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3 years ago

Stop lamps must be visible within____ to the rear

Physics

2 answers:

FromTheMoon [43]3 years ago

5 0

Answer:300ft to the rear

Explanation:

According to the legislative information code conduct of article 3 vehicle code on rear lighting which states that all

stoplamps shall be plainly visible and understandable from a distance of 300 feet from the rear of the vehicle both during normal sunlight and at nighttime, except that stoplamps on a vehicle of a size required to be equipped with clearance lamps shall be visible from a distance of 500 feet from the rear of the vehicle during those times.

Charra [1.4K]3 years ago

3 0

The answer is 300 feet. The stop lamp or lamps on the rear of a vehicle must show a red light that is set in motion upon application of the service or foot brake and, in a vehicle manufactured or assembled on or after January 1, 1964, must be visible from a distance of not less than 300 feet to the rear in normal sunlight. Take note, if the vehicle is manufactured or assembled January 1, 1964, the stop lamp or lamps must be visible from a distance of not less than 100 feet. Also, the stop lamp may be combined with one or more other rear lamps.

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A car (mass 1200 kg, speed 100 km/h) and a truck (mass 2800 kg, speed 50 km/h) are moving in the same direction along a highway.

Sloan [31]

Answer:

Speed of the wreck after the collision is 65 km/h

Explanation:

When a car hits truck and sticks together, the collision would be totally inelastic. Since the both the vehicles locked together, they have the same final velocity.

Mass of car $m_{1}=1200 kg$

Mass of truck $m_{2}=2800 kg$

Initial speed of the car $u_{1}=100 km /h$

Initial speed of the truck $u_{2}=50 km /h$

The final velocity of the wreck will be

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

since final speed are same, $v_{1}=v_{2}=v$

$m_{1}u_{1}+m_{2}u_{2}=(m_{1}+m_{2})v$

$1200\times 100 +2800 \times 50 =(1200+2800)v\\v=\frac{260000}{4000} \\v=65 km/h$

5 0

3 years ago

Consider a cloudless day on which the sun shines down across the United States. If 2073 kJ of energy reaches a square meter ( m

Mnenie [13.5K]

The total amount of energy per hour is $2.039\cdot 10^{16} kJ$

Explanation:

In this problem we are told that the amount of energy reaching a square meter in the United States per hour is

$E_1 = 2073 kJ$

The total surface area of the United States is

$A=9.834\cdot 10^6 km^2$

And converting into squared metres,

$A=9.834\cdot 10^6 \cdot 10^6 = 9.834\cdot 10^{12} m^2$

Therefore, the total energy reaching the entire United States per hour is given by:

$E=AE_1 = (9.834\cdot 10^{12})(2073)=2.039\cdot 10^{16} kJ$

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5 0

3 years ago

I AM........ INEVITABLE

Archy [21]

Answer:

that's nice very nice super duper nicer

5 0

3 years ago

Please help!!!!!!!! 25 points

Archy [21]

Answer:

B: beaks, finches.

Explanation:

Beaks, finches. This is because he noticed that fruit-eating finches had beaks similar to parrots, while finches that ate insects had narrow beaks. The difference in their beaks is due to adapting to different environments which causes them to evolve into different species of finches.

7 0

3 years ago

Read 2 more answers

he block is released, and it slides 2.0 m (from the point at which it is released) across a horizontal surface before friction s

alex41 [277]

Answer:

0.245

Explanation:

When the block is released, the initial elastic potential energy stored in the spring is entirely converted into kinetic energy of the block.

Therefore, we can calculate the initial speed of the block:

$\frac{1}{2}kx^2 = \frac{1}{2}mv^2$

where the term on the left is the potential energy and where the term on the right is the kinetic energy, and where

k = 4500 N/m is the spring constant

x = 8.0 cm = 0.08 m is the compression of the spring

m = 3.0 kg is the mass of the block

v is the initial velocity

Solving for v,

$v=\sqrt{\frac{kx^2}{m}}=\sqrt{\frac{(4500)(0.08)^2}{3.0}}=3.1 m/s$

Then, after the block is released, all its kinetic energy is converted into thermal energy as the block slows down, due to friction. Therefore, the work done by friction is equal to the initial kinetic energy of the block.

The force of friction is

$F=\mu mg$