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3 years ago

5

A top fuel drag racer reaches a maximum speed of 330 mph at the end of the 1/4-mile (402 m) racetrack. Part A - Assuming that th

e acceleration is constant during the race.
a. Calculate the average value of the acceleration of the top fuel car.
b. What is the ratio of that acceleration to the acceleration due to gravity?

1 answer:

Aloiza [94]3 years ago

4 0

Answer:

a. 27 m/s2

b. 2.75:1

Explanation:

a.

Since it is assumed that the race car starts from rest, initial velocity, u = 0 m/s

At the end of 402 m, distance, s = 402 m; and

Final velocity, v = 330 m/h

Converting this to m/s,

$v=\dfrac{330\times1609}{3600}=147.5$ m/s

Using the equation of motion

$v^2=u^2+2as$

$a=\dfrac{v^2-u^2}{2s}$

$a=\dfrac{147.5^2-0^2}{2\times402}=27$

b.

If acceleration due to gravity is taken as 9.8 m/s2, then

$a:g=27:9.8=2.75:1$

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Answer:

0.34 sec

Explanation:

Low point of spring ( length of stretched spring ) = 5.8 cm

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Determine the oscillation period

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Kx = Mg

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x = 2.9 * 10^-2 m

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3 years ago

A cylindrical resistor element on a circuit board dissipates 1.2 W of power. The resistor is 2 cm long, and has a diameter of 0.

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Answer:

(a) The resistor disspates 103680 joules during a 24-hour period.

(b) The heat flux of the resistor is approximately 4340.589 watts per square meter.

(c) The fraction of heat dissipated from the top and bottom surfaces is 0.045.

Explanation:

(a) The amount of heat dissipated ( $Q$ ), measured in joules, by the cylindrical resistor is the power multiplied by operation time ( $\Delta t$ ), measured in hours. That is:

$Q = \dot Q \cdot \Delta t$ (1)

If we know that $\dot Q = 1.2\,W$ and $\Delta t = 86400\,s$ , then the amount of heat dissipated by the resistor is:

$Q = (1.2\,W)\cdot (86400\,s)$

$Q = 103680\,J$

The resistor disspates 103680 joules during a 24-hour period.

(b) The heat flux ( $Q'$ ), measured in watts per square meter, is the heat transfer rate divided by the area of the cylinder ( $A$ ), measured in square meters:

$Q' = \frac{\dot Q}{A}$ (2)

$Q' = \frac{\dot Q}{\frac{\pi}{2}\cdot D^{2}+\pi\cdot D \cdot h }$ (3)

Where:

$D$ - Diameter, measured in meters.

$h$ - Length, measured in meters.

If we know that $\dot Q = 1.2\,W$ , $D = 4\times 10^{-3}\,m$ and $h = 2\times 10^{-2}\,m$ , the heat flux of the resistor is:

$Q' = \frac{1.2\,W}{\frac{\pi}{2}\cdot (4\times 10^{-3}\,m)^{2}+\pi\cdot (4\times 10^{-3}\,m)\cdot (2\times 10^{-2}\,m) }$

$Q' \approx 4340.589\,\frac{W}{m^{2}}$

The heat flux of the resistor is approximately 4340.589 watts per square meter.

(c) Since heat is uniformly transfered, then the fraction of heat dissipated from the top and bottom surfaces ( $r$ ), no unit, is the ratio of the top and bottom surfaces to total surface:

$r = \frac{\frac{\pi}{2}\cdot D^{2}}{A}$ (3)

If we know that $A \approx 2.765\times 10^{-4}\,m^{2}$ and $D = 4\times 10^{-3}\,m$ , then the fraction is:

$r = \frac{\frac{\pi}{2}\cdot (4\times 10^{-3}\,m)^{2} }{2.765\times 10^{-4}\,m^{2}}$

$r = 0.045$

The fraction of heat dissipated from the top and bottom surfaces is 0.045.

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