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Harlamova29_29 [7]

3 years ago

13

An object of mass 80 kg is released from rest from a boat into the water and allowed to sink. While gravity is pulling the objec

t down, a buoyancy force of StartFraction 1 Over 40 EndFraction times the weight of the object is pushing the object up (weight = mg). If we assume that water resistance exerts a force on the object that is proportional to the velocity of the object, with proportionality constant 10 N-sec/m, find the equation of motion of the object. After how many seconds will the velocity of the object be 60 m/sec? Assume that the acceleration due to gravity is 9.81 m divided by sec squared.

1 answer:

Karolina [17]3 years ago

8 0

Answer:

$v=+(9.56475-\frac{v}{8})\times t$

$t=57.9352\ s$

Explanation:

Given:

mass of the object, $m=80\ kg$

force of buoyancy acting on the object, $F_B=\frac{m.g}{40} =19.62\ N$

initial velocity of the object, $u=0\ m.s^{-1}$

Given that the water resistance acts on the object is proportional to $10\ N.s.m^{-1}$

So, the net force acting on the object:

$F=m.g-F_B+10\times v$

$F=784.8-19.62+10.v$

$F=765.18-10.v$ .........................................(1)

Now the acceleration of the object can be given as:

$a=\frac{F}{m}$

$a=\frac{765.18-10v}{80}$

$a=9.56475-\frac{v}{8}$

From the standard equation of motion:

$v=u+at$

$v=$ final velocity

$u=$ initial velocity

$t=$ time

$v=+(9.56475-\frac{v}{8})\times t$ .................................(2)

Now the velocity of the object is, $v=60\ m.s^{-1}$

$60=+(9.56475-\frac{60}{8})\times t$

$t=57.9352\ s$

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Read 2 more answers

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The given parameters;

<em>Current flowing in the wire, I = 4.00 mA</em>
<em>Initial diameter of the wire, d₁ = 4 mm = 0.004 m</em>
<em>Final diameter of the wire, d₂ = 1 mm = 0.001 m</em>
<em>Length of wire, L = 2.00 m</em>
<em>Density of electron in the copper, n = 8.5 x 10²⁸ /m³</em>

<em />

The initial area of the copper wire;

$A_1 = \frac{\pi d^2}{4} = \frac{\pi \times (0.004)^2}{4} =1.257\times 10^{-5} \ m^2$

The final area of the copper wire;

$A_2 = \frac{\pi d^2}{4} = \frac{\pi (0.001)^2}{4} = 7.86\times 10^{-7} \ m^2$

The initial drift velocity of the electrons is calculated as;

$v_d_1 = \frac{I}{nqA_1} \\\\v_d_1 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 1.257\times 10^{-5}} \\\\v_d_1 = 2.34 \times 10^{-8} \ m/s$

The final drift velocity of the electrons is calculated as;

$v_d_2 = \frac{I}{nqA_2} \\\\v_d_2 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 7.86\times 10^{-7}} \\\\v_d_2 = 3.74\times 10^{-7} \ m/s$

The change in the mean drift velocity is calculated as;

$\Delta v = v_d_2 -v_d_1\\\\\Delta v = 3.74\times 10^{-7} \ m/s \ -\ 2.34 \times 10^{-8} \ m/s = 3.506\times 10^{-7} \ m/s$

The time of motion of electrons for the initial wire diameter is calculated as;

$t_1 = \frac{L}{v_d_1} \\\\t_1 = \frac{2}{2.34\times 10^{-8}} \\\\t_1 = 8.547\times 10^{7} \ s$

The time of motion of electrons for the final wire diameter is calculated as;

$t_2 = \frac{L}{v_d_1} \\\\t_2= \frac{2}{3.74 \times 10^{-7}} \\\\t_2 = 5.348 \times 10^{6} \ s$

The average acceleration of the electrons is calculated as;

$a = \frac{\Delta v}{\Delta t} \\\\a = \frac{3.506 \times 10^{-7} }{(8.547\times 10^7)- (5.348\times 10^6)} \\\\a = 4.38\times 10^{-15} \ m/s^2$

Thus, the change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

Learn more here: brainly.com/question/22406248

7 0

3 years ago