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4 years ago

Stop lamps must be visible within____ to the rear

Physics

2 answers:

FromTheMoon [43]4 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]4 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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GREYUIT [131]

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3 years ago

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3 years ago

Martha (50 kg) is attracted to Stewart (70 Kg) who sits 4 m away. What is the gravitational attraction between them? G=6.67 x 10

kodGreya [7K]

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1.46× 10^-8 N

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8 0

3 years ago

A ball falls off the desk and hits the floor. Which

Yuliya22 [10]

Answer:

The potential energy is transformed into kinetic energy

Explanation:

This particular case is defined as the principle of energy conservation since energy is not created or destroyed only transforms. When you have potential energy it can be transformed into kinetic energy or vice versa. In this problem, we have the case of a ball that sits on a desk and then falls to the ground. In this way the ground will be taken as a reference point, this is a point at which the potential energy will be equal to zero in such a way that when the ball is on the desktop that is above the reference line its potential energy will be maximum. As the ball drops its potential energy decreases, as the height relative to the ground (reference point) decreases. In contrast its kinetic energy increases and increases as it approaches the ground. So when it hits the ground it will have maximum kinetic energy and will be equal to the potential energy for when the ball was on the desk.

Therefore:

$E_{p} = potential energy [J] = E_{k} = kinetic energy [J]where:\\E_{p} =m*g*h\\m =mass [kg]\\g=gravity[m/s^2]\\h=elevation[m]\\E_{k} = \frac{1}{2} *m*v^{2} \\where:\\v=velocity [m/s]\\\frac{1}{2} *m*v^{2} = m*g*h$

8 0

3 years ago

In each case, lifting or pushing, why must you exert a force to move the object? Q1-2: How much more effort is required to lift

musickatia [10]

Answer:

Newton's first law applies. An object at rest will stay that way until a force is applied.
Any amount of effort can be applied to any amount of mass (in the ideal case). The question is not sufficiently specific.

Explanation:

A force is required to move an object because the object will stay at rest until a force is applied.

The effort required to lift or push two masses instead of one depends on the desired effect. For the same kinetic energy, no more effort is required. For the same momentum, half the effort is required for two masses. For the same velocity, double the effort is required.

4 0

3 years ago