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3 years ago

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A 1500 kg car carrying four 90 kg people travels over a "washboard" dirt road with corrugations 3.7 m apart. The car bounces wit

h maximum amplitude when its speed is 20 km/h. When the car stops, and the people get out, by how much does the car body rise on its suspension?

1 answer:

umka21 [38]3 years ago

7 0

Answer:

Car body rise on its suspension by 0.0309 m

Explanation:

We have given mass of the car m = 1500 kg

Mass of each person = 90 kg

Speed of the car $v=20km/hr=20\times \frac{5}{18}=5.555m/sec$

Distance traveled by car d = 3.7 m

So time period $T=\frac{distance}{speed}=\frac{4}{5.55}=0.72sec$

Frequency $f=\frac{1}{T}=\frac{1}{0.72}=1.388Hz$

Angular frequency is $\omega =2\pi f=2\times 3.14\times 1.388=8.722rad/sec$

Angular frequency is equal to $\omega =\sqrt{\frac{k}{m}}$

$8.722 =\sqrt{\frac{k}{1500}}$

k = 114109.92 N/m

Now weight of total persons will be equal to spring force

$4mg=kx$

$4\times 90\times 9.8=114109.92\times x$

x = 0.0309 m

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When a golf club hits a 0.0459 kg ball at rest, it exerts a 2380 N force for 0.00100 s. What is the speed of the ball afterwards

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Answer:

51.85m/s

Explanation:

Given parameters:

Mass of ball = 0.0459kg

Force = 2380N

Time taken = 0.001s

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Speed of the ball afterwards = ?

Solution:

To solve this problem, we use Newton's second law of motion:

F = m x $\frac{v - u}{t}$

F is the force

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Answer:

The difference between the two is, well for one

Spectrum: The entire range that the "waves" could be such, as visible light, x-ray's and so on.

Waves: These are different because they aren't telling you or showing the entire spectrum just which they length that they are.

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Explanation:

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An object has the acceleration graph shown in (Figure 1). Its velocity at t=0s is vx=2.0m/s. Draw the object's velocity graph fo

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Answer:

Explanation:

We may notice that change in velocity can be obtained by calculating areas between acceleration lines and horizontal axis ("Time"). Mathematically, we know that:

$v_{b}-v_{a} = \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt$

$v_{b} = v_{a}+ \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt$

Where:

$v_{a}$ , $v_{b}$ - Initial and final velocities, measured in meters per second.

$t_{a}$ , $t_{b}$ - Initial and final times, measured in seconds.

$a(t)$ - Acceleration, measured in meters per square second.

Acceleration is the slope of velocity, as we know that each line is an horizontal one, then, velocity curves are lines with slopes different of zero. There are three region where velocities should be found:

Region I (t = 0 s to t = 4 s)

$v_{4} = 2\,\frac{m}{s} +\int\limits^{4\,s}_{0\,s} {\left(-2\,\frac{m}{s^{2}} \right)} \, dt$

$v_{4} = 2\,\frac{m}{s}+\left(-2\,\frac{m}{s^{2}} \right) \cdot (4\,s-0\,s)$

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Region II (t = 4 s to t = 6 s)

$v_{6} = -6\,\frac{m}{s} +\int\limits^{6\,s}_{4\,s} {\left(1\,\frac{m}{s^{2}} \right)} \, dt$

$v_{6} = -6\,\frac{m}{s}+\left(1\,\frac{m}{s^{2}} \right) \cdot (6\,s-4\,s)$

$v_{6} = -4\,\frac{m}{s}$

Region III (t = 6 s to t = 10 s)

$v_{10} = -4\,\frac{m}{s} +\int\limits^{10\,s}_{6\,s} {\left(2\,\frac{m}{s^{2}} \right)} \, dt$

$v_{10} = -4\,\frac{m}{s}+\left(2\,\frac{m}{s^{2}} \right) \cdot (10\,s-6\,s)$

$v_{10} = 4\,\frac{m}{s}$

Finally, we draw the object's velocity graph as follows. Graphic is attached below.

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