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Natalka [10]
3 years ago
9

Sprinters push off from the ball of their foot, then bend their knee to bring their foot up close to the body as they swing thei

r leg forward for the next stride. Why is this an effective strategy for running fast?
Physics
1 answer:
Nataliya [291]3 years ago
7 0

There are two interchangeable concepts that can be applied to this theory. The first one is the moment of Inertia and the second the perioricity. When the runners put their leg forward they reduce their moment of inertia allowing their center of mass to maintain the speed they carry and not waste energy. In this way when they swing, the leg begins to act as a kind of pendulum, and remember that the pendulum is directly proportional to the length, that is,

T = 2 \pi \sqrt {l / g}

In this way, by bringing their legs closer they reduce the distance and make the period shorter.

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Batteries can supply a steady flow of electrons.<br> True<br> False
Alecsey [184]

The answer is TRUE, batteries CAN supply a steady flow of electrons.

6 0
2 years ago
A rod has a radius of 10 mm is subjected to an axial load of 15 N such that the axial strain in the rod is ????௫ = 2.75*10-6, de
EleoNora [17]

Answer:

Knowing we only have one load applied in just one direction we have to use the Hooke's law for one dimension

ex = бx/E

бx = Fx/A = Fx/πr^{2}

Using both equation and solving for the modulus of elasticity E

E = бx/ex = Fx / πr^{2}ex

E = \frac{15}{pi (10 * 10^{-3})^{2} * 2.75 * 10^{-6}    } = 17.368 * 10^{9} Pa = 17.4 GPa

Apply the Hooke's law for either y or z direction (circle will change in every direction) we can find the change in radius

ey = \frac{1}{E} (бy - v (бx + бz)) = -\frac{v}{E}бx

= \frac{vFx}{Epir^{2} } = \frac{0.23 * 15}{pi (10 * 10^{-3)^{2} } * 17.362 * 10^{9}  } = -0.63 *10^{-6}

Finally

ey = Δr / r

Δr = ey * r = 10 * -0.63* 10^{-6} mm = -6.3 * 10^{-6} mm

Δd = 2Δr = -12.6 * 10^{-6} mm

Explanation:

5 0
3 years ago
A projectile is launched into the air with an initial speed of 40 m/s and a launch angle of 20° above the horizontal. The projec
miv72 [106K]

Explanation:

V=40m/s

Vy=V.sina=40.sin20=40 . 0.342=13.68m/s

Vx=V.cosa=40.cos20=40 . 0.766=30.64m/s

Projectile travels during 5 seconds and the ramge becomes:

x=V.t=30.64 . 5=153.2m

7 0
2 years ago
A piston of volume 0.1 m3 contains two moles of a monatomic ideal gas at 300K. If it undergoes an isothermal process and expands
seropon [69]

Answer:

the work is done by the gas on the environment -is W= - 3534.94 J (since the initial pressure is lower than the atmospheric pressure , it needs external work to expand)

Explanation:

assuming ideal gas behaviour of the gas , the equation for ideal gas is

P*V=n*R*T

where

P = absolute pressure

V= volume

T= absolute temperature

n= number of moles of gas

R= ideal gas constant = 8.314 J/mol K

P=n*R*T/V

the work that is done by the gas is calculated through

W=∫pdV=  ∫ (n*R*T/V) dV

for an isothermal process T=constant and since the piston is closed vessel also n=constant during the process then denoting 1 and 2 for initial and final state respectively:

W=∫pdV=  ∫ (n*R*T/V) dV =  n*R*T  ∫(1/V) dV = n*R*T * ln (V₂/V₁)

since

P₁=n*R*T/V₁

P₂=n*R*T/V₂

dividing both equations

V₂/V₁ = P₁/P₂

W= n*R*T * ln (V₂/V₁)  = n*R*T * ln (P₁/P₂ )

replacing values

P₁=n*R*T/V₁ = 2 moles* 8.314 J/mol K* 300K / 0.1 m3= 49884 Pa

since P₂ = 1 atm = 101325 Pa

W= n*R*T * ln (P₁/P₂ ) = 2 mol * 8.314 J/mol K * 300K * (49884 Pa/101325 Pa) = -3534.94 J

5 0
3 years ago
Forces that act in equal and opposite directions on an object
Akimi4 [234]
These are known as balanced forces because they will not change the motion of the object, and it will remain at rest unless forces become unbalanced- meaning they would be unequal and not opposing. 
5 0
2 years ago
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