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cluponka [151]
3 years ago
6

A turn signal lamp must be visible in normal sunlight at a distance of at least ______ feet from the front and rear of the vehic

le if the vehicle is at least 80 inches wide, and at least _____ feet from front and rear of the vehicle if the vehicle is less than 80 inches wide.
Physics
1 answer:
kaheart [24]3 years ago
8 0

Answer:

500; 300 feet.

Explanation:

A turn signal lamp can be defined as an amber blinking lamp which indicates a driver's intent to change direction. It is extremely important that drivers gives a turn signal (flashing light) on the side toward which he or she is turning either left or right.

Simply stated, the turn signal lamp indicate the driver's intent to turn either leftward or rightward by displaying flashing lights to the front and rear of his or her vehicle.

A turn signal lamp must be visible in normal sunlight at a distance of at least 500 feet from the front and rear of the vehicle if the vehicle is at least 80 inches wide, and at least 300 feet from front and rear of the vehicle if the vehicle is less than 80 inches wide according to the transportation traffics code.

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A rock is thrown horizontally from a high building at 33.8 m/s. What is the magnitude of its velocity 4.25 s later?
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<h2>Answer:53.63ms^{-2}</h2>

Explanation:

The equations of motion used in this question is v=u+at

When a object is projected horizontally from a sufficiently height,the x-component of acceleration remains zero because there is no force that drags the object in x direction.

But,due to gravity,the object accelerates downward at a rate of 9.8ms^{-2}.

In X-Direction,

Given that initial velocity=u_{x}=33.8ms^{-1}

Using v=u+at,

v_{x}=33.8+(0)4.25=33.8ms^{-1}

In Y-Direction,

Given that initial velocity=u_{x}=0ms^{-1}

Using v=u+at,

v_{y}=0+(9.8)4.25=41.65ms^{-1}

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3 years ago
This problem follows up on a discussion from lecture. A wind turbine with an efficiency of 45% for converting wind energy into e
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Answer:

4.1 m

Explanation:

10 kW = 10000 W

20mi/h = 20*1.6 km/mi = 32 km/h = 32 * 1000 (m/km) *(1/3600) hr/s = 8.89 m/s

The power yielded by the wind turbine can be calculated using the following formula

P = \frac{1}{2} \rho v^3 A C_p

where \rho = 1.2 kg/m^3 is the air density, v = 8.89 m/s is the wind speed, A is the swept area and C_p = 0.45 is the efficiency

10000 = 0.5 * 1.2 * 8.89^3 * A * 0.45

10000 = 190A

A = 10000 / 190 = 52.7 m^2

The swept area is a circle with radius r being the blade length

\pi r^2 = A = 52.7

r^2 = 52.7 / \pi = 16.79

r = \sqrt{16.79} = 4.1 m

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