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laiz [17]
3 years ago
5

The purpose of biofeedback is to help subjects __________.

Physics
2 answers:
padilas [110]3 years ago
8 0

The purpose of biofeedback is to help subjects<u> "control their internal processes".</u>


Biofeedback is a kind of treatment that utilizes sensors joined to your body to gauge key body capacities. Biofeedback is expected to enable you to take in more about how your body functions. This data may assist you with developing better power over certain body capacities and address wellbeing concerns.  

This feedback causes you center around rolling out inconspicuous improvements in your body, for example, loosening up specific muscles, to accomplish the outcomes you need, for example, decreasing agony. Generally, biofeedback enables you to utilize your musings to control your body, frequently to enhance a wellbeing condition or physical performance.

anzhelika [568]3 years ago
6 0
A is the answer my friend.
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1/3 the weight than it is on earth, duh
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If a photon has a frequency of 1.15 × 1015 hertz, what is the energy of the photon? Given: Planck's constant is 6.63 × 10-34 jou
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A charge Q is uniformly spread over one surface of a very large nonconducting square elastic sheet having sides of length d. At
GuDViN [60]

Answer:

E/4

Explanation:

The formula for electric field of a very large (essentially infinitely large) plane of charge is given by:

E = σ/(2ε₀)

Where;

E is the electric field

σ is the surface charge density

ε₀ is the electric constant.

Formula to calculate σ is;

σ = Q/A

Where;

Q is the total charge of the sheet

A is the sheet's area.

We are told the elastic sheet is a square with a side length as d, thus ;

A = d²

So;

σ = Q/d²

Putting Q/d² for σ in the electric field equation to obtain;

E = Q/(2ε₀d²)

Now, we can see that E is inversely proportional to the square of d i.e.

E ∝ 1/d²

The electric field at P has some magnitude E. We now double the side length of the sheet to 2L while keeping the same amount of charge Q distributed over the sheet.

From the relationship of E with d, the magnitude of electric field at P will now have a quarter of its original magnitude which is;

E_new = E/4

3 0
3 years ago
. One long wire carries a current of 30 A along the entire x axis. A second-long wire carries a current of 40 A perpendicular to
d1i1m1o1n [39]

Answer:

magnitude of net magnetic field at given point is

B = 5 \times 10^{-6} T

Explanation:

As we know that magnetic field due to a long current carrying wire is given as

B = \frac{\mu_o i}{2\pi r}

here we we will find the magnetic field due to wire which is along x axis is given as

i = 30 A

r = 2 m

now we have

B_1 = \frac{4\pi \times 10^{-7} (30)}{2\pi (2m)}

B_1 = 3\times 10^{-6} T into the plane

Now similarly magnetic field due to another wire which is perpendicular to xy plane is given as

i = 40 A

r = 2 m

now we have

B_2 = \frac{4\pi \times 10^{-7} (40)}{2\pi (2m)}

B_2 = 4\times 10^{-6} T along + x direction

Since the two magnetic field is perpendicular to each other

So here net magnetic field is given as

B = \sqrt{B_1^2 + B_2^2}

B = 5 \times 10^{-6} T

5 0
3 years ago
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