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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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Heat engines were first envisioned and built during the industrial revolution. Explain the thermodynamics of a heat engine comme

Artyom0805 [142]

Heat engines were developed during industrial revolution.

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The source of a heat engine is present at a higher temperature as compared to the sink. Due to the temperature difference, the heat will flow from source to sink through working substance.

Let us consider $T_{1}\ and\ T_{2}$ are the temperature of source and sink.

As the source is at higher temperature as compared to sink, heat will flow from source to sink.

$Let\ Q_{1}\ and\ Q_{2}$ are the heat provided by source and heat rejected to sink.

Hence, the work done by the working substance will be -

$W\ =\ Q_{1}-Q_{2}$

The efficiency of a heat engine is defined as the ratio of output to the input energy.

Here output = workdone [W]

Hence, the efficiency of a heat engine is calculated as -

$Efficiency\ [\eta]=\frac{W}{Q_{1}}$

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$=\ 1-\frac{Q_{2}} {Q_{1}}$

This is the expression for the efficiency of heat engine.

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2 years ago

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A sailboat starts from rest and accelerates at a rate of 0.21 m/s^2 over a distance of 280 m. find the magnitude of the boat's f

sasho [114]

We use the kinematic equations,

$v=u+at$ (A)

$S= ut + \frac{1}{2} at^2$ (B)

Here, u is initial velocity, v is final velocity, a is acceleration and t is time.

Given, $u=0$ , $a=0.21 \ m/s^2$ and $s= 280 m$ .

Substituting these values in equation (B), we get

$280 \ m = 0 +\frac{1}{2} (0.21 m/s^2) t^2 \\\\ t^2 = \frac{280 \times 2}{0.21 } \\\\ t= 51.63 \ s$ .

Therefore from equation (A),

$v = 0 + (0.21) \times (51.63 s)= 10.84 \ m/s$

Thus, the magnitude of the boat's final velocity is 10.84 m/s and the time taken by boat to travel the distance 280 m is 51.63 s

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2 years ago

An electric eel (Electrophorus electricus) can produce a shock of up to 600 V and a current of 1 A for a duration of 2 ms, which

Irina-Kira [14]

Answer:

$2\times 10^{-3}\ C$

6000

1.2 J

$3.33\times 10^{-6}\ F$

Explanation:

I = Current = 1 A

t = Time = 2 ms

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V = Voltage = 100 mV

Charge is given by

$Q=It\\\Rightarrow Q=1\times 2\times 10^{-3}\\\Rightarrow Q=2\times 10^{-3}\ C$

The charge flowing through the electrocytes in that amount of time is $2\times 10^{-3}\ C$

The maximum potential is given by

$V_m=nV\\\Rightarrow n=\dfrac{V_m}{V}\\\Rightarrow n=\dfrac{600}{100\times 10^{-3}}\\\Rightarrow n=6000$

The number of electrolytes is 6000

Energy is given by

$E=Pt\\\Rightarrow E=V_mIt\\\Rightarrow E=600\times 1\times 2\times 10^{-3}\\\Rightarrow E=1.2\ J$

The energy released when the electric eel delivers a shock is 1.2 J

Equivalent capacitance is given by

$C_e=\dfrac{Q}{V_m}\\\Rightarrow C_e=\dfrac{2\times 10^{-3}}{600}\\\Rightarrow C_e=3.33\times 10^{-6}\ F$

The equivalent capacitance of all the electrocyte cells in the electric eel is $3.33\times 10^{-6}\ F$

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2 years ago

Which statement correctly describes the relationship between frequency and wavelength?

Len [333]

The relationship between the frequency and wavelength of a wave is given by the equation:

v=λf, where v is the velocity of the wave, λ is the wavelength and f is the frequency.

If we divide the equation by f we get:

λ=v/f

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So the second statement is true: As the frequency of a wave increases, the shorter the wavelength is.

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A student releases a block of mass m from rest at the top of a slide of height h1. The block moves down the slide and off the en

Nina [5.8K]

Answer:

B) d = √ ( 4 h₂ ( H - 2h₂))

Explanation:

A) 1) If the height of the slide is very small, there is no speed to leave the table, therefore do not recreate almost any horizontal distance

2) If the height is very small downwards, it touches the earth a little and the horizon is small,

B) to find an equation for horizontal distance (d)

We must maximize the speed at the bottom of the slide let's use energy

Starting point Higher

Em₀ = U = m g h₁

Final point. Lower (slide bottom)

Emf = K + U = ½ m v² + m gh₂

As there is no friction the energy is conserved

mgh₁ = ½ m v² + mgh₂

v² = 2 g (h₁-h₂)

This is the speed with which the block leaves the table, bone is the horizontal speed (vₓ)

The distance traveled when leaving the table can be searched with kinematics, projectile launch

x = v₀ₓ t

y = t - ½ g t²

The height is the height of the table (y = h₂), as it comes out horizontally the vertical speed is zero

t = √ 2h₂ / g

We substitute in the other equation

d = √ (2g (h₁-h₂)) √ 2h₂ / g

d = √ (4 h₂ (h₁-h₂))

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h₁ = H -h₂

d = √ ( 4 h₂ ( H - 2h₂))

Explanation:

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