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1 year ago

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3. A 64 lb weight stretches a spring 4 ft in equilibrium. The weight is initially displaced 6 inches above equilibrium and given

a downward velocity of 6 ft/sec. Find its displacement for t > 0 if the medium resists the motion with a force equal to 3 times the speed in ft/sec (that is, the damping constant c = 3). Note that there is no external force in this case.

1 answer:

Damm [24]1 year ago

6 0

Answer:

Explanation:

the solution is given in the attached pictures

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An object on a number line moved from x = 15 cm to x = 165 cm and then

olasank [31]

Answer:

v_avg = 2.9 cm/s

Explanation:

The average velocity of the object is the sum of the distance of all its trajectories divided the time:

$v_{avg}=\frac{x_{all}}{t}$

x_all is the total distance traveled by the object. In this case you have that the object traveled in the first trajectory 165cm-15cm = 150cm, and in the second one, 165cm - 25cm = 140cm

Then, x_all = 150cm + 140cm = 290cm

The average velocity is, for t = 100s

$v_{avg}=\frac{290cm}{100s}=2.9\frac{cm}{s}$

hence, the average velocity of the object in the total trajectory traveled is 2.9 cm/s

3 0

1 year ago

Read 2 more answers

Two small aluminum spheres, each of mass 0.0250 kilograms, areseparated by 80.0 centimeters.

sineoko [7]

Answer:

Total number of electrons

$N = 7.25 \times 10^{24}$

electrons removed from each sphere

$N = 5.27 \times 10^{15}$

Fraction of electrons transferred is given as

$f = 7.27 \times 10^{-10}$

Explanation:

As we know that moles is defined as

$n =\frac{mass}{molar mass}$

$n = \frac{0.0250}{0.026982}$

$n = 0.93$

so number of atoms of Al in each sphere is given as

$N = 0.93(6.02 \times 10^{23})$

$N = 5.58 \times 10^{23}$

Now number of electrons in each atom is given as

atomic number = number of electrons in each atom = 13

total number of electrons in each sphere is

$N = 13 \times (5.58 \times 10^{23})$

$N = 7.25 \times 10^{24}$

Also we know that force of attraction between them is given as

$F= \frac{kq_1q_2}{r^2}$

$1.00 \times 10^4 = \frac{(9\times 10^9)q^2}{0.80^2}$

$q = 8.4 \times 10^{-4} C$

now we have

$q = Ne$

$8.4 \times 10^{-4} = N(1.6 \times 10^{-19}$

$N = \frac{(8.4 \times 10^{-4})}{1.6 \times 10^{-19}}$

$N = 5.27 \times 10^{15}$

Fraction of electrons transferred is given as

$f = \frac{5.27 \times 10^{15}}{7.25 \times 10^{24}}$

$f = 7.27 \times 10^{-10}$

6 0

1 year ago

Two long conducting cylindrical shells are coaxial and have radii of 20 mm and 80 mm. The electric potential of the inner conduc

xxMikexx [17]

Answer: 14.52*10^6 m/s

Explanation: In order to explain this problem we have to consider the energy conservation for the electron within the coaxial cylidrical wire.

the change in potential energy for the electron; e*ΔV is equal to energy kinetic gained for the electron so:

e*ΔV=1/2*m*v^2 v^=(2*e*ΔV/m)^1/2= (2*1.6*10^-19*600/9.1*10^-31)^1/2=14.52 *10^6 m/s

3 0

1 year ago

A seasoned mini golfer is trying to make par on a tricky number five hole. The golfer can complete the hole by hitting the ball

liberstina [14]

Answer:5.17 m/s

Explanation:

Given

let u be the speed at cliff initial point

range over cliff is 1.45 m

and range of projectile is given by

$R=\frac{u^2\sin 2\theta }{g}$

$1.45=\frac{u^2\sin 90}{9.8}$

u=3.77 m/s

Conserving Energy

$E_{bottom}=E_{initial\ point\ at\ cliff}$

Kinetic energy=Kinetic energy +Potential energy gained

Let v be the initial velocity

$\frac{mv^2}{2}=mgh+\frac{mu^2}{2}$

$v^2=u^2+2gh$

$v=\sqrt{u^2+2gh}$

$v=\sqrt{3.77^2+2\time 9.8\times 0.64}$

$v=\sqrt{26.75}=5.17 m/s$

5 0

1 year ago

A certain fuse "blows" if the current in it exceeds 1.0 A, at which instant the fuse melts with a current density of What is the

Alborosie

Answer:

<em>0.45 mm</em>

Explanation:

The complete question is

a certain fuse "blows" if the current in it exceeds 1.0 A, at which instant the fuse melts with a current density of 620 A/ cm^2. What is the diameter of the wire in the fuse?

A) 0.45 mm

B) 0.63 mm

C.) 0.68 mm

D) 0.91 mm

Current in the fuse is 1.0 A

Current density of the fuse when it melts is 620 A/cm^2

Area of the wire in the fuse = I/ρ

Where I is the current through the fuse

ρ is the current density of the fuse

Area = 1/620 = 1.613 x 10^-3 cm^2

We know that 10000 cm^2 = 1 m^2, therefore,

1.613 x 10^-3 cm^2 = 1.613 x 10^-7 m^2

Recall that this area of this wire is gotten as

A = $\frac{\pi d^{2} }{4}$

where d is the diameter of the wire

1.613 x 10^-7 = $\frac{3.142* d^{2} }{4}$

6.448 x 10^-7 = 3.142 x $d^{2}$

$d^{2}$ = $\sqrt{ 2.05*10^-7}$

d = 4.5 x 10^-4 m = <em>0.45 mm</em>

8 0

1 year ago