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3 years ago

Explain why isokinetic exercises cannot generally be performed by individuals at home.

2 answers:

6 0

Iso-kinetic exercises cannot be performed at home by an individual because they require specific machines that enable to exercise at a constant speed with professional guidance.

<u>Explanation :</u>

Iso-kinetic exercises are that kind of training regimes that require a constant pace to strengthen your muscles with the help of machines with significant specifications.

The machines involved in iso-kinetic exercises control your pace with variable resistance throughout the session and hence, it’s not a cup of tea for an individual to trace his/her performances and the respective improvements.

So, before going for a iso-kinetic schedule, be sure to have a word with your physiotherapist to train well with the machines under the specifications that can better suit your body and its requirements.

Rzqust [24]3 years ago

3 0

Answer:

Iso-kinetic exercises are that kind of training regimes that require a constant pace to strengthen your muscles with the help of machines with significant specifications.

The machines involved in iso-kinetic exercises control your pace with variable resistance throughout the session and hence, it’s not a cup of tea for an individual to trace his/her performances and the respective improvements.

So, before going for a iso-kinetic schedule, be sure to have a word with your physiotherapist to train well with the machines under the specifications that can better suit your body and its requirements.

Explanation:

got it right on test

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As steam is slowly injected into a turbine, the angular acceleration of the rotor is observed to increase linearly with the time

gavmur [86]

Answer:

217.04 rad/s

Explanation:

We are given that

Initial velocity, $\omega_0=0$

t=10 s

Number of rev=20

We have to find the angular velocity at t=24 s

$\theta_1=2\pi\times 20=40\pi rad$

$\alpha=kt$

$\frac{d\omega}{dt}=kt$

$\int d\omega=\int_{0}^{t} ktdt$

$\omega=\frac{kt^2}{2}$

$\frac{d\theta}{dt}=\frac{kt^2}{2}$

$\int d\theta=\int_{0}^{t}\frac{kt^2}{2} dt$

$\theta=\frac{kt^3}{6}$

Substitute the values

$40\pi=\frac{k(10)^3}{6}$

$k=\frac{40\pi\times 6}{(10)^3}=\frac{24\pi}{100}$

Substitute the value of k and t=24

$\omega=\frac{24\pi\times (24)^2}{2\times 100}=217.04 rad/s$

8 0

4 years ago

A vertical piston-cylinder device contains water and is being heated on top of a range. During the process, 63 Btu of heat is tr

jolli1 [7]

Answer:

The water would have a change in the energy of 50 BTU

Explanation:

Under the assumption that the piston-cylinder system is a closed system, then the conservation law of the system would be: $E_{in}-E_{out}=E_{change}$ , where $E_{in}$ is all the energy that goes into the system, $E_{out}$ is all the energy that leaves the system and, $E_{change}$ is the change of the energy of the system.
For this case the energy that is feeded to the system is heat: 63 BTU.
There is two ways that energy is leaving this system: 8 BTU of heat as losses through the side walls and, 5 BTU of work done by the vapor on the piston.
Then the change of energy of the system is: $E_{change}=E_{in}-E_{out}=63 BTU-(8BTU+5BTU)=50BTU$ .

6 0

3 years ago

Which of the following is NOT an effective pre-exercise energizing snack?

maks197457 [2]

Cookies would not be an effective pre-exercise energizing snack, as their high sugar content would be absorbed quickly by the body, not sustaining the athlete for the duration of their event.

4 0

4 years ago

Read 2 more answers

Help mee! Physics i think!

xxTIMURxx [149]

Answer:

Use the form of equation:

Q=mL

You have the specific latent heat of vaporization L = 2.260*10^{6}

And Q, the heat energy supplied, which equals 1695 KJ = 1695*10^{3} J

So you can get the mass by substitution in the formula below.

5 0

3 years ago

Suppose that the period of a particular ideal mass-spring system is 5 s . What would be the period of the system if the mass wer

yan [13]

Answer: T2 = 7.07s

Explanation: The period of a loaded spring of spring constant k and mass m is given by

T= 2π √m/k

With 2π constant and k, it can be seen with little algebra that

T² is proportional to mass m

Hence (T1)²/m1 = (T2) ²/m2

Where T1 = 5, T2 =?, let m1 = m hence m2 = 2m.

By substituting, we have that

5²/m = (T2) ²/2m

25 / m = (T2) ²/2m

25 × 2m = (T2) ² × m

25 × 2 = (T2) ²

50 = (T2) ²

T2 = √50

T2 = 7.07s

5 0

4 years ago