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lbvjy [14]
2 years ago
9

g 2. The _____ spans the distance from the _____ to the location of the applied force. moment arm; pivot point moment of inertia

; center of mass torque; center of mass None of the above
Physics
1 answer:
Alik [6]2 years ago
7 0

Answer:

The correct answer to the following question will be Option A (moment arm; pivot point).

Explanation:

  • The moment arm seems to be the duration seen between joint as well as the force section trying to act mostly on the joint. Each joint that is already implicated in the workout seems to have a momentary arm.
  • The moment arm extends this same distance from either the pivot point to just the position of that same pressure exerted.
  • The pivotal point seems to be the technical indicators required to fully measure the appropriate demand trends alongside different time-frames.

The other three choices are not related to the given situation. So that option A is the appropriate choice.

You might be interested in
Honeybees can see light in the ________ range of the electromagnetic spectrum.
konstantin123 [22]
Humans can see wavelengths in the visible part of the electromagnetic spectrum. That is the range of approximately 400 - 700 nm. Honeybees can see visible light and about 100 nm more in the ultraviolet part of the electromagnetic spectrum. That is approximately 300 - 700 nm. 
4 0
2 years ago
The Slowing Earth The Earth's rate of rotation is constantly decreasing, causing the day to increase in duration. In the year 20
NNADVOKAT [17]

Answer:

The average angular acceleration of the Earth, α  = 6.152 X 10⁻²⁰ rad/s²

Explanation:

Given data,

The period of 365 rotation of Earth in 2006, T₁ = 365 days, 0.840 sec

                                                                                  = 3.1536 x 10⁷ +0.840

                                                                                 = 31536000.84 s

The period of 365 rotation of Earth in 2006, T₀ = 365 days

                                                                               = 31536000 s

Therefore, time period of one rotation on 2006, Tₐ = 31536000.84/365

                                                                                   = 86400.0023 s

The time period of rotation is given by the formula,

                                <em>Tₐ = 2π /ωₐ</em>

                                 ωₐ = 2π / Tₐ

Substituting the values,

                                  ωₐ = 2π /  365.046306        

                                      = 7.272205023 x 10⁻⁵ rad/s

Therefore, the time period of one rotation on 1906, Tₓ = 31536000/365

                                                                                    = 86400 s

Time period of rotation,

                                   Tₓ = 2π /ωₓ

                                    ωₓ = 2π / T

                                           =  2π /86400

                                          = 7.272205217  x 10⁻⁵ rad/s

The average angular acceleration

                                   α  = (ωₓ -   ωₐ) /  T₁

             = (7.272205217  x 10⁻⁵ - 7.272205023 x 10⁻⁵) / 31536000.84

                                    α  = 6.152 X 10⁻²⁰ rad/s²

Hence the average angular acceleration of the Earth, α  = 6.152 X 10⁻²⁰ rad/s²

3 0
2 years ago
A 10-turn coil of wire having a diameter of 1.0 cm and a resistance of 0.50 Ω is in a 1.0 mT magnetic field, with the coil orien
n200080 [17]

Answer:

The voltage across the capacitor is 1.57 V.

Explanation:

Given that,

Number of turns = 10

Diameter = 1.0 cm

Resistance = 0.50 Ω

Capacitor = 1.0μ F

Magnetic field = 1.0 mT

We need to calculate the flux

Using formula of flux

\phi=NBA

Put the value into the formula

\phi=10\times1.0\times10^{-3}\times\pi\times(0.5\times10^{-2})^2

\phi=7.85\times10^{-7}\ Tm^2

We need to calculate the induced emf

Using formula of induced emf

\epsilon=\dfrac{d\phi}{dt}

Put the value into the formula

\epsilon=\dfrac{7.85\times10^{-7}}{dt}

Put the value of emf from ohm's law

\epsilon =IR

IR=\dfrac{7.85\times10^{-7}}{dt}

Idt=\dfrac{7.85\times10^{-7}}{R}

Idt=\dfrac{7.85\times10^{-7}}{0.50}

Idt=0.00000157=1.57\times10^{-6}\ C

We know that,

Idt=dq

dq=1.57\times10^{-6}\ C

We need to calculate the voltage across the capacitor

Using formula of charge

dq=C dV

dV=\dfrac{dq}{C}

Put the value into the formula

dV=\dfrac{1.57\times10^{-6}}{1.0\times10^{-6}}

dV=1.57\ V

Hence, The voltage across the capacitor is 1.57 V.

5 0
2 years ago
In Young's double slit experiment, 402 nm light gives a fourth-order bright fringe at a certain location on a flat screen. What
Alex777 [14]

Answer:

λ₂ = 357.3 nm

Explanation:

The expression for double-slit interference is

          d sin θ = m λ                 constructive interference

          d sin θ = (m + ½) λ        destructive interference.

The initial data corresponds to a constructive interference, they indicate that we are in the fourth order (m = 4), let's look for the separation of the slits

         d sin θ = m λ₁

       

now ask for destructive interference for m = 4

        d sin θ = (m + ½) λ₂

we match these two expressions

         m λ₁ = (m + ½) λ₂

         λ₂ = ( m / m + ½) λλ₁  

let's calculate

         λ₂ =\frac{4}{(4.000 +0.5) \ 401}

        λ₂ = 357.3 nm

7 0
1 year ago
Because the sensitivity of the human ear varies over the audio spectrum, how do many sound systems handle very high frequencies
antoniya [11.8K]

Answer:

Option 2.

Amplify; Amplify

Explanation:

Amplification of sound simply means increasing the loudness of the sound by modifying its electro-acoustic signals. To make very high and very low frequencies of sound audible by the human ear,  the sounds need to be amplified by the sound system amplifier.

Sound systems modify high frequencies by amplifying them, and also, they also modify low frequencies by amplifying them. This changes their wave form and brings them into the hearing range of humans.

3 0
1 year ago
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