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Ghella
1 month ago
12

A fisherman has caught a very large, 5.0kg fish from a dock that is 2.0m above the water. He is using lightweightfishing line th

at willbreak under a tension of 54N or more. He is eager to the fish to the dock in the shortest time possible. If the fish is at rest at the water's surface, what is the least amount of time in which the fisherman can raise the fish to the dock without losing it?
Physics
1 answer:
Ostrovityanka [3.2K]1 month ago
7 0
The time taken, t = 2 seconds, can be deduced from the scenario involving the fish being lifted.
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Consider an electron with charge −e and mass m orbiting in a circle around a hydrogen nucleus (a single proton) with charge +e.
kicyunya [3294]

Result:

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

Explanation:

The electromagnetic attraction between the electron and the proton in the nucleus is equivalent to the centripetal force:

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

where

k represents the Coulomb constant

e denotes the charge of the electron

e denotes the charge of the proton in the nucleus

r signifies the distance from the electron to the nucleus

v indicates the velocity of the electron

is the mass of the electron

Rearranging for v, we determine

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

Inside a hydrogen atom, the distance separating the electron from the nucleus is roughly

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

while the mass of the electron is

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

and the charge is

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

By plugging in the values into the formula, we achieve

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
1 month ago
If the diameter of a hole is 4.500 mm, what should be the diameter of a rivet at 23.0 ∘C, if its diameter is to equal that of th
kicyunya [3294]

Answer:

Explanation:

at 23 degrees Celsius, the diameter measures 4.511 mm

GIVEN DATA:

diameter of hole  = 4.500 mm

T_1 = 23.0 degrees Celsius

T_2 = - 78.0 degrees Celsius

the expansion coefficient of aluminum is 2.4*10^{-5} (degrees Celsius)^{-1}

the diameter at 23 degrees Celsius is stated as

d = d_o (1+ \alpha \delta T)

 = 4.5 (1+2.4*10^{-5} *(23-(-78)))

   = 4.511 mm

the diameter of the rivet after temperature change is given as

d= d_o + \Delta d

 = d_o(1+ \alpha \delta T)

= 0.4500 *(1+2.4*10^{-5} *(23-(-78)))

= 0.4511 cm

8 0
1 month ago
A suspension system is being designed for a 1300 kg vehicle. When the vehicle is empty, its static deflection is measured as 2.5
Keith_Richards [3271]

Answer: damping coefficient = 1.5×10^5Ns/m

Explanation:

Refer to the attached file for the solution

3 0
1 month ago
Two events are observed in a frame of reference S to occur at the same space point, with the second event occurring after a time
Sav [3153]

Answer:

The distinction between the locations of the two events as perceived in S is calculated to be 4.42 x 10^8 m

Explanation:

The time dilation concept has been utilized to address this question, as detailed in the attachment.

5 0
29 days ago
When a vertical beam of light passes through a transparent medium, the rate at which its intensity I decreases is proportional t
Keith_Richards [3271]

Response:

The intensity of light 18 feet underwater is about 0.02%

Clarification:

Employing Lambert's law

Let dI / dt = kI, where k is a proportionality factor, I represents the intensity of incident light, and t indicates the thickness of the medium

Then dI / I = kdt

Taking logarithms,

ln(I) = kt + ln C

I = Ce^kt

At t=0, I=I(0) implies C=I(0)

I = I(0)e^kt

At t=3 & I=0.25I(0), we find 0.25=e^3k

Solving for k gives k = ln(0.25)/3

k = -1.386/3

k = -0.4621

I = I(0)e^(-0.4621t)

I(18) = I(0)e^(-0.4621*18)

I(18) = 0.00024413I(0)

The intensity of light 18 feet underwater is about 0.2%

3 0
2 months ago
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