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AfilCa [17]
4 years ago
6

A large truck and a small car have a head-on collision. which scenario is true?the truck exerts a greater force on the car than

the car exerts on the truck.the car exerts a greater force on the truck than the truck exerts on the car.the force that the truck exerts on the car and the force that the car exerts on the truck are the same magnitude.
Physics
2 answers:
Marianna [84]4 years ago
8 0

Answer:

the force that the truck exerts on the car and the force that the car exerts on the truck are the same magnitude.

Explanation:

As we know that when truck and car collide with each other then due to this collision the two will exert force on each other

This force will follow Newton's III law

This is law of Action - Reaction as per which force of action must be equal and opposite to the law of reaction.

So we have

\vec F_a + \vec F_b = 0

\vec F_a = -\vec F_b

so two forces are equal in magnitude but opposite in direction

so correct answer is

the force that the truck exerts on the car and the force that the car exerts on the truck are the same magnitude.

zalisa [80]4 years ago
5 0
A. The truck exerts a greater force on the car than the car exerts on the truck. 
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This question is incomplete, the complete question is;

An engineer is tasked with developing a model to study a cylindrical heat exchanger in a steam system. The prototype cylinder is 2.54 cm in diameter and the steam properties are: velocity = 30 m/s; density = 0.6 kg/m³; and absolute viscosity = 1.197 X 10⁻⁵ N-s/m², respectively. The model is going to be tested in a water tunnel where the water temperature is 20°C and the velocity is 3 m/s

a) what is the prototype Reynolds number, based on using the cylinder diameter as the characteristic length?

b) what should be the diameter of the model be to ensure dynamic similitude?

Answer:

a) the prototype Reynolds number is 38195.488

b) 0.01278 m or 1.278 cm should be the diameter of the model be to ensure dynamic similitude

Explanation:

Given that;

1 prototype ; d = 2.54 cm = 0.0254 m, Vp = 30 m/s, Sp = 0.6 kg/m³, Up = 1.197 × 10⁻⁵ N-s/m²

Model{ water at 20°C}; dm = ?, Vm = 3 m/s, Pm = 998.23 kg/m³, Um = 1.002 × 10⁻³ N-s/m²

a) what is the prototype Reynolds number,

to calculate prototype Reynolds number we use the expression;

(Re)p = SpVpdP / Up

we substitute our value

(Re)p =  (0.6 × 30 × 0.0254) / 1.197 × 10⁻⁵

(Re)p = 38195.488

Therefore the prototype Reynolds number is 38195.488

b)

what should be the diameter of the model be to ensure dynamic similitude?

i.e dm = ?

so dynamic similarity [ viscous flow]

(Re)m = (Re)p

[PmVmdm / Um] = [SpVpdP / Up]

we substitute

[998.23 × 3 × dm / 1.002 × 10⁻³] = (0.6 × 30 × 0.0254) / 1.197 × 10⁻⁵

2994.69dm / 1.002 × 10⁻³ = 38195.488

2994.69dm = 38.2718

dm = 38.2718 / 2994.69

dm = 0.01278 m or 1.278 cm

Therefore 0.01278 m or 1.278 cm should be the diameter of the model be to ensure dynamic similitude

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3 years ago
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Answer:

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