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Thepotemich [5.8K]

2 years ago

10

(1 point) Suppose a spring with spring constant 7 N/m is horizontal and has one end attached to a wall and the other end attache

d to a 3 kg mass. Suppose that the friction of the mass with the floor (i.e., the damping constant) is 2 N⋅s/m. Set up a differential equation that describes this system. Let x to denote the displacement, in meters, of the mass from its equilibrium position, and give your answer in terms of x,x′,x′′. Assume that positive displacement means the mass is farther from the wall than when the system is at equilibrium.

1 answer:

dlinn [17]2 years ago

5 0

Answer:

x'' + (2/3)x' + $\frac{7}{3}$ x=0

Explanation:

Given:

spring constant, k= 7 N/m ; mass = 3kg ;and damping constant, c=2 N-s/m

a)

Differential Equation:

m( $d^{2} x/dt^{2})= -kx-cdx/dt$ ........... (1)

Substitute the values of m,k, and c in (1).

3( $d^{2} x/dt^{2})= -7x-(2dx/dt$ )

3( $d^{2} x/dt^{2}) + 7x + (2dx/dt$ )=0

i.e , 3x'' +2x' + 7x =0

By dividing the equation with ;

x'' + (2/3)x' + $\frac{7}{3} x$ =0

The differential equation for the system is

x'' + (2/3)x' + $\frac{7}{3}$ x=0

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The driver of a car slams on the brakes when he sees a tree blocking the road. The car slows uniformly with acceleration of −5.2

sergejj [24]

Answer:

The car strikes the tree with a final speed of 4.165 m/s

The acceleration need to be of -5.19 m/seg2 to avoid collision by 0.5m

Explanation:

First we need to calculate the initial speed $V_{0}$

$x=V_{0} *t+\frac{1}{2} *a*(t^{2} )\\62.5m=V_{0} *4.15s+\frac{1}{2} *-5.25\frac{m}{s^{2} } *(4.15^{2} )\\V_{0}=25.953\frac{m}{s}$

Once we have the initial speed, we can isolate the final speed from following equation:

$V_{f} =V_{0} +a*t$ $V_{f}= 4.165 \frac{m}{s}$

Then we can calculate the aceleration where the car stops 0.5 m before striking the tree.

To do that, we replace 62 m in the first formula, as follows:

$x=V_{0} *t+\frac{1}{2} *a*(t^{2} )\\62m=25.953\frac{m}{s}*4.15s+\frac{1}{2} *-a\frac{m}{s^{2} } *(4.15^{2} )\\a=-5.19\frac{m}{s^{2} }$

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2 years ago

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Levi observed properties of four different waves and recorded observations about each one in his chart. A 2-column table with 4

makvit [3.9K]

Answer:

Wave W is a sound wave, Waves X and Y are light waves, and it is impossible to tell what kind of wave Wave Z is.

Explanation:

W travels fastest through metal

X travels fastest through air,

Y travels more slowly through water than air

Z travels more slowly at cool temperatures

W appears to be sound wave as sound travels fastest through metal .

X appears to be light wave as light travels fastest in air .

Y also appears to be light wave as speed of light is reduced when it passes from air to water .

Z It is impossible to tell anything about the nature of Z wave .

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Second choice . . . False, because electrons are not involved in the atomic mass.

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2 years ago

Two insulated copper wires of similar overall diameter have very different interiors. One wire possesses a solid core of copper,

balandron [24]

Answer with Explanation:

We are given that

Radius of solid core wire=r=2.28 mm= $2.28\times 10^{-3} m$

$1mm=10^{-3} m$

Radius of each strand of thin wire=r'=0.456 mm= $0.456\times 10^{-3} m$

Current density of each wire= $J=3750 A/m^2$

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Where $\pi=3.14$

Using the formula

Cross section area of copper wire has solid core = $3.14\times (2.28\times 10^{-3})^2=16.3\times 10^{-6} m^2$

Current density = $J=\frac{I}{A}$

Using the formula

$3750=\frac{I}{16.3\times 10^{-6}}$

$I=3750\times 16.3\times 10^{-6}=0.061 A$

Total number of strands=19

Area of strand wire= $A'=19\times 3.14\times (0.456\times 10^{-3})^2=12.4\times 10^{-6} m^2$

$J'=\frac{I'}{A'}$

$3750=\frac{I'}{19\times 3.14(0.456\times 10^{-3})^2}$

$I'=3750\times 19\times 3.14(0.456\times 10^{-3})^2$

$I'=0.047 A$

b.Resistivity of copper wire= $\rho=1.69\times 10^{-8}\Omega-m$

Length of each wire =6.25 m

Resistance, R= $\frac{\rho l}{A}$

Using the formula

Resistance of solid core wire= $R=\frac{1.69\times 10^{-8}\times 6.25}{16.3\times 10^{-6}}=6.5\times 10^{-3}\Omega$

Resistance of strand wire= $R'=\frac{1.69\times 10^{-8}\times 6.25}{12.4\times 10^{-6}}=8.5\times 10^{-3}\Omega$

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2 years ago

In an experiment to measure the wavelength of light using a double slit, it is found that the fringes are too close together to

Mandarinka [93]

Answer:

halve the slit separation

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As we know that

In YDS experiment, the equation of fringe width is as follows

$\beta = \frac{\lambda D}{d}$

where,

D denotes the separation in the middle of screen and slits

d denotes the distance in the middle of two slits

And to increase the Δx we have to decrease the d i.e, the distance between the two slits

Hence, the first option is correct

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2 years ago