How do you find the velocity of an object?

Mechanical waves require a medium to transfer their energy. What are examples of mechanical waves

NikAS [45]

The answer is A hopes this helps!

8 0

3 years ago

A mass weighing 11 lb stretches a spring 4in. The mass is pulled down an additional 3 in and is then set in motion with an initi

Shtirlitz [24]

Answer:

a) $x(t) = 2.845\cdot \cos \left(1.732\cdot t -0.473\pi \right)$ , b) $T = 3.628\,s$ , $A = 2.845\,ft$ , $\phi = -0.473\pi$

Explanation:

a) The system mass-spring is well described by the following equation of equilibrium:

$\Sigma F = k\cdot x - m\cdot g = m\cdot a$

After some handling in physical and mathematical definition, the following non-homogeneous second-order linear differential equation:

$\frac{d^{2}x}{dt^{2}}+\frac{k}{m}\cdot x = g$

The solution of this equation is:

$x (t) = A\cdot \cos \left(\sqrt{\frac{k}{m} } \cdot t + \phi\right)$

The velocity function is:

$v(t) = \sqrt{\frac{k}{m} }\cdot A\cdot \sin \left(\sqrt{\frac{k}{m} }\cdot t +\phi \right)$

Initial conditions are:

$x(0\,s) = 0.25\,ft, v(0\,s) = -5\,\frac{ft}{s}$

Equations at $t = 0\,s$ are:

$0.25\,ft = A\cdot \cos \phi\\-5\,\frac{ft}{s} =\sqrt{\frac{k}{m} }\cdot A\cdot \sin \phi$

The spring constant is:

$k = \frac{11\,lbf}{0.333\,ft}$

$k = 33\,\frac{lbf}{ft}$

After some algebraic handling, amplitude and phase angle are found:

$\phi = -0.473\pi$

$A = 2.845\,ft$

The position can be described by this function:

$x(t) = 2.845\cdot \cos \left(1.732\cdot t -0.473\pi \right)$

b) The period of the motion is:

$T = \frac{2\pi}{\sqrt{\frac{k}{m} } }$

$T = 3.628\,s$

The amplitude is:

$A = 2.845\,ft$

The phase of the motion is:

$\phi = -0.473\pi$

8 0

4 years ago

The ability to distinguish between acceleration and velocity will be critical to your understanding of many other concepts in th

Sauron [17]

Answer:

Object could only be moving with increasing speed.

Explanation:

Let us consider the general formula of acceleration:

a = (Vf - Vi)/t

Vf = Vi + at -------- equation 1

where,

Vf = Final Velocity

Vi = Initial Velocity

a = acceleration

t = time

<u>FOR POSITIVE ACCELERATION:</u>

Vf = Vi + at

since, both acceleration and time are positive quantities. Hence, it means that the final velocity of the object shall be greater than the initial velocity of the object.

Vf > Vi

It clearly shows that if an object moves with positive acceleration. <u>It could only be moving with increasing speed.</u>

Solving the same equation for negative acceleration shows that the final velocity will be less than initial velocity and object will be moving with decreasing speed.

And for the constant velocity final and initial velocities are equal and thus, acceleration will be zero.

7 0

3 years ago

In a compound microscope, the objective has a focal length of 1.0 cm, the eyepiece has a focal length of 2.0 cm, and the tube le

expeople1 [14]

Answer:

The value is $m \approx 310$