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Romashka-Z-Leto [24]
2 years ago
15

As you sit in a fishing boat, you noticed that 22 waves pass the boat every 60 seconds . If the distance from one crest to the n

ext is 0.5 m, what is the speed of these waves?
Physics
1 answer:
Zinaida [17]2 years ago
8 0

Answer:

0.1835m/s

Explanation:

The formula for calculating the speed of wave is expressed as;

v = fλ

f is the frequency - The number of oscillations completed in one seconds

If 22 waves pass the boat every 60 seconds,

number of wave that passes in 1 seconds = 22/60 = 0.367 waves

Therefore the frequency f of the wave is 0.367Hertz

λ (wavelength) is the distance between successive crest and trough of a wave

λ = 0.5m

Substitute the given values into the formula

v = fλ

v = 0.367 * 0.5

v = 0.1835

Hence the speed of the waves is 0.1835m/s

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How long would it take 2.0x10^20 electrons to pass through a point in a conductor if the current was 10.0A?
inysia [295]

1 coulomb of electric charge is carried by  6.25 x 10^18 electrons

1 Ampere = 1 coulomb per second
10 A = 10 coulombs per second

(2.0 x 10^20 electrons) x (coul / 6.25 x 10^18 electrons) / (10 coul/sec) =

         (2.0 x 10^20) / (6.25 x 10^18 x 10)    sec  =  <em>3.2 seconds</em>


6 0
3 years ago
Consider an electron with charge −e and mass m orbiting in a circle around a hydrogen nucleus (a single proton) with charge +e.
alexandr1967 [171]

Answer:

v=\sqrt{k\frac{e^2}{m_e r}}, 2.18\cdot 10^6 m/s

Explanation:

The magnitude of the electromagnetic force between the electron and the proton in the nucleus is equal to the centripetal force:

k\frac{(e)(e)}{r^2}=m_e \frac{v^2}{r}

where

k is the Coulomb constant

e is the magnitude of the charge of the electron

e is the magnitude of the charge of the proton in the nucleus

r is the distance between the electron and the nucleus

v is the speed of the electron

m_e is the mass of the electron

Solving for v, we find

v=\sqrt{k\frac{e^2}{m_e r}}

Inside an atom of hydrogen, the distance between the electron and the nucleus is approximately

r=5.3\cdot 10^{-11}m

while the electron mass is

m_e = 9.11\cdot 10^{-31}kg

and the charge is

e=1.6\cdot 10^{-19} C

Substituting into the formula, we find

v=\sqrt{(9\cdot 10^9 m/s) \frac{(1.6\cdot 10^{-19} C)^2}{(9.11\cdot 10^{-31} kg)(5.3\cdot 10^{-11} m)}}=2.18\cdot 10^6 m/s

7 0
3 years ago
An electron enters a region of uniform electric field with an initial velocity of 64 km/s in the same direction as the electric
8090 [49]

Answer:

1.) 11 km/s

2.) 9.03 × 10^-5 metres

Explanation:

Given that an electron enters a region of uniform electric field with an initial velocity of 64 km/s in the same direction as the electric field, which has magnitude E = 48 N/C.

Electron q = 1.6×10^-19 C

Electron mass = 9.11×10^-31 Kg

(a) What is the speed of the electron 1.3 ns after entering this region?

E = F/q

F = Eq

Ma = Eq

M × V/t = Eq

Substitute all the parameters into the formula

9.11×10^-31 × V/1.3×10^-9 = 48 × 1.6×10^-19

V = 7.68×10^-18 /7.0×10^-22

V = 10971.43 m/s

V = 11 Km/s approximately

(b) How far does the electron travel during the 1.3 ns interval?

The initial velocity U = 64 km/s

S = ut + 1/2at^2

S = 64000×1.3×10^-6 + 1/2 × 8.4×10^12 × ( 1.3×10^-9)^2

S =8.32×10^-5 + 7.13×10^-6

S = 9.03 × 10^-5 metres

3 0
2 years ago
I need help
Gala2k [10]

The energy of the wave will decrease.

The energy of a wave is given as

E = h f

where E = energy of waver

h = plank's constant

f = frequency of the wave.

From the formula , we see that the energy of the wave is directly proportional to the frequency of the wave. hence as the frequency of the wave decrease, the energy of the wave will decrease.

8 0
3 years ago
Below is a single strand of DNA, what is the complementary base pair for the strand?
GREYUIT [131]
The complementary base pair is:
TTC, CTG, AGT, CTA.
5 0
2 years ago
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