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3 years ago

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I need help, please help an object starts from rest at time t=0 and moves in the with constant acceleration. The object travels

3m from time t=1 to time t=2. what is the acceleration of the object?

1 answer:

vichka [17]3 years ago

4 0

<span>a=8(m/<span>sec2</span>).. i answered this earlier i think.</span>

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mass weighing 16 pounds stretches a spring 8 3 feet. The mass is initially released from rest from a point 2 feet below the equi

valina [46]

Answer:

The answer is

" $x(t)= e^\frac{-t}{2}((\frac{-4}{3})\cos\frac{\sqrt{47}}{2}t- \frac{-64\sqrt{47}}{141} \sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}(\cos(3t)+ \sin (3t))$ ".

Explanation:

Taking into consideration a volume weight = 16 pounds originally extends a springs $\frac{8}{3}$ feet but is extracted to resting at 2 feet beneath balance position.

The mass value is =

$W=mg\\m=\frac{w}{g}\\m=\frac{16}{32}\\m= \frac{1}{2} slug\\$

The source of the hooks law is stable,

$16= \frac{8}{3} k \\\\8k=16 \times 3 \\\\k=16\times \frac{3}{8} \\\\k=6 \frac{lb}{ft}\\\\$

Number $\frac{1}{2}$ times the immediate speed, i.e .. Damping force

$\frac{1}{2} \frac{d^2 x}{dt^2} = -6x-\frac{1}{2}\frac{dx}{dt}+10 \cos 3t \\\\\frac{1}{2} \frac{d^2 x}{dt^2}+ \frac{1}{2}\frac{dx}{dt}+6x =10 \cos 3t \\ \\\frac{d^2 x}{dt^2} +\frac{dx}{dt}+12x=20\cos 3t \\\\$

The m^2+m+12=0 and m is an auxiliary equation,

$m=\frac{-1 \pm \sqrt{1-4(12)}}{2}\\\\m=\frac{-1 \pm \sqrt{47i}}{2}\\\\\ m1= \frac{-1 + \sqrt{47i}}{2} \ \ \ \ or\ \ \ \ \ m2 =\frac{-1 - \sqrt{47i}}{2}$

Therefore, additional feature

$x_c (t) = e^{\frac{-t}{2}}[C_1 \cos \frac{\sqrt{47}}{2}t+ C_2 \sin \frac{\sqrt{47}}{2}t]$

Use the form of uncertain coefficients to find a particular solution.

Assume that solution equation,

$x_p = Acos(3t)+B sin(3t) \\x_p'= -3A sin (3t) + 3B cos (3t)\\x_p}^{n= -9 Acos(3t) -9B sin (3t)\\$

These values are replaced by equation ( 1):

$\frac{d^2x}{dt}+\frac{dx}{dt}+ 12x=20 \cos(3t) -9 Acos(3t) -9B sin (3t) -3Asin(3t)+3B cos (3t) + 12A cos (3t) + 12B sin (3t)\\\\3Acos 3t + 3B sin 3t - 3Asin 3t + 3B cos 3t= 20cos(3t)\\(3A+3B)cos3t -(3A-3B)sin3t = 20 cos (3t)\\$

Going to compare cos3 t and sin 3 t coefficients from both sides,

The cost3 t is 3A + 3B= 20 coefficients

The sin 3 t is 3B -3A = 0 coefficient

The two equations solved:

$3A+3B = 20 \\\frac{3B -3A=0}{}\\6B=20\\B= \frac{20}{6}\\B=\frac{10}{3}\\$

Replace the very first equation with the meaning,

$3B -3A=O\\3(\frac{10}{3})-3A =0\\A= \frac{10}{3}\\$

equation is

$x_p\\\\\frac{10}{3} cos (3 t) + \frac{10}{3} sin (3t)$

The ultimate plan for both the equation is therefore

$x(t)= e^\frac{-t}{2} (c_1 cos \frac{\sqrt{47}}{2}t)+c_2\sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}\cos (3t)+\frac{10}{3}\sin (3t)$

Initially, the volume of rest x(0)=2 and x'(0) is extracted by rest i.e.

Throughout the general solution, replace initial state x(0) = 2,

Replace x'(0)=0 with a general solution in the initial condition,

$x(t)= e^\frac{-t}{2} [(c_1 cos \frac{\sqrt{47}}{2}t)+c_2\sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}\cos (3t)+\frac{10}{3}\sin (3t)]\\\\$

$x(t)= e^\frac{-t}{2} [(-\frac{\sqrt{47}}{2}c_1\sin\frac{\sqrt{47}}{2}t)+ (\frac{\sqrt{47}}{2}c_2\cos\frac{\sqrt{47}}{2}t)+c_2\cos\frac{\sqrt{47}}{2}t) +c_1\cos\frac{\sqrt{47}}{2}t +c_2\sin\frac{\sqrt{47}}{2}t + \frac{-1}{2}e^{\frac{-t}{2}} -10 sin(3t)+10 cos(3t) \\\\$

$c_2=\frac{-64\sqrt{47}}{141}$

$x(t)= e^\frac{-t}{2}((\frac{-4}{3})\cos\frac{\sqrt{47}}{2}t- \frac{-64\sqrt{47}}{141} \sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}(\cos(3t)+ \sin (3t))$

5 0

3 years ago

A 0.400-kg object is swung in a circular path and in a vertical plane on a 0.500-m-length string. If the angular speed at the bo

Talja [164]

Answer:

T = 16.72 N

Explanation:

When the object is swung in a circular path, and in a vertical plane, there are two forces external to the object acting on it at any time: the gravity (which is always downward) and the tension in the string (which always points towards the center of the circle).

At the bottom of the circle, the tension is directly upward, so these two forces, are opposite each other, and the difference between them is the centripetal force , which at this point, keeps the object swinging in a circle.

This is the point of the trajectory where T is maximum.

We can apply Newton's 2nd Law, choosing an axis vertical (y-axis) being the upward direction the positive one, as follows:

T- m*g = m*a

The acceleration, at the bottom of the circle, is only normal (as there are no forces in the horizontal direction) , and is equal to the centripetal acceleration, as follows:

ac = v² / r = ω²*r⇒ T- m*g = m*ω²*r

Replacing by the givens, we can solve for T as follows:

T = m* (ω²*r+g) = 0.4 kg*((8.00)² rad/sec²*0.5m)+9.8 m/s²) = 16.72 N

5 0

3 years ago

A wheel of radius R, mass M, and moment of inertia I is mounted on a frictionless, horizontal axles. A light cord wrapped around

Alex_Xolod [135]

Answer:

$\alpha =\frac{m*g*R}{I-m*R^2}$

$a = \frac{m*g*R^2}{I-m*R^2}$

$T=\frac{I*m*g}{I-m*R^2}$

Explanation:

By analyzing the torque on the wheel we get:

$T*R=I*\alpha$ Solving for T: $T=I/R*\alpha$

On the object:

$T-m*g = -m*a$ Replacing our previous value for T:

$I/R*\alpha-m*g = -m*a$

The relation between angular and linear acceleration is:

$a=\alpha*R$

So,

$I/R*\alpha-m*g = -m*\alpha*R$

Solving for α:

$\alpha =\frac{R*m*g}{I+m*R^2}$

The linear acceleration will be:

$a =\frac{R^2*m*g}{I+m*R^2}$

And finally, the tension will be:

$T =\frac{I*m*g}{I+m*R^2}$

These are the values of all the variables: α, a, T

8 0

3 years ago

A heated iron is pressed against a shirt, warming it. This is an

Nady [450]

Answer:

Convection? I'm pretty sure that's it

6 0

3 years ago

A player kicks a soccer ball from ground level and sends it flying at an angle of 30 degrees at a speed of 26 m/s. What is the m

Akimi4 [234]

Answer:

8.6 m

Explanation:

The motion of a soccer ball is a motion of a projectile, with a uniform motion along the horizontal (x-) direction and an accelerated motion along the vertical (y-) direction, with constant acceleration $a=g=-9.8 m/s^2$ towards the ground (we take upward as positive direction, so acceleration is negative).

The initial velocity along the vertical direction is

$v_{y0} = v_0 sin \theta = (26 m/s)(sin 30^{\circ})=13 m/s$

Now we can consider the motion along the vertical direction only. the vertical velocity at time t is given by:

$v_y(t)=v_{y0} +at$

At the point of maximum height, $v_y(t)=0$ , so we can find the time t at which the ball reaches the maximum height:

$0=v_{y0}+at\\t=-\frac{v_{y0}}{a}=-\frac{13 m/s}{-9.8 m/s^2}=1.33 s$

And now we can use the equation of motion along the y-axis to find the vertical position of the ball at t=1.33 s, which corresponds to the maximum height of the ball:

$y(t)=v_{y0}t + \frac{1}{2}at^2=(13 m/s)(1.33 s)+\frac{1}{2}(-9.8 m/s^2)(1.33 s)^2=8.6 m$

4 0

3 years ago