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3 years ago

A vehicle should never be parked closer to a fire hydrant than:

2 answers:

8 0

A vehicle should never be parked closer to a fire hydrant than: 15 feet
Fire hydrants are reserved by the government as the source of water for the fire department to put out the fire during a burning accident.
If you put your vehicle too close to the hydrant, it could potentially block the process during an emergency situation and could cost us humans' life.

Thepotemich [5.8K]3 years ago

5 0

A vehicle should not be parked closer to fire hydrant more than 15 feet because it may hinder the fire department’s vehicle to use that water in case of emergency. 

Further Explanation:

The fire hydrants are made at several public points all across the city. These fire hydrants are used to prevent the fire in the public places by sudden supply of the water from the nearest hydrant point.

If any vehicle is parked closer to the fire hydrant, it would hinder the fire department vehicle to reach the fire hydrant point and fill water from it.

Therefore, in order to keep the path for the fire department vehicle in case of emergency, the government has made a rule that no vehicle would be parked in the range of 15 feet from the fire hydrant so that the access to the fire hydrant is easy and unblocked in case of any small or big emergency.

Thus, a vehicle should not be parked closer to fire hydrant more than 15 feet because it may hinder the fire department’s vehicle to use that water in case of emergency.

Learn More:

1. Forces of attraction limit the motion of particles most in brainly.com/question/947434

2. The amount of kinetic energy an object has depends on its brainly.com/question/137098

3. Which of the following are units for expressing rotational velocity, commonly denoted by brainly.com/question/2887706

Answer Details:

Grade: High School

Subject: Physics

Chapter: Basic Science

Keywords: Fire hydrant, vehicle, never be parked, 15 feet, fire department, vehicle, fire emergency, parking, unblocked, emergency situation.

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The Electric field is zero at a distance 2.492 cm from the origin.

Let A be point where the charge $15\times10^-6$ C is placed which is the origin.

Let B be the point where the charge $9\times 10^-6$ C is placed. Given that B is at a distance 1 cm from the origin.

Both the charges are positive. But charge at origin is greater than that of B. So we can conclude that the point on the x-axis where the electric field = 0 is after B on x - axis.

i.e., at distance 'x' from B.

Using Coulomb's law, $\frac{kQ_A^2}{d_A^2} = \frac{kQ_B^2}{d_B^2}$ ,

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$d_B=x cm$

k is the Coulomb's law constant.

On substituting the values into the above equation, we get,

$\frac{(15\times10^-6)^2}{(1+x)^2} =\frac{(9\times10^-6)^2}{x^2}$

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1 year ago

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magnitude: 21.6; direction: 33.7 degrees

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