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slava [35]
3 years ago
10

Which activity is ideal for fast-twitch muscles?

Engineering
1 answer:
ryzh [129]3 years ago
7 0
A. Running. According to healthline.com, “Fast twitch muscles are optimized for short, intense activities, such as sprinting, powerlifting, and jumping”
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Crank OA rotates with uniform angular velocity 0  4 rad/s along counterclockwise. Take OA= r= 0.5
mafiozo [28]
CRABK DAT SOUIJA BOI LIKE OOOOO WATCH ME WATCH ME OOOO LASAGNA Can you lend me 700$ because I used my toaster as a bath heater and now my legs are gone plz I need money for bandaids
3 0
4 years ago
5. A non-cold-worked brass specimen of average grain size 0.01 mm has a yield strength of 150 MPa. Estimate the yield strength o
Papessa [141]

Answer:

97.17 MPa

Explanation:

Given:-

- The nominal strength of the grain, σ0  = 25 MPa

- The average grain size of the brass specimen, d* = 0.01 m

- The yield strength of the non-cold worked specimen, σy = 150 MPa

- Conditions of cold-working: T = 500°C , t = 1000 s

Find:-

Estimate the yield strength of this alloy after cold - working process

Solution:-

- The nominal strength of the grain is a function of yield strength of the material, grain yield factor ( Ky ) and the grain size.

- the following relation is used to determine the grain strength:

                             σ0  = σy  - ( Ky / √( d ) )

- We will use the above relation to determine the grain yield factor ( Ky ) for the alloy as follows. Note: here we will use the average value of grain size:

                            Ky = ( σy  - σy )*√( d* )

                            Ky = ( 150 - 25 ) * √0.01

                            Ky = 12.5 MPa - √mm

- Now we will use the cold working conditions of T = 500 C and time of the process is t = 1000 s. We will look up the elongated size of the grain after the cold-working process in lieu with its yield factor ( Ky ). Use figure 7.25.

- The cold-worked grain size with the given conditions can be read off from the figure 7.25. The new size comes out to be d = 0.03 mm.

- We will again use the nominal grain strength relation expressed initially. And compute for the new yield strength of the cold-worked alloy.

                            σ0  = σy  - ( Ky / √( d ) )

                            σy = σ0 + ( Ky / √( d ) )

                            σy = 25MPa + ( 12.5 / √( 0.03 mm ) )

                            σy = 97.17 MPa

- We see that the yield strength of the alloy decreases after cold-working process. This happens because the cold working process leaves with inter-granular strain ( dislocation of planes ) in the material structure which results from the increase in grain size.

6 0
3 years ago
Create a function (prob3_5) that will take inputs of vectors x and y in feet, scalar N, scalars L and W in feet and scalars T1 a
Shalnov [3]

Answer:

clear, clc

prob3_5([1,2,3],[6,5,7],12,11,22,55,76)

function T=prob3_5(x,y,N,L,W,T1,T2)

w=zeros(1,length(x));

for n=1:2:N

for i=1:length(x)

w(i)=w(i)+(2/pi)*(2/n)*sin(n*pi*x(i)/L).*sinh(n*pi*y(i)/L)/sinh(n*pi*W/L);

end

end

T=(T2-T1)*w+T1;

end

Explanation:

Please input the commands into MATLAB

3 0
3 years ago
The cars of a roller-coaster ride have a speed of 19.0 km/h as they pass over the top of the circular track. Neglect any frictio
Dmitrij [34]

Complete Question

The cars of a roller-coaster ride have a speed of 19.0 km/h as they pass over the top of the circular track. Neglect any friction and calculate their speed v when they reach the horizontal bottom position. At the top position, the radius of the circular path of their mass centers is 21 m, and all six cars have the same mass.V = -18 m What is v?X km/h

Answer:

v=23.6m/s

Explanation:

Velocity v_c=18.0km/h

Radius r=21m

initial velocity uu=19=>5.27778

Generally the equation for Angle is mathematically given by

\theta=\frac{v_c}{2r}

\theta=\frac{18}{2*21}

\theta=0.45

\theta=25.7831 \textdegree

Generally

Height of mass

h=\frac{rsin\theta}{\theta}

h=\frac{21sin25.78}{0.45}

h=20.3m

Generally the equation for Work Energy is mathematically given by

0.5mv_0^2+mgh=0.5mv^2

Therefore

v=\sqrt{u^2+2gh}

v=\sqrt{=5.27778^2+2*9.81*20.3}

v=23.6m/s

3 0
3 years ago
A 3.7 g mass is released from rest at C which has a height of 1.1 m above the base of a loop-the-loop and a radius of 0.2 m . Th
Dmitrij [34]

Answer:

Normal force = 0.326N

Explanation:

Given that:

mass released from rest at C = 3.7 g = 3.7 × 10⁻³ kg

height of the mass = 1.1 m

radius = 0.2 m

acceleration due to gravity = 9.8 m/s²

We are to determine the normal force pressing on the track at A.

To to that;

Let consider the conservation of energy relation; which says:

mgh = mgr + 1/2 mv²

gh = gr + 1/2 v²

gh - gr = 1/2v²

g(h-r) = 1/2v²

v² = 2g(h-r)

However; the normal force will result to a centripetal force; as such, using the relation

N =mv²/r

replacing the value for v² = 2g(h-r) in the above relation; we have:

Normal force = 2mg(h-r)/r

Normal force = 2 × 3.7 × 10⁻³ × 9.8 ( 1.1 - 0.2 )/ 0.2

Normal force = 0.065268/0.2

Normal force = 0.32634 N

Normal force = 0.326N

6 0
3 years ago
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