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bekas [8.4K]
3 years ago
13

The landing gear of an airplane can be idealized as the spring-mass-damper system shown in fig. 3.52. if the runway surface is d

escribed determine the values of k and c that limit the amplitude of vibration of the airplane (x) to 0.1 m. assume
Physics
1 answer:
Serhud [2]3 years ago
5 0
The land of airplane gear of an airplane can be idealized as the spring-mass-damper system shown in fig. 3.52. if the runway surface is described
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During the contraction of the heart, 65 cm3 blood is ejected from the left ventricle into the aorta with a velocity of approxima
vova2212 [387]

Answer:

The value  F  =  0.1396 \ N

Explanation:

From the question we are told that

   The volume blood  ejected is  V  =  65 \ cm^3  = 65*10^{-6} \  m^3

    The velocity of the blood ejected is  v  = 103 \  cm/s  = \frac{103}{100} = 1.03 \ m/s

    The density of blood is  \rho = 1060 \  kg/m^3

     The heart beat is R = 59 \  bpm(beats \  per \  minute) = \frac{59}{60}= 0.9833\ bps

The average force exerted by the blood on the wall of the aorta is mathematically represented as

      F  =  2 * \rho  *  V  *  R *  v

=>    F  =  2 * 1060  *  65*10^{-6}  *  0.9833 *  1.03

=>    F  =  0.1396 \ N

8 0
3 years ago
Human sense impressions are subjective and qualitative
Fittoniya [83]

Answer:trueee

Explanation:

7 0
3 years ago
What happens to the atomic number of an atom when the number of neutrons in the nucleus of that atom increases? a It decreases b
kupik [55]

Answer:

It remains the same

Explanation:

It remains the same. This is because the number of protons doesn't change and the number of protons determines the atomic number.

8 0
2 years ago
If a book has a mass of 3 kg, what is the book's weight in N?
sesenic [268]

Answer:

29.4 N

Explanation:

F = ma

F = (3 kg) (9.8 m/s²)

F = 29.4 N

5 0
3 years ago
A student standing on a stationary skateboard tosses a textbook with a mass of mb = 1.35 kg to a friend standing in front of him
nataly862011 [7]

Answer:

a) u_c=0\ m.s^{-1}       &        m_c.v_c=m_b.v_b\times \cos\theta

b) v_c=0.0566\ m.s^{-1}

c) p_e=2.9218\ kg.m.s^{-1}

Explanation:

Given:

mass of the book, m_b=1.35\ kg

combined mass of the student and the skateboard, m_c=97\ kg

initial velocity of the book, v_b=4.61\ m.s^{-1}

angle of projection of the book from the horizontal, \theta=28^{\circ}

a)

velocity of the student before throwing the book:

Since the student is initially at rest and no net force acts on the student so it remains in rest according to the Newton's first law of motion.

u_c=0\ m.s^{-1}

where:

u_c= initial velocity of the student

velocity of the student after throwing the book:

Since the student applies a force on the book while throwing it and the student standing on the skate will an elastic collision like situation on throwing the book.

m_c.v_c=m_b.v_b\times \cos\theta

where:

v_c= final velcotiy of the student after throwing the book

b)

m_c.v_c=m_b.v_b\times \cos\theta

97\times v_c=1.35\times 4.61\cos28

v_c=0.0566\ m.s^{-1}

c)

Since there is no movement of the student in the vertical direction, so the total momentum transfer to the earth will be equal to the momentum of the book in vertical direction.

p_e=m_b.v_b\sin\theta

p_e=1.35\times 4.61\times \sin28^{\circ}

p_e=2.9218\ kg.m.s^{-1}

6 0
3 years ago
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