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grigory
1 month 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:
Yuliya22 [3.3K]1 month ago
3 0
Contact me for the complete response, if it’s not too late.
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A cliff diver on an alien planet dives off of a 32 meter tall cliff and lands in a sea of hydrochloric acid 1.20 seconds later.
inna [3103]

Answer:

44.4m/s^2

Explanation:

Utilize the equation...S = ut + 1/2at^2

where...S = 32m...u = 0m/s....t = 1.20s

32 = (0)(1.20) + 0.5(1.20^2)a

; The acceleration due to gravity is 44.4m/s^2

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4. Susan observed that different kinds and amounts of fossils were present in a cliff behind her house. She wondered why changes
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The expected measurements should range as follows: 5, 10, 15, 20, and 25 meters.
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A standard 14.16-inch (0.360-meter) computer monitor is 1024 pixels wide and 768 pixels tall. Each pixel is a square approximate
inna [3103]

Answer:

0.0031 m

Explanation:

y = Length of pixel = 281 μm

L = Distance to screen = 1.3 m

\lambda = Wavelength = 550 nm

d = Pupil diameter

\theta = Angle

We have the expression

tan\theta=\dfrac{y}{L}\\\Rightarrow \theta=tan^{-1}\dfrac{y}{L}\\\Rightarrow \theta=tan^{-1}\dfrac{281\times 10^{-6}}{1.3}

We have the expression

sin\theta=1.22\dfrac{\lambda}{d}\\\Rightarrow d=\dfrac{1.22\lambda}{sin\theta}\\\Rightarrow d=\dfrac{1.22\times 550\times 10^{-9}}{sin\left(tan^{-1}\frac{281\times 10^{-6}}{1.3}\right)}\\\Rightarrow d=0.0031\ m

The pupil diameter calculates to 0.0031 m

4 0
1 month ago
Which of the following four circuit diagrams best represents the experiment described in this problem?
inna [3103]

There's an absence of circuit diagrams.  

Initially, this causes worry for a moment, until we remember that we have no understanding of the experiment mentioned in the problem either, rendering such worries unnecessary.

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1 month ago
A solid sphere is released from the top of a ramp that is at a height h1 = 2.30 m. It rolls down the ramp without slipping. The
Yuliya22 [3333]

Answer:

The horizontal distance d that the ball covers before it lands is 1.72 m.

Explanation:

Given,

Height of ramp h_{1}=2.30\ m

Height of bottom of ramp h_{2}=1.69\ m

Diameter = 0.17 m

We need to determine the horizontal distance d the ball travels before landing.

We need to calculate the time

Using the equation of motion

h_{2}=ut+\dfrac{1}{2}gt^2

t=\sqrt{\dfrac{2h_{2}}{g}}

t=\sqrt{\dfrac{2\times1.69}{9.8}}

t=0.587\ sec

Next, we can find the ball's velocity

Using the kinetic energy formula

K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2

K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}\times(\dfrac{2}{5}mr^2)\times(\dfrac{v}{r})^2

K.E=\dfrac{7}{10}mv^2

By applying the conservation of energy

K.E=mg(h_{1}-h_{2})

\dfrac{7}{10}mv^2=mg(h_{1}-h_{2})

v^2=\dfrac{10}{7}\times g(h_{1}-h_{2})

We substitute the values into the equation

v=\sqrt{\dfrac{10\times9.8\times(2.30-1.69)}{7}}

v=2.922\ m/s

Next, we determine the horizontal distance d the ball travels before landing

Using the distance formula

d =vt

Where. d = distance

t = time

v = velocity

We substitute the values into the formula

d=2.922\times 0.587

d=1.72\ m

Thus, the horizontal distance d that the ball travels before it lands is 1.72 m.

8 0
1 month ago
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