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anyanavicka [17]
3 years ago
6

When Babe Ruth hit a homer over the 7.5-m-high right-field fence 95 m from home plate, roughly what was the minimum speed of the

ball when it left the bat? Assume the ball was hit 1.0 m above the ground and its path initially made a 38 degree angle with the ground.
Physics
1 answer:
Sholpan [36]3 years ago
4 0
<span>95m /2 = 47.5m
 
47.5m = Vocos38t

</span><span>Vy = Voy -9.8t

0 = Vosin38 - 9.8t
 0 = Vosin38 - 9.8(47.5 / Vocos38)
0 = Vo^2(0.62) - 590.7
</span>
<span>Vo = 30.87 m/s
</span>
Hope this helps
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Answer:

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Firstly removing off one strip and it leaves electrons behind, so the strip becomes positively charged.

2. The roll however is not negatively charged because it is "earthed " by the hand holding it, thus excess negatives repel each other away through the hand.

3.Tearing off the next strip and once more it leaves electrons behind, the new strip is also positively charged and will repel the first strip.

4. Then, tear two strips apart and one will leave electrons behind on the other. Meaning that one strip is positive and the other is negative and they will attract each other.

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a child is swinging on swing, describe what happens to both the kinectic energy and potential enegry of the child as she swings
Igoryamba

Answer:

The K.E is maximum when the child is at the vertical position and the P.E is maximum at the extreme deviated position from the vertical.

Explanation:

  • A child is swinging on swing up and down has both kinetic and potential energy.
  • The total mechanical energy of the system is conserved throughout the system. At any instant the total mechanical energy is given by,

                                      E = K.E + P.E

  • The K.E is maximum when the child is at the vertical position.
  • The P.E is maximum at the extreme deviated position from the vertical.
  • And when K.E is maximum P.E becomes minimum and vice versa as per the law of conservation of energy.
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An oscillator consists of a block attached to a spring (k = 500 N/m). At some time t, the position (measured from the system's e
Alex_Xolod [135]

Answer:

a) \omega = 10.407\,\frac{rad}{s}, b) m = 4.617\,kg, c) A = 1.355\,m

Explanation:

a) The system have a simple armonic motion, whose position function is:

x(t) = A\cdot \cos (\omega\cdot t + \phi)

The velocity function is determined by deriving the position function in terms of time:

v(t) = -\omega \cdot A \cdot \sin(\omega\cdot t + \phi)

The acceleration function is found by deriving again:

a(t) = -\omega^{2} \cdot A \cdot \cos (\omega\cdot t + \phi)

Let assume that t = 0\,s. The following nonlinear system is built:

A\cdot \cos \phi = 0.660\,m

-\omega \cdot A \cdot \sin \phi = -12.3\,\frac{m}{s}

-\omega^{2}\cdot A \cdot \sin \phi = -128\,\frac{m}{s^{2}}

System can be reduced by divinding the second and third expressions by the first expression:

\omega \cdot \tan \phi = 18.636\,\frac{1}{s}

\omega^{2}\cdot \tan \phi = 193.94\,\frac{1}{s^{2}}

Now, the last expression is divided by the first one:

\omega = 10.407\,\frac{rad}{s}

b) The mass of the block is:

m = \frac{k}{\omega^{2}}

m = \frac{500\,\frac{N}{m} }{(10.407\,\frac{rad}{s})^{2} }

m = 4.617\,kg

c) The phase angle is:

\phi = \tan^{-1} \left(\frac{18.636\,\frac{1}{s} }{\omega}  \right)

\phi \approx 0.338\pi

The amplitude is:

A = \frac{0.660\,m}{\cos 0.338\pi}

A = 1.355\,m

8 0
4 years ago
Read 2 more answers
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