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bezimeni [28]
3 years ago
10

The_method is the safest turning technique to use as it does not

Physics
2 answers:
masha68 [24]3 years ago
4 0

Answer: A.

Explanation:

Ilya [14]3 years ago
3 0

Answer:

The correct answer is A. The push/pull also known as hand-to-hand technique

Explanation:

A. <em>PUSH/ PULL STEERING (HAND-TO-HAND)</em>

This is the correct answer. The push/pull turn is also known as the hand-to-hand can be done such that your hands never cross over the steering wheel. Since the hands do not cross over the steering wheel, there is less chance of an injury to the face, hands or arms induced by your hands or arms in the event of a frontal crash due to an airbag. This technique has three methods;

The 10 and 2 position

The 9 and 3 position

The 8 and 4 position

Of the three methods, the 9 and 3 position is the preferred method of steering. Two and 10 o’clock is not recommended because it can be dangerous in vehicles with smaller steering wheels and equipped with airbags.

B. <em>SECURE TURN</em>

This option is wrong. It is not safe as regards to airbag deployment

C. <em>HAND-OVER-HAND STEERING</em>

This option is wrong. This method of steering is used when turning at low speeds with limited visibility at an intersection or when parking the vehicle or recovering from a skid. In hand-over-hand steering, the driver’s arms will pass across the airbag deployment zone. In the event of an airbag deployment, the driver arms will be thrown back into his face.

D. <em>THE THREE POINT TURN</em>

This option is wrong. It is a method you use when you want to execute a U-turn and you can only do it in limited place when the road is free of traffic. It should not be done near a main road, near a short curve, near an intersection or near the crest of a hill

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This question involves the concepts of the law of conservation of energy, kinetic energy, and potential energy.

The height of the hill is "166.76 m".

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According to the law of conservation of energy at the highest point of the roller coaster ride, that is, the hill, the whole (maximum) kinetic energy of the roller coaster is converted into its potential energy:

Maximum\ Kinetic\ Energy\ Lost = Maximum\ Potential\ Energy\ Gain\\\\&#10;\frac{1}{2}mv_{max}^2=mgh\\\\&#10;h=\frac{v_{max}^2}{2g}

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Therefore,

h=\frac{(57.2\ m/s)^2}{2(9.81\ m/s^2)}\\\\&#10;

<u>h = 166.76 m</u>

Learn more about the law of conservation of energy here:

brainly.com/question/101125

7 0
3 years ago
Consider that a clay model of a tiger has a mass of 0.195 kg and glides on ice at a speed of 0.75 m/s. It hits another clay mode
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Answer:

The final velocity after the collision is 0.27 m/s.    

Explanation:

Given that,

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Initially, second model is at rest, v' = 0

We need to find the final velocity after the collision. It is a case of inelastic collision. Using the conservation of linear momentum as :

mv+m'v'=(m+m')V\\\\V=\dfrac{mv+m'v'}{(m+m')}\\\\V=\dfrac{0.195\times 0.75}{(0.195 +0.335 )}\\\\V=0.27\ m/s

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Explanation:

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