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Nadusha1986 [10]
3 years ago
15

A bowling ball is traveling at 3 m/s. It starts to roll up a ramp. How high above the ground will the ball be when it stops roll

ing? Neglect friction and assume the ramp is plenty long enough to do this. (For this problem your initial kinetic energy can be set equal to your final potential energy. 1/2mv2 = mgh).
Physics
2 answers:
Zepler [3.9K]3 years ago
8 0

Answer:

FREE FREE FREE FREEE FREEEEEEEEEEE FREEEEEEEEEEEEEEEE FREEEEEEEEEEEEEEEEEE

Explanation:

FREE FREE FREE FREEE FREEEEEEEEEEE FREEEEEEEEEEEEEEEE FREEEEEEEEEEEEEEEEEE

vivado [14]3 years ago
5 0

Answer:

h= 0.45 m

Explanation:

PE= 1/2 mv^2             KE= mgh

v= 3m/s

vf= 0 m/s

h=?

PE= 1/2(1kg)(3m/s)^2

PE= 4.5 J

4.5 J/ 1kg(9.8 m/s^2)

h=0.45 m

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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
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Answer:

<h2>Uses of plane mirrors</h2>

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Explanation:

Hope it is helpful....

6 0
3 years ago
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GuDViN [60]

The equation has one zero, and the system has three solutions ( 1 real and 2 complex solutions).

<h3>Solution of the polynomial equation</h3>

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Answer:

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Answer:

D is the answer  I think (0 w 0 )

Explanation:

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