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neonofarm [45]
3 years ago
9

The position of the front bumper of a test car under microprocessor control is given by x(t) = 2.31 m + (4.90 m/s2)t2 - (0.100 m

/s6)t6. (a) find its position at the instants when the car has zero veloc
Physics
1 answer:
vovangra [49]3 years ago
3 0
The equation of the car is given by the equation,

                          x(t) = 2.31 + 4.90t² - 0.10t⁶

If we are going to differentiate the equation in terms of x, we get the value for velocity.

                  dx/dt = 9.8t - 0.6t⁵

Calculate for the value of t when dx/dt = 0.

                 dx/dt = 0 = (9.8 - 0.6t⁴)(t)

The values of t from the equation is approximately equal to 0 and 2. 

If we substitute these values to the equation for displacement,

(0)   , x = 2.31 + 4.90(0²) - 0.1(0⁶) = 2.31

(2)    , x = 2.31 + 4.90(2²) - 0.1(2⁶) = 15.51

Thus, the positions at the instants where velocity is zero are 2.31 and 15.51 meters. 
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The time taken to fill 200 cm³ is 1 hour.
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Therefore,
\frac{200 \,  cm^{3}}{v \, cm^{3}/s}  = 3600 \, s \\ v =  \frac{200}{3600} =0.0556 \, cm^{3}

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8 0
3 years ago
The speed of sound in air is 345 m/s. A tuning fork vibrates above the open end of a sound resonance tube. If sound waves have w
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To develop this problem it is necessary to apply the concept of Frequency based on speed and wavelength.

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v = 345m/s

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6 0
3 years ago
Why do the graphs differ?​
melisa1 [442]

Well first graph represents rectangular hyperbola

vu = c^2 ( c is constant)

AS 1/v + 1/u = 1/f

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it is of the form y = - x + k

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The gage pressure in a liquid at a depth of 3 m is read to be 39 kPa. Determine the gage pressure in the same liquid at a depth
ioda

Answer: 117 kPa

Explanation:

For the liquid at depth 3 m, the gauge pressure is equal to = P₁=39 kPa

For the liquid at depth 9m, the gauge pressure is equal to= P₂

Now we are given the condition that the liquid is same. That must imply that the density must be same throughout the depth.

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So \frac{39}{P_2}= \frac{3}{9}

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5 0
3 years ago
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