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viktelen [127]
3 years ago
10

So, why can a properly executed karate kick break a concrete block without fracturing bones [16]? first, bone is a very strong m

aterial. its ultimate compressive strength is approximately 40 times larger than concrete. second, contact is made with the edge of the foot. this concentrates the force into a small area of the target and reduces the likelihood of bending a bone to the point of fracture. third, the collision with the target is essentially inelastic and extends over several milliseconds, so the peak force, though large, does not exceed the strength of the bone. [8] george
b. benedek and felix m. h. villars, physics with illustrative examples from medicine and biology, vol. 1. (menlo park: addison-wesley publishing co., 1974). [16] s. r. wilk, r.
e. mcnair, and m. s. feld, am. j. phys. 51, 783 (1983). if a (cross-sectional area of the tibia) ~ 2.5 cm2


c


m


2


, compute how far a 67 kg person can fall and land stiff-legged on both legs without breaking a bone. assume f is split evenly between two legs. (hint: f≤2σ a


f


≤


2


σ


 


a


)
Physics
1 answer:
Sav [38]3 years ago
3 0
PM me for full answer, please. If it's not too late.
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A small object begins a free-fall from a height of =81.5 m at 0=0 s . After τ=2.20 s , a second small object is launched vertica
-BARSIC- [3]

Answer:

33.2 m

Explanation:

For the first object:

y₀ = 81.5 m

v₀ = 0 m/s

a = -9.8 m/s²

t₀ = 0 s

y = y₀ + v₀ t + ½ at²

y = 81.5 − 4.9t²

For the second object:

y₀ = 0 m

v₀ = 40.0 m/s

a = -9.8 m/s²

t₀ = 2.20 s

y = y₀ + v₀ t + ½ at²

y = 40(t−2.2) − 4.9(t−2.2)²

When they meet:

81.5 − 4.9t² = 40(t−2.2) − 4.9(t−2.2)²

81.5 − 4.9t² = 40t − 88 − 4.9 (t² − 4.4t + 4.84)

81.5 − 4.9t² = 40t − 88 − 4.9t² + 21.56t − 23.716

81.5 = 61.56t − 111.716

193.216 = 61.56t

t = 3.139

The position at that time is:

y = 81.5 − 4.9(3.139)²

y = 33.2

7 0
3 years ago
A 2.00 kg box slides on a rough, horizontal surface, hits a spring with a speed of 1.90 m/s and compresses it a distance of 10.0
oksian1 [2.3K]

Answer:

Explanation:

Given

mass of box m=2\ kg

speed of box v=1.9\ m/s

distance moved by the box x=10\ cm

coefficient of kinetic friction \mu _k=0.66

Friction  force f_r=\mu_kN

f_r=0.66\times mg

f_r=0.66\times 2\times 9.8=12.936 \N

Kinetic Energy of box will be utilize to overcome friction and rest is stored in spring in the form of elastic potential energy

\frac{1}{2}mv^2=f_r\cdot x+\frac{1}{2}kx^2

\frac{1}{2}\times 2\times 1.9^2=12.936\times 0.1+\frac{1}{2}\times k\times (0.1)^2

3.61-1.2936=0.005\times k

k=463.28\ N/m

3 0
3 years ago
Read 2 more answers
A 0.10-kg ball, traveling horizontally at 25 m/s, strikes a wall and rebounds at 19 m/s. What is the 7) magnitude of the change
IRISSAK [1]

Answer:

Change in momentum will be -4.4 kgm/sec

So option (A) is correct option

Explanation:

Mass of the ball is given m = 0.10 kg

Initial velocity of ball v_1=25m/sec

And velocity after rebound v_2=-19m/sec

We have to find the change in momentum

So change in momentum is equal to =m(v_2-v_1)=0.1\times (-19-25)=-4.4kgm/sec ( here negative sign shows only direction )

So option (A) will be correct answer

5 0
3 years ago
Imagine standing on a bathroom scale and reading your weight. now lift one foot and read your weight again. does the scale read
7nadin3 [17]
Less because now there is less force on the scale and you are unbalanced. Therefore the scale will have a lower number then before.

8 0
3 years ago
Read 2 more answers
A scientist classifies some plants into two groups
gizmo_the_mogwai [7]

Answer:

I'm pretty sure the answer is C . since deciduous trees are trees that loose leaves seasonally and coniferous trees are trees that don't loose leaves seaosnally and survive through the winter.

8 0
3 years ago
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