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3 years ago

When you walk across the ground and push on it with your feet...

Physics

2 answers:

DerKrebs [107]3 years ago

4 0

Answer:

The ground pushes back on your feet with equal force

Explanation:

When you walk across the ground and push on it with your feet, the ground pushes back on your feet with an equal and opposite force.

This interpretation and knowledge is gotten from Newton's third law of motion.

It states that "action and reaction force are equal and opposite in nature".

The force applied to a body responds with an opposite force in the other direction.
Therefore, the reaction force is of equal magnitude but directed in another direction.

stich3 [128]3 years ago

3 0

Answer:

the answer is the second one: the ground pushes back on your feet with equal force

Explanation:

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Answer:

0.046

Explanation:

displacement = velocity/ time

d = 6m/s / 130s

d = 0.046m

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3 years ago

While unrealistic, we will examine the forces on a leg when one falls from a height by approximating the leg as a uniform cylind

Leno4ka [110]

Answer:

Part A: 7.75 m/s

Part B: 2330.8 kN

Part C: 24.03 kN

Part D: 4.8 kN

Part E: $1.7\times 10^{9} Dyn/cm^{2}$

Part F: Option D

Bending his legs increases the time over which the ground applies force, thus decreasing the force applied by the ground.

Explanation:

Part A

From the fundamental kinematic equation

$v^{2}=u^{2}+2gh$ where v is the velocity of the man just before hitting the ground, g is acceleration due to gravity, u is initial velocity, h is the height.

Since the initial velocity is zero hence

$v^{2}=2gh$

$v=\sqrt 2gh$

Substituting 10 m/s2 for g and 3 m for h we obtain

$v=\sqrt 2\times 10\times 3 =\sqrt 60= 7.745967\approx 7.75 m/s$

Part B 

Force exerted by the leg is given by

F=PA where P is pressure, F is force, A is the cross-section of the bone

$A=\frac {\pi d^{2}}{4}$

Substituting 2.3 cm which is equivalent to 0.023m for d and $1.7\times10^{8} N/m2$ for P we obtain the force as

$F=PA=1.7\times10^{8}*\frac {\pi (0.023)^{2}}{4}= 2330818.276\approx 2330.8 kN$

Part C

The fundamental kinematic equation is part (a) can also be written as

$v^{2}=u^{2}+2a\triangle x$ and making a the subject then

$a=\frac {v^{2}-u^{2}}{2\triangle x}$ where a is acceleration and $\triangle x$ is the change in length

Substituting the value obtained in part a, 7.75 m/s for v, u is zero and 1cm which is equivalent to 0.01 m for $\triangle x$ then

$a=\frac {7.75^{2}-0^{2}}{2\times 0.01}= 3003.125 m/s^{2}$

Force exerted on the man is given by

$F=ma=80\times 3003.125= 240250 N\approx 24.03 kN$

Part D

The fundamental kinematic equation is part (a) can also be written as

$v^{2}=u^{2}+2a\triangle h$ and making a the subject then

$a=\frac {v^{2}-u^{2}}{2\triangle h}$ where a is acceleration and $\triangle h$ is the change in height

Also, force exerted on the man is given by $F=ma=m\times \frac {v^{2}-u^{2}}{2\triangle h}$

Substituting 80 Kg for m, 50 cm which is equivalent to 0.5m for \triangle h and other values as used in part c

$F=ma=m\times \frac {v^{2}-u^{2}}{2\triangle h}=80\times \frac {7.75^{2}-0^{2}}{2\times 0.5}= 4805 N\approx 4.8 kN$

Part E

$P=1.7\times 10^{8}=1.7\times 10^{8}\times (\frac {10^{5} Dyn}{10^{4} cm^{2}}=1.7\times 10^{9} Dyn/cm^{2}$

Part F

Bending his legs increases the time over which the ground applies force, thus decreasing the force applied by the ground

7 0

3 years ago

The period

Cloud [144]

Answer: p = - {1/2} , q = {1/2}

Explanation: The frequency of oscillation of a pendulum is given as

F = 1/2π *√{l/g}

Where √ is square root

l is lenght

g is acceleration due to gravity

But

F = 1/T

Where T is the period of Oscillation

Making T subject of formula we have

T= 1/F

T = 2π√{g/l}

Here the power on l is -[1/2]= p

Also,

Power on g is 1/2 =q

All because of the square root.

5 0

3 years ago