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1 month ago

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If you are anchored in a fixed spot, and a set of six waves pass underneath you during a 60 second time interval, what is the wa

ve frequency (in waves/second) of the waves?

1 answer:

Ostrovityanka [3.2K]1 month ago

5 0

Answer:

Explanation:

Within a duration of 60 seconds, six waves are observed.

With a total of 6 waves,

this equates to 3 wavelengths.

As a result,

the period for each wavelength is calculated as 60 divided by 3.

Thus, period = 20 seconds.

According to the frequency-period relationship,

f = 1 / T

f = 1 / 20

f = 0.05 Hz

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Answer:

The force is $F = 8*10^{-12} \ N$

Explanation:

According to the inquiry, we understand that

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The kinesin's velocity is $v = 800 nm/s = 800*10^{-9} m/s$

The power generated by the ATP in one second can be expressed mathematically as

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After substituting the values

$P = 80 * 0.8*10^{-19 }$

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$F = \frac{6.4*0^{-18}}{800 *10^{-9}}$

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a) 19440 km/h²; b) 10 seconds.

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Three point charges are arranged on a line. Charge q3=+5.00x10^-9C and is at the origin. Charge q2=-3.00x10^-9C and is at x=+4.0

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Answer:

The result is 0.750 NC

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20 days ago

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A rope connects boat A to boat B. Boat A starts from rest and accelerates to a speed of 9.5 m/s in a time t = 47 s. The mass of

ValentinkaMS [3465]

Answer: 339.148N

Explanation:

Given data:

Time (t) = 47s

Initial speed (U) = 0m/s

Final speed (V) = 9.5m/s

Mass of B = 540kg

Frictional force on B = 230N

Since both boats are linked, movement of A causes B to move as well.

What is the acceleration of boat A?

Applying the motion formula:

V = u + at

9.5 = 0 + a * 47

a = 9.5 / 47

a = 0.2021 m/s²

To determine the force necessary to accelerate boat B, as both boats experience the same force:

F = Mass * acceleration

F = 540 * 0.2021 = 109.14N

Given that there is a frictional force of 230N acting on boat B, the overall force (Tension) becomes:

Tension = frictional force + applied force = (109.14 + 230)N = 339.148N

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1 month ago

Wire A has the same length and twice the radius of wire B. Both wires are made of the same material and carry the same current.

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Answer:

$V_A= \frac{I_A}{n_A e A_A}= \frac{I}{ne 4A_B}= \frac{1}{4} \frac{I}{neA_B}$

$V_B= \frac{I_B}{n_B e A_B}= \frac{I}{ne A_B}$

Consequently, we find that:

$V_A = \frac{1}{4} V_B$

Thus, the most suitable answer would be:

a. vA = vB/4

Explanation:

In this situation, we can establish the following conditions:

$L_A = L_B =L$ both wires share the same length

both wires carry an identical current $I_A = I_B =I$

Both wires are constructed of the same material, indicating that the electron density (n) remains constant across both wires

$n_A = n_B =n$

We also know that $r_A = 2 r_B$ where r signifies the radius.

Given that wires are cylindrical in shape, we can determine the area for each case:

$A_A= \pi r^2_A = \pi (2r_B)^2 = 4 \pi r^2_B= 4 A_B$

$A_B = \pi r^2_B$

Thus, we conclude that

$A_A = 4 A_B$

Now we are aware that the drift velocity of an electron in a wire can be described by:

$v_d = \frac{I}{neA}$

Where I denotes the current, n is the electron density, e represents the electron charge, and A signifies the area.

By substituting, we arrive at:

$V_A= \frac{I_A}{n_A e A_A}= \frac{I}{ne 4A_B}= \frac{1}{4} \frac{I}{neA_B}$

$V_B= \frac{I_B}{n_B e A_B}= \frac{I}{ne A_B}$

So we observe that:

$V_A = \frac{1}{4} V_B$

Thus, the most fitting answer is:

a. vA = vB/4

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24 days ago