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nikklg [1K]
3 years ago
5

The knee extensors insert on the tibia at an angle of 30 degrees (from the longitudinal axis of the tibia), at a distance of 3 c

m from the axis of rotation at the knee. How much force must the knee extensors exert to produce an angular acceleration at the knee of 1 rad/s2 , given a mass of the lower leg and foot of 4.5 kg, and a radius of gyration of 23 cm
Physics
1 answer:
Reika [66]3 years ago
7 0

Answer:

the knee extensors must exert 15.87 N

Explanation:

Given the data in the question;

mass m = 4.5 kg

radius of gyration k = 23 cm = 0.23 m

angle ∅ = 30°

∝ = 1 rad/s²

distance of 3 cm from the axis of rotation at the knee r = 3 cm = 0.03 m

using the expression;

ζ = I∝

ζ = mk²∝

we substitute

ζ = 4.5 × (0.23)² × 1

ζ  = 0.23805 N-m

so

from; ζ = rFsin∅

F = ζ / rsin∅

we substitute

F = 0.23805 / (0.03 × sin( 30 ° )

F = 0.23805 / (0.03 × 0.5)

F F = 0.23805 / 0.015

F = 15.87 N

Therefore, the knee extensors must exert 15.87 N

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An aircraft is at a standstill. It accelerates to 140kts in 28 seconds. The aircraft weighs 28,000 lbs. How many feet of runway
stiv31 [10]

Answer:

runway use is 3307.8 feet

Explanation:

given data

velocity = 140 kts = 140 × 0.5144 m/s = 72.016 m/s

time = 28 seconds

weight = 28000 lbs

to find out

How many feet of runway was used

solution

we will use here first equation of motion for find acceleration

v = u + at     ..............1

here v is velocity given and u is initial velocity that is 0 and a is acceleration and t is time

put here value in equation 1

72.016 = 0 + a(28)

a = 2.572 m/s²

and

now apply third equation of motion

s = ut + 0.5×a×t²    .......................2

here s is distance and u is initial speed and t is time and a is acceleration

put here all value in equation 2

s = 0 + 0.5×2.572×28²  

s = 1008.24 m = 3307.8 ft

so  runway use is 3307.8 feet

5 0
4 years ago
Which statements about radiocarbon dating are true?
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8 0
3 years ago
Read 2 more answers
(a) An ideal gas initially at pressure p0 undergoes a free expansion until its volume is 3.80 times its initial volume. What the
trapecia [35]

Answer:

a)P/Po=0.263

b)γ=1.33 So gas is triatomic

c)\dfrac{KE_f}{KE_i}=1.56

Explanation:

a)

initial pressure = Po

Initial volume = Vo

Final volume = 3.8 Vo

Lets take final pressure is P

we know that for free expansion process

PV= Constant

Po x Vo = P x 3.8 Vo

P=0.263 Po

So

P/Po=0.263

b)

Now gas is compressed in adiabatic manner

Final pressure = 1.56 Po

                       =1.56 Po

We know that for adiabatic process

P_1V_1^{\gamma}=P_2V_2^{\gamma}

\dfrac{V_2}{V_1}=\left(\dfrac{P_1}{P_2}\right)^{\dfrac{1}{\gamma}}

0.263P_o(3.8V_o)^{\gamma}=1.56P_o\times V_o^{\gamma}

γ=1.33 So gas is triatomic

c)

We know that average kinetic energy given as

KE=\dfrac{3}{2}KT

\dfrac{KE_f}{KE_i}=\dfrac{T_f}{T_i}\

\dfrac{KE_f}{KE_i}=\dfrac{P_fV_f}{P_iV_i}

\dfrac{KE_f}{KE_i}=\dfrac{1.56P_oV_o}{P_oV_o}

\dfrac{KE_f}{KE_i}=1.56

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trasher [3.6K]

It was important because it tells why the denser material had sunken and the lighter is floating on the surface.

<u>Explanation:</u>

Eventually, after around 500 million years, our young planet's temperature warmed to the liquefying purpose of iron—about 1,538° Celsius (2,800° Fahrenheit). This crucial crossroads in Earth's history is known as the iron fiasco. The iron fiasco permitted more prominent, progressively quick development of Earth's liquid, rough material.

We also know that at one point all the material in the planet was molten because the denser material had sunk and the lighter was floating.

3 0
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timurjin [86]
It's true charged particles are always acted on by an electric field.when the velocity of the electron is parallel to the magnetic field, the magnetic force vanishes.
4 0
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