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DENIUS [597]
3 years ago
5

The maximum energy a bone can absorb without breaking is surprisingly small. Experimental data show that both leg bones, togethe

r, of a healthy human adult can absorb about 200 J before breaking. From what maximum height could a 75 kg person jump and land rigidly upright on both feet without breaking their legs
Physics
1 answer:
wolverine [178]3 years ago
5 0

Answer:

0.21m

Explanation:

Note that the 200 J before breaking legs is Kinetic energy

So if potential energy= kinetic energy

Then 200J = mgh

h= 200/75x 9.8

= 0.21m so the maximum height will be 0.21m

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vladimir1956 [14]
I think through convection I forget. Plus its enclosed (and, well closed)
4 0
3 years ago
Billy picks up a 40 lb. dumbbell (mass = 18.14 kg). The center of his hand, where the dumbbell is held, is 56 cm (0.56 m) from t
umka21 [38]

Answer:

<h2>Force due to biceps is given as</h2><h2>F = 1991.05 N</h2>

Explanation:

For balancing the force we know that

Torque due to weight hold on his hand = torque due to force applied by biceps

So here we will have

mg \times L = F \times d

so we have

18.14 \times 9.8 \times 0.56 = F \times (0.05)

F = 1991.05 N

8 0
3 years ago
The center of gravity of a loaded truck depends on how the truck is packed. If it is 4.0 m high and 2.4 m wide, and its CG is 2.
Vitek1552 [10]

The slope of the road can be given as the ratio of the change in vertical

distance per unit change in horizontal distance.

  • The maximum steepness of the slope where the truck can be parked without tipping over is approximately <u>54.55 %</u>.

Reasons:

Width of the truck = 2.4 meters

Height of the truck = 4.0 meters

Height of the center of gravity = 2.2 meters

Required:

The allowable steepness of the slope the truck can be parked without tipping over.

Solution:

Let, <em>C</em> represent the Center of Gravity, CG

At the tipping point, the angle of elevation of the slope = θ

Where;

tan\left(\theta \right) = \dfrac{\overline{AM}}{\overline{CM}}

The steepness of the slope is therefore;

\mathrm{The \ steepness \  of  \ the  \ slope}= \dfrac{\overline{AM}}{\overline{CM}} \times 100

Where;

\overline{AM} = Half the width of the truck = \dfrac{2.4 \, m}{2} = 1.2 m

\overline{CM} = The elevation of the center of gravity above the ground = 2.2 m

\mathrm{The \ steepness \  of  \ the  \ slope}= \dfrac{1.2}{2.2} \times 100 \approx 54.55\%

tan\left(\theta \right) = \mathbf{\dfrac{2.2}{1.2}} = \dfrac{11}{6}

Elevation \ of \ the \ road \ \theta = arctan\left( \dfrac{6}{11} \right)  \approx 28.6 ^{\circ}

The maximum steepness of the slope where the truck can be parked is <u>54.55 %</u>.

Learn more here:

brainly.com/question/20793607

3 0
2 years ago
NEED THIS DONE ASAP HELP
ExtremeBDS [4]

Answer: Exercise Physiology

Explanation:

4 0
3 years ago
A battery supplies 0.75 A to three resistors connected in parallel. The current through the first resistor is 0.24 A and the cur
Anastasy [175]
Total current flow in circuit=0.75
Current flow through third resistor=0.75-(0.24+0.22)
=0.29
7 0
3 years ago
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