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deff fn [24]
3 years ago
6

Automobiles must be able to sustain a frontal impacl The automobile design must allow low speed impacts with little sustained da

mage, while allowing the vehicle front end structure to deform and absorb impact energy at higher speeds. Consider a frontal impact test of a 1000 kg mass vehicle. (a) For a low speed test at 2.5 rnls, compute the energy in the vehicle just prior to impacl If the bumper is a pure elastic element, what is the effective design stiffness required to limit the bumper maximum deflection during impact to 4 em
Physics
1 answer:
valentinak56 [21]3 years ago
3 0

Answer: the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm is 3906250 N/m

Explanation:

Given that;

mass of vehicle m = 1000 kg

for a low speed test; V = 2.5 m/s

bumper maximum deflection = 4 cm = 0.04 m

First we determine the energy of the vehicle just prior to impact;

W_v = 1/2mv²

we substitute

W_v = 1/2 × 1000 × (2.5)²

W_v = 3125 J

now, the the effective design stiffness k will be:

at the impact point, energy of the vehicle converts to elastic potential energy of the bumper;

hence;

W_v = 1/2kx²

we substitute

3125 = 1/2 × k (0.04)²

3125 = 0.0008k

k = 3125 / 0.0008

k = 3906250 N/m

Therefore, the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm is 3906250 N/m

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Answer:

a)   298.5 nm ,  522.4 nm  and b)  radiation frequency does not change

Explanation:

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On the other hand, when the light reaches another medium its average speed within the medium changes, it is now less than the speed of light in a vacuum (c) for this to happen as we saw that the frequency is constant there must be a change in the wavelength of the radiation that is characterized by the ratio

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   λₙ = 700 / 1.34

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4 0
3 years ago
A 5.0 kg block hangs from the ceiling by a mass-less rope. A Second block with a mass of 10.0 kg is attached to the first block
gayaneshka [121]

The tension in the first and second rope are; 147 Newton and 98 Newton respectively.

Given the data in the question

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To find the Tension in each of the ropes, we make use of the equation from Newton's Second Laws of Motion:

F = m\ *\ a

Where F is the force, m is the mass of the object and a is the acceleration ( In this case the block is under gravity. Hence ''a" becomes acceleration due to gravity  g = 9.8m/s^2 )

For the First Rope

Total mass hanging on it; m_T = m_1 + m_2 = 5.0kg + 10.0kg = 15.0kg

So Tension of the rope;

F = m\ * \ g\\\\F = 15.0kg \ * 9.8m/s^2\\\\F = 147 kg.m/s^2\\\\F = 147N

Therefore, the tension in the first rope is 147 Newton

For the Second Rope

Since only the block of mass 10kg is hang from the second, the tension in the second rope will be;

F = m\ * \ g\\\\F = 10.0kg \ * 9.8m/s^2\\\\F = 98 kg.m/s^2\\\\F = 98N

Therefore, the tension in the second rope is 98 Newton

Learn More, brainly.com/question/18288215

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2 years ago
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Answer:

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