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3 years ago

Which describes an interaction within the musculoskeletal system?

Physics

2 answers:

rosijanka [135]3 years ago

7 0

Answer:

The musculoskeletal system involves the complex interactions of muscles, bones, and connective tissues.

When a muscle contracts, a bone will move. When a bone contracts, a muscle will move. When a ligament contracts, a tendon will move.

STatiana [176]3 years ago

6 0

Answer:

A. when a muscle contracts , a bone will move

Explanation

thats the correct answer to this question on edg2021 :)

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What current flows through a 2.54cm diameter rod of pure silicon that is 20cm long when 1000V is applied?

vfiekz [6]

Answer: $0.0039\ A$

Explanation:

Given

Diameter of the rod $d=2.54\ cm$

length of rod is $l=20\ cm$

Resistivity of silicon is $\rho=6.4\times 10^2\ \Omega-m$

cross-section of the rod $A$

$\Rightarrow A=\dfrac{\pi d^2}{4}\\\\\Rightarrow A=\dfrac{3.142\times 2.54^2\times 10^{-4}}{4}\\\\\Rightarrow A=5.067\times 10^{-4}\ m^2$

Resistance of rod is R

$\Rightarrow R=\dfrac{\rho l}{A}$

$\Rightarrow R=\dfrac{640\times 0.20}{5.067\times 10^{-4}}\\\\\Rightarrow R=25.26\times 10^4\ \Omega$

Current is given by

$\Rightarrow I=\dfrac{V}{R}\\\\\Rightarrow I=\dfrac{1000}{25.26\times 10^4}\\\\\Rightarrow I=0.0039\ A$

3 0

3 years ago

In a nuclear physics experiment, a proton (mass 1.67×10^(−27)kg, charge +e=+1.60×10^(−19)C) is fired directly at a target nucleu

Arte-miy333 [17]

The given question is incomplete. The complete question is as follows.

In a nuclear physics experiment, a proton (mass $1.67 \times 10^(-27)$ kg, charge +e = $+1.60 \times 10^(-19)$ C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed $2.50 \times 10^6$ m/s. The proton comes momentarily to rest at a distance $5.31 \times 10^(-13)$ m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are $5.31 \times 10^{-13}$ m apart?

Explanation:

The given data is as follows.

Mass of proton = $1.67 \times 10^{-27}$ kg

Charge of proton = $1.6 \times 10^{-19} C$

Speed of proton = $2.50 \times 10^{6} m/s$

Distance traveled = $5.31 \times 10^{-13} m$

We will calculate the electric potential energy of the proton and the nucleus by conservation of energy as follows.

$(K.E + P.E)_{initial}$ = $(K.E + P.E)_{final}$

$(\frac{1}{2} m_{p}v^{2}_{p}) = (\frac{kq_{p}q_{t}}{r} + 0)$

where, $\frac{kq_{p}q_{t}}{r} = U = Electric potential energy$

U = $(\frac{1}{2}m_{p}v^{2}_{p})$

Putting the given values into the above formula as follows.

U = $(\frac{1}{2}m_{p}v^{2}_{p})$

= $(\frac{1}{2} \times 1.67 \times 10^{-27} \times (2.5 \times 10^{6})^{2})$

= $5.218 \times 10^{-15} J$

Therefore, we can conclude that the electric potential energy of the proton and nucleus is $5.218 \times 10^{-15} J$ .

4 0

3 years ago

A charged particle moving along the x-axis enters a uniform magnetic field pointing along the z-axis. Because of an electric fie

Furkat [3]

Answer:

Explanation:

Let the charge particle have charge equal to +q .

force due to electric field will be along the field that is along y - axis . To balance it force by magnetic force must be along - y axis. ( negative of y axis )

force due to magnetic field = q ( v x B ) , v is velocity and B is magnetic field.

F = q ( v i x B k ) , ( velocity is along x direction and magnetic field is along z axis. )

= (Bqv) - j

= - Bqv j

The force will be along - ve y - direction .

If we take charge as negative or - q

force due to electric field will be along - y axis .

magnetic force = F = -q ( v i x B k )

= + Bqv j

magnetic force will be along + y axis

So it is difficult to find out the nature of charge on the particle from this experiment.

5 0

3 years ago

A certain pair of slits are separated by a distance d. Monochromatic coherent light falls on this pair of slits and the interfer

DanielleElmas [232]

Answer:

The new separation distance between adjacent bright fringes will be <u>4 mm</u>

Explanation:

Since, the distance between adjacent bright fringes is given by the formula:

Δx₁ = λL/d = 2 mm -------- eqn (1)

where,

Δx = Distance between adjacent bright fringes

λ = wavelength of light = constant for both cases

L = Distance between the slits and the screen

d = slit separation

Now, for the second case:

Slit Separation = d/2

Therefore,

Δx₂ = λL/(d/2)

Δx₂ = 2(λL/d)

using eqn (1), we get:

Δx₂ = 2 Δx₁

Δx₂ = 2(2 mm)

5 0

3 years ago

A child slides down a hill on a toboggan with acceleration of 1.8 m/s2. if she starts from rest, how far has she traveled in:

Stells [14]

A child slides down a hill on a toboggan with acceleration of 1.8 m/s2. if she starts from rest, how far has she traveled in: 2 seconds

Answer:

2.4 m

Explanation:

From the question above,

Applying equation of motion,

s = ut+at²/2....................... Equation 1

Where t = time, u = initial velocity, a = acceleration, s = distance.

make s the subject of the equation,

Given: a = 1.8 m/s², t = 2 seconds, u = 0 m/s (from rest)

Substitute these value into equation 1

s = 0(2)+1.8(2²)/2

s = 1.2(4)/2

s = 2.4 m.

Hence she has traveled 2.4 m

4 0

3 years ago