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Serjik [45]
2 years ago
8

To understand the meaning of the variables that appear in the equations for rotational kinematics with constant angular accelera

tion. Rotational motion with a constant nonzero acceleration is not uncommon in the world around us. For instance, many machines have spinning parts. When the machine is turned on or off, the spinning parts tend to change the rate of their rotation with virtually constant angular acceleration. Many introductory problems in rotational kinematics involve motion of a particle with constant, nonzero angular acceleration. The kinematic equations for such motion can be written as
Physics
1 answer:
Elis [28]2 years ago
5 0

Answer:

   ω = ω₀ + α t

      ω² = ω₀² + 2 α  θ

      θ = θ₀ + ω₀ t + ½ α t²

Explanation:

Rotational kinematics can be treated as equivalent to linear kinematics, for this change the displacement will change to the angular displacement, the velocity to the angular velocity and the acceleration to the angular relation, that is

     x → θ

     v → ω

     a → α

with these changes the three linear kinematics relations change to

      ω = ω₀ + α t

      ω² = ω₀² + 2 α  θ

      θ = θ₀ + ω₀ t + ½ α t²

where it should be clarified that to use these equations the angles must be measured in radians

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How long will it take to fall 50 m?
Ad libitum [116K]

Answer:

down below

Explanation:

Since we aren't the given the time, lets say that an object to 25 seconds to fall 50 meters. We can use the formula [ s = d/t ] to solve.

s = 50/25

s = 2

Therefore, the object was falling at a rate of 2 meters per second.

Best of Luck!

6 0
2 years ago
What is photographic film
Vesnalui [34]

Answer: its a strip of transparent film, one side coated with gelatin emulsion containing microscopically small light-sensitive silver halide crystals

Explanation:

3 0
3 years ago
An electron and a proton are held on an x axis, with the electron at x = + 1.000 m
mixas84 [53]

Answer:

  r2 = 1 m

therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m

Explanation:

For this exercise we must use conservation of energy

the electric potential energy is

          U = k \frac{q_1q_2}{r_{12}}

for the proton at x = -1 m

          U₁ =- k \frac{e^2 }{r+1}

for the electron at x = 1 m

          U₂ = k \frac{e^2 }{r-1}

starting point.

        Em₀ = K + U₁ + U₂

        Em₀ = \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1}

final point

         Em_f = k e^2 ( -\frac{1}{r_2 +1} + \frac{1}{r_2 -1})

   

energy is conserved

        Em₀ = Em_f

        \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e^2 (- \frac{1}{r_2 +1} + \frac{1}{r_2 -1})              

       

        \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e²(  \frac{2}{(r_2+1)(r_2-1)} )

we substitute the values

½ 9.1 10⁻³¹ 450 + 9 10⁹ (1.6 10⁻¹⁹)² [ - \frac{1}{20+1} + \frac{1}{20-1} ) = 9 109 (1.6 10-19) ²( \frac{2}{r_2^2 -1} )

          2.0475 10⁻²⁸ + 2.304 10⁻³⁷ (5.0125 10⁻³) = 4.608 10⁻³⁷ ( \frac{1}{r_2^2 -1} )

          2.0475 10⁻²⁸ + 1.1549 10⁻³⁹ = 4.608 10⁻³⁷     \frac{1}{r_2^2 -1}

          \frac{2.0475 \ 10^{-28} }{1.1549 \ 10^{-37} } = \frac{1}{r_2^2 -1}

          r₂² -1 = (4.443 10⁸)⁻¹

           

          r2 = \sqrt{1 + 2.25 10^{-9}}

          r2 = 1 m

therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m

4 0
3 years ago
Consider the accuracy of recording muscle electrical activity using electrodes placed on the skin surface, compared to directly
OverLord2011 [107]

Answer: The common difference between surface EMG and intramuscular EMG is that that former is non-invasive while the later is an invasive method

Explanation:

Electromyography (EMG) is used clinically for the examination of muscle excitations (muscle electrical activity) in both normal or abnormal conditions. There are two forms of EMG includes:

--> Surface EMT and

--> Intramuscular EMT

Surface EMT is a non invasive method of examination of muscle excitations for superficial and easily accessible muscles.

Intramuscular EMT is the invasive method of examination of muscle excitations usually for deep muscles.

The difference between the two forms of EMT includes:

- surface EMT is non- invasive while intramuscular EMT is invasive

- surface EMT is used to access superficial muscle while intramuscular EMT is used to access deep muscles.

- surface EMT requires less skill and time to carry out while intramuscular EMT requires special skills and takes more time while carrying out the procedure.

8 0
3 years ago
Physics question, help please!
vova2212 [387]

The change in potential energy when the block falls to ground is -480J.

The maximum change in kinetic energy of the ball is 480 J.

The initial kinetic energy of the ball is 0 J.

The final  kinetic energy of the ball is 0.148J.

The initial potential energy of the ball is 0.187 J.

The final  potential energy of the ball is 0 J.

The work done by the air resistance is 0.039 J.

<h3>Change in potential energy when the block falls to ground</h3>

ΔP.E = -mgh

ΔP.E = -Wh

ΔP.E = - 40 x 12

ΔP.E = -480 J

<h3>Maximum change in kinetic energy of the ball</h3>

ΔK.E = - ΔP.E

ΔK.E = - (-480 J)

ΔK.E = 480 J

<h3>Initial kinetic energy of the ball</h3>

K.Ei = 0.5mv²

where;

  • v is zero since it is initially at rest

K.Ei = 0.5m(0) = 0

<h3>Final kinetic energy</h3>

K.Ef =  0.5mv²

K.Ef = 0.5(0.0091)(5.7)²

K.Ef = 0.148 J

<h3>Initial potential energy of the ball</h3>

P.Ei = mghi

P.Ei = 0.0091 x 9.8 x 2.1

P.Ei = 0.187 J

<h3>Final potential energy</h3>

P.Ef = mghf

P.Ef = 0.0091 x 9.8 x 0

P.Ef = 0

<h3>Work done by the air resistance</h3>

W = ΔE

W = P.E - K.E

W = 0.187 J - 0.148 J

W = 0.039 J

Learn more about potential energy here: brainly.com/question/1242059

#SPJ1

<h3 />
7 0
1 year ago
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