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Dmitriy789 [7]
3 years ago
6

_______ is the best way to restore weak muscles.

Physics
2 answers:
Lelechka [254]3 years ago
5 0
Rehabilitation my dude
Alex3 years ago
3 0

Rehabilitation is your answer b

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At the starting gun a runner accelerates from rest at 1.9 m/s2 for 2.2s. What is the runner’s speed 2s after she starts running?
sergeinik [125]

Answer:

v_{f} = 3.8 m/s

Explanation:

When the runner start the running, she is at rest. So the initial velocity of the runner is zero (0 m/s). And then she accelerates at  a rate of 1.9 m/s² for 2.2 s. But we need to find the final velocity after t = 2 s when she starts the running. So, we take time equal to 2 second.

Given data:

Initial velocity = v_{i} = 0 m/s

Acceleration = a = 1.9 m/s²   (for 2.2 s after starting the race)

Time = t = 2 s

Final velocity = v_{f} = ?

According to the First equation of motion

v_{f} =  v_{i} + at

v_{f} = 0 + (1.9)(2)

v_{f} = 3.8 m/s

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State three factors that determine the strength of electromagnetic​
Sunny_sXe [5.5K]

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Factors that affect the strength of electromagnets are the nature of the core material, strength of the current passing through the core, the number of turns of wire on the core and the shape and size of the core.Nov 28, 2020

Explanation:

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Consider a machine of mass 70 kg mounted to ground through an isolation system of total stiffness 30,000 N>m, with a measured
kvv77 [185]

Answer:

a)0.0229 m

b)0.393 rad

c)1.57

d)707.6 N

e)0.298 m/s

Explanation:

Given:

  • Mass of the machine, m=70 kg
  • Stiffness of the system, k=30000 N/m
  • Damping ratio=0.2
  • Damping force, F=450 N
  • Angular velocity \omega=13\ \rm rad/s

a)We know that the amplitude X at steady state is given by

X=\dfrac{\dfrac{F_0}{m}}{\sqrt{\omega_n^2-\omega^2)^2 +(2\rho \omega_n\omega)^2}}\\

Where

  • \omega_n=\sqrt{\dfrac{k}{m}}\\\\=\sqrt{\dfrac{30000}{70}}\\\\=20.7\ \rm rad/s
  • \omega=13\ \rm rad/s
  • \rho=0.2
  • F_0=450\ \rm N
  • m=70\ \rm kg

X=\dfrac{\dfrac{450}{70}}{\sqrt{20.7^2-13^2)^2 +(2\times 0.2\times20.7\times13)^2}}\\[tex]X=0.0229\ \rm m

b) The phase shift of the motion is given by

\tan\phi=\dfrac{2\rho \omega_n \omega }{\omega_n^2-\omega^2}\\\\\dfrac{2\times0.2\times20.7\times13 }{20.7^2-\13^2}\\\\\phai=0.393\\

c)Transmissibility ratio is given by

T.R.=\sqrt{\dfrac{1+(2\rho r)^2}{(1-r^2)^2+{(2\rho r)^2}}}\\\\T.R.=\sqrt{\dfrac{1+(2\times0.2\times0.628)^2}{(1-0.628^2)^2+{(2\times0.2\times0/628)^2}}}\\\\=1.57

d)The magnitude of the force transmitted to the ground is

F_T=(T.R)\times F_0\\\\=450\times1.57\\\\=707.6\ \rm N

e)The maximum velocity is given by V_{max}

V_{max}=\omega A_0\\\\=13\times 0.0229\\\\=0.298\ \rm m/s

6 0
3 years ago
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