200 kg/m)
T = 1250 N
(d) The x term has a negative coefficient, so the wave moves to the right (positive x-direction).
(e) The maximum transverse speed is Aω.
(0.0800 m) (75.0 rad/s)
6.00 m/s
(f) Plug in the values and find y.
y = (0.0800 m) cos((75.0 rad/s) (2.00 s) − (0.300 m⁻¹) (1.00 m))
y = 0.0365 m
Folk song
(Word cap filler)
A. 10 rations = 1 deca-ration.
b. 2000 mockingbirds = 2 x 10³ = 2 kilo-mockingbirds.
c. 10⁻⁵ phones = 1 micro-phones.
d. 10⁻⁹ goats = 1 nano-goats.
e. 1018 miners = 1.018 x 10³ = 1.018 kilo-miners.