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3 years ago

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A mass of 5 kg stretches a spring 20 cm. The mass is acted on by an external force of 10 sin t 6 N (newtons) and moves in a medi

um that imparts a viscous force of 6 N when the speed of the mass is 3 cm/s. If the mass is set in motion from its equilibrium position with an initial velocity of 3 cm/s, formulate the initial value problem describing the motion of the mass. (Use g = 9.8 m/s2 for the acceleration due to gravity. Let u(t), measured positive downward, denote the displacement in meters of the mass from its equilibrium position at time t seconds. Use up for u' and upp for u''.

1 answer:

nordsb [41]3 years ago

5 0

Answer:

u" + 40u' + 49u = 2 sin(t/6)

upp + 40up + 49u = 2 sin(t/6)

Explanation:

Step 1: Data given

mass = 5 kg

L = 20 cm = 0.2 m

F = 10 sin(t/6)N

Fd(t) = - 6 N

u(0) = 0.03 m/s

u(0) = 0

u'(0) = 3 cm/s

Step 2:

ω =kL

k = ω/L = m*g /L = (5*9.8)/0.2 = 245 kg/s²

Since Fd(t) = -γu'(t) we know:

γ =- Fd(t) / u'(t) = 6N/ 0.03 m/s = 200 Ns/m

The initial value problem which describes the motion of the mass is given by

5u" + 200u' + 245u = 10 sin(t/6) u(0) = 0 ; u'(0) = 0.03

This is equivalent to:

u" + 40u' + 49u = 2 sin(t/6) u(0) = 0 ; u'(0) = 0.03

upp + 40up + 49u = 2 sin(t/6)

With u in m and t in s

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3 years ago

A(n) 71.1 kg astronaut becomes separated from the shuttle, while on a space walk. She finds herself 70.2 m away from the shuttle

denpristay [2]

Answer:

10.347 minutes.

Explanation:

According to F = ma, she exerts force on camera of the magnitude

F = 0.67Kg*12m/ $s^{2}$ = 8.04N, assuming it took her one second to accelerate camera to 12m/s, then by newtons third law, which says every action has equal and opposite reaction , the camera exerts the same amount of force on the astronaut which gives her acceleration of a = $\frac{8.04N}{70.2Kg} = 0.1130801680m/s^2$ .

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at this velocity , the astronaut has to cover the distance of 70.2 meters, it will take her 620.7985075s = 10.347 min to get to the shuttle (using S = vt).

4 0

3 years ago

I need answers and solvings to these questions

den301095 [7]

1) The period of a simple pendulum depends on B) III. only (the length of the pendulum)

2) The angular acceleration is C) $15.7 rad/s^2$

3) The frequency of the oscillation is C) 1.6 Hz

4) The period of vibration is B) 0.6 s

5) The diameter of the nozzle is A) 5.0 mm

6) The force that must be applied is B) 266.7 N

Explanation:

1)

The period of a simple pendulum is given by

$T=2\pi \sqrt{\frac{L}{g}}$

where

T is the period

L is the length of the pendulum

g is the acceleration of gravity

From the equation, we see that the period of the pendulum depends only on its length and on the acceleration of gravity, while there is no dependence on the mass of the pendulum or on the amplitude of oscillation. Therefore, the correct option is

B) III. only (the length of the pendulum)

2)

The angular acceleration of the rotating disc is given by the equation

$\alpha = \frac{\omega_f - \omega_i}{t}$

where

$\omega_f$ is the final angular velocity

$\omega_i$ is the initial angular velocity

t is the time elapsed

For the compact disc in this problem we have:

$\omega_i = 0$ (since it starts from rest)

$\omega_f = 300 rpm \cdot \frac{2\pi rad/rev}{60 s/min}=31.4 rad/s$ is the final angular velocity

t = 2 s

Substituting, we find

$\alpha = \frac{31.4-0}{2}=15.7 rad/s^2$

3)

For a simple harmonic oscillator, the acceleration and the displacement of the system are related by the equation

$a=-\omega^2 x$

where

a is the acceleration

x is the displacement

$\omega$ is the angular frequency of the system

For the oscillator in this problem, we have the following relationship

$a=-100 x$

which implies that

$\omega^2 = 100$

And so

$\omega = \sqrt{100}=10 rad/s$

Also, the angular frequency is related to the frequency $f$ by

$f=\frac{\omega}{2\pi}$

Therefore, the frequency of this simple harmonic oscillator is

$f=\frac{10}{2\pi}=1.6 Hz$

4)

When the mass is hanging on the sping, the weight of the mass is equal to the restoring force on the spring, so we can write

$mg=kx$

where

m is the mass

$g=9.8 m/s^2$ is the acceleration of gravity

k is the spring constant

x = 8.0 cm = 0.08 m is the stretching of the spring

We can re-arrange the equation as

$\frac{k}{m}=\frac{g}{x}=\frac{9.8}{0.08}=122.5$

The angular frequency of the spring is given by

$\omega=\sqrt{\frac{k}{m}}=\sqrt{122.5}=11.1 Hz$

And therefore, its period is

$T=\frac{2\pi}{\omega}=\frac{2\pi}{11.1}=0.6 s$

5)

According to the equation of continuity, the volume flow rate must remain constant, so we can write

$A_1 v_1 = A_2 v_2$

where

$A_1 = \pi r_1^2$ is the cross-sectional area of the hose, with $r_1 = 5 mm$ being the radius of the hose

$v_1 = 4 m/s$ is the speed of the petrol in the hose

$A_2 = \pi r_2^2$ is the cross-sectional area of the nozzle, with $r_2$ being the radius of the nozzle

$v_2 = 16 m/s$ is the speed in the nozzle

Solving for $r_2$ , we find the radius of the nozzle:

$\pi r_1^2 v_1 = \pi r_2^2 v_2\\r_2 = r_1 \sqrt{\frac{v_1}{v_2}}=(5)\sqrt{\frac{4}{16}}=2.5 mm$

So, the diameter of the nozzle will be

$d_2 = 2r_2 = 2(2.5)=5.0 mm$

6)

According to the Pascal principle, the pressure on the two pistons is the same, so we can write

$\frac{F_1}{A_1}=\frac{F_2}{A_2}$

where

$F_1$ is the force that must be applied to the small piston

$A_1 = \pi r_1^2$ is the area of the first piston, with $r_1= 2 cm$ being its radius

$F_2 = mg = (1500 kg)(9.8 m/s^2)=14700 N$ is the force applied on the bigger piston (the weight of the car)

$A_2 = \pi r_2^2$ is the area of the bigger piston, with $r_2= 15 cm$ being its radius

Solving for $F_1$ , we find

$F_1 = \frac{F_2A_1}{A_2}=\frac{F_2 \pi r_1^2}{\pi r_2^2}=\frac{(14700)(2)^2}{(15)^2}=261 N$

So, the closest answer is B) 266.7 N.

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5 0

3 years ago

A cat is running at 24 m/s. It then accelerates at 7 m/s2. How long will it take the cat to reach a speed of 49 m/s?

kozerog [31]

Answer:

t should be 3.57 second

Explanation:

Formula used is v = u+at

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

Substitute each of the info given into the formula and calculate.

49 = 24 + (7)t

t = 3.57s

8 0

3 years ago

Mademuasel [1]

According to Newton's Second Law of Motion :

The Force acting on an Object is equal to Product of Mass of the Object and Acceleration produced due to the Force.

$:\implies$ Force acting = Mass of the Object × Acceleration

Given : Force = 50 newton and Mass of the Object = 10 kg

Substituting the respective values in the Formula, we get :

$:\implies$ 50 N = 10 kg × Acceleration

$:\implies \mathsf{Acceleration = \dfrac{50\;N}{10\;kg}}$

$:\implies$ Acceleration of the Object = 5 m/s²

4 0

3 years ago