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Svetllana [295]
2 years ago
12

A mass weighing 4 lb stretches a spring 2 in. Suppose that the mass is given an additional 6-in displacement in the positive dir

ection and then released. The mass is in a medium that exerts a viscous resistance of 6 lb when the mass has a velocity of 3 ft/s. Under the assumptions discussed in this section, formulate the initial value problem that governs the motion of the mass.
Physics
1 answer:
givi [52]2 years ago
6 0

Answer:

\frac{1}{8} y'' + 2y' + 24y=0

Explanation:

The standard form of the 2nd order differential equation governing the motion of mass-spring system is given by

my'' + \zeta y' + ky=0

Where m is the mass, ζ is the damping constant, and k is the spring constant.

The spring constant k can be found by

w - kL=0

mg - kL=0

4 - k\frac{1}{6}=0

k = 4\times 6 =24

The damping constant can be found by

F = -\zeta y'

6 = 3\zeta

\zeta = \frac{6}{3} = 2

Finally, the mass m can be found by

w = 4

mg=4

m = \frac{4}{g}

Where g is approximately 32 ft/s²

m = \frac{4}{32} = \frac{1}{8}

Therefore, the required differential equation is

my'' + \zeta y' + ky=0

\frac{1}{8} y'' + 2y' + 24y=0

The initial position is

y(0) = \frac{1}{2}

The initial velocity is

y'(0) = 0

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A 10-turn coil of wire having a diameter of 1.0 cm and a resistance of 0.50 Ω is in a 1.0 mT magnetic field, with the coil orien
n200080 [17]

Answer:

The voltage across the capacitor is 1.57 V.

Explanation:

Given that,

Number of turns = 10

Diameter = 1.0 cm

Resistance = 0.50 Ω

Capacitor = 1.0μ F

Magnetic field = 1.0 mT

We need to calculate the flux

Using formula of flux

\phi=NBA

Put the value into the formula

\phi=10\times1.0\times10^{-3}\times\pi\times(0.5\times10^{-2})^2

\phi=7.85\times10^{-7}\ Tm^2

We need to calculate the induced emf

Using formula of induced emf

\epsilon=\dfrac{d\phi}{dt}

Put the value into the formula

\epsilon=\dfrac{7.85\times10^{-7}}{dt}

Put the value of emf from ohm's law

\epsilon =IR

IR=\dfrac{7.85\times10^{-7}}{dt}

Idt=\dfrac{7.85\times10^{-7}}{R}

Idt=\dfrac{7.85\times10^{-7}}{0.50}

Idt=0.00000157=1.57\times10^{-6}\ C

We know that,

Idt=dq

dq=1.57\times10^{-6}\ C

We need to calculate the voltage across the capacitor

Using formula of charge

dq=C dV

dV=\dfrac{dq}{C}

Put the value into the formula

dV=\dfrac{1.57\times10^{-6}}{1.0\times10^{-6}}

dV=1.57\ V

Hence, The voltage across the capacitor is 1.57 V.

5 0
2 years ago
A circular coil of wire of 200 turns and diameter 2.0 cm carries a current of 4.0 A. It is placed in a magnetic field of 0.70 T
DerKrebs [107]

Answer:

0.087976 Nm

Explanation:

The magnetic torque (τ) on a current-carrying loop in a magnetic field is given by;

τ = NIAB sinθ     --------- (i)

Where;

N = number of turns of the loop

I = current in the loop

A = area of each of the turns

B = magnetic field

θ = angle the loop makes with the magnetic field

<em>From the question;</em>

N = 200

I = 4.0A

B = 0.70T

θ = 30°

A = π d² / 4        [d = diameter of the coil = 2.0cm = 0.02m]

A = π x 0.02² / 4 = 0.0003142m²         [taking π = 3.142]

<em>Substitute these values into equation (i) as follows;</em>

τ = 200 x 4.0 x 0.0003142 x 0.70 sin30°

τ = 200 x 4.0 x 0.0003142 x 0.70 x 0.5

τ = 200 x 4.0 x 0.0003142 x 0.70      

τ = 0.087976 Nm

Therefore, the torque on the coil is 0.087976 Nm

3 0
2 years ago
Calculate the mass of gold that occupies 5.0 × 10−3 cm3 . the density of gold is 19.3 g/cm3
ANEK [815]

Answer;

= 0.0965 g

Explanation;

Mass is given by multiplying density by volume

Mass = density * volume  

Mass = 19.3 g/cm³×0.005 cm³  

Mass = 0.0965 g

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2 years ago
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The kinetic energy of the car is A.) 0.466 J
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