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Helen [10]
3 years ago
10

The ball will oscillate along the z axis between z=d and z=−d in simple harmonic motion. What will be the angular frequency ω of

these oscillations? Use the approximation d≪a to simplify your calculation; that is, assume that d2+a2≈a2.
Physics
1 answer:
Alinara [238K]3 years ago
5 0

Any kind of frequency, including the angular kind, is closely involved with
time. Still, for some unknown reason,you've given us no time information
whatsoever ... a peculiar decision on your part, since we can be sure that
it's right there, inexorably intertwined with the part of the question that you
DID copy and share with us. 

Furthermore and moreover, for one with no prior experience with simple
harmonic motion, the many symbols in this question such as ' d ', ' a ',
' << ', ' d₂ ', and ' a₂ ' would be of no help at all to guide him toward a
solution.  On the contrary, he would conclude that the question itself
had been posted by some alien life form.

To sum up:  Come back and post the drawing that goes along with the
question, make sure you have presented all of the information that the
question includes, and then we'll talk.


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AlekseyPX

Answer:

160 kg

12 m/s

Explanation:

m_1 = Mass of first car = 120 kg

m_2 = Mass of second car

u_1 = Initial Velocity of first car = 14 m/s

u_2 = Initial Velocity of second car = 0 m/s

v_1 = Final Velocity of first car = -2 m/s

v_2 = Final Velocity of second car

For perfectly elastic collision

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}\\\Rightarrow m_2v_2=m_{1}u_{1}+m_{2}u_{2}-m_{1}v_{1}\\\Rightarrow m_2v_2=120\times 14+m_2\times 0-(120\times -2)\\\Rightarrow m_2v_2=1920\\\Rightarrow m_2=\frac{1920}{v_2}

Applying in the next equation

v_2=\frac{2m_1}{m_1+m_2}u_{1}+\frac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\frac{2\times 120}{120+\frac{1920}{v_2}}\times 14+\frac{m_2-m_1}{m_1+m_2}\times 0\\\Rightarrow \left(120+\frac{1920}{v_2}\right)v_2=3360\\\Rightarrow 120v_2+1920=3360\\\Rightarrow v_2=\frac{3360-1920}{120}\\\Rightarrow v_2=12\ m/s

m_2=\frac{1920}{v_2}\\\Rightarrow m_2=\frac{1920}{12}\\\Rightarrow m_2=160\ kg

Mass of second car = 160 kg

Velocity of second car = 12 m/s

5 0
3 years ago
A launched hopper reach to 1.20 m maximum height. How much is it’s launch velocity?
garri49 [273]

The launch velocity is 4.8 m/s

Explanation:

We can solve this problem by applying the law of conservation of energy. In fact, the mechanical energy of the hopper (equal to the sum of the potential energy + the kinetic energy) is conserved. So we can write:

U_i +K_i = U_f + K_f

where:

U_i is the initial potential energy, at the bottom

K_i is the initial kinetic energy, at the bottom

U_f is the final potential energy, at the top

K_f is the final kinetic energy, at the top

We can rewrite the equation as:

mgh_i + \frac{1}{2}mu^2 = mgh_f + \frac{1}{2}mv^2

where:

m is the mass of the hopper

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

h_i = 0 is the initial height

u is the launch speed of the hopper

h_f = 1.20 m is the maximum altitude reached by the hopper

v = 0 is the final speed (which is zero when the hopper reaches the maximum height)

Solving the equation for u, we find the launch speed of the hopper:

u=\sqrt{2gh_g}=\sqrt{2(9.8)(1.20)}=4.8 m/s

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3 years ago
The average diameter of one tennis ball in a package of three is 6.8 cm. Which of the following is the combined volume of all th
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We want to find the combined volume of 3 tennis balls. We will get that the combined volume is 493.7 cm^3

First, remember that for a sphere of diameter D, the volume is:

V = \frac{4}{3}*3.14*(\frac{D}{2})^3

Where 3.14 is pi.

Here we know that the average diameter of a tennis ball is 6.8cm, then we can replace that in the above equation to find the volume (in average) of a single tennis ball:

V = \frac{4}{3}*3.14*(\frac{6.8cm}{2})^3 = 164.5 cm^3

Now, in 3 balls of tennis, the combined volume will be 3 times the above one, this is:

3*V = 3*164.5cm^3 = 493.7 cm^3

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The object will move if the forces are unbalanced.

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