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svlad2 [7]
2 years ago
6

A stand shows a pendulum in 5 positions. A has the pendulum bob at a 45 degree angle to the left. C has the pendulum bob straigh

t down, and B is halfway between A and C. E is has the pendulum bob at a 45 degree angle to the right, and D is halfway between C and E.
Determine how potential and kinetic energy changes at each position of the pendulum as the ball swings from A to E.

Position A:

Position B:

Position C:

Position D:

Position E:
Physics
2 answers:
bazaltina [42]2 years ago
7 0

Answer:

position A : is all potential

position B : is losing potential  and gaining kinetic energy

position c : is all kinetic

position D : is losing kinetic and gaining potential

position E : is all potential

kotegsom [21]2 years ago
6 0

Answer:

plz brainliest

Explanation:

position A : is all potential

position B : is losing potential  and gaining kinetic energy

position c : is all kinetic

position D : is losing kinetic and gaining potential

position E : is all potential

thx have a great day

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A piano wire has a length of 81 cm and a mass of 2.0
choli [55]
<span>Frequency = 394 Hz
 Length of the string L = 81 cm = 0.81 m
 Mass of the string = 0.002 kg
 Tension T = ?
 Wave length of the string is two times the length.
  n x lambda = 2L, we also have lambda = vt = v / f, t is time period and given n = 1.
  Therefore L = v / 2f => v = 2fL
 Deriving form force equation, force here is tension T so
  v = squareroot of (TL/m) hence
   2fL = squareroot of (TL/m) => 4 x f^2 x L^2 = (T x L) / m => T = 4 x f^2 x L x m
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2 years ago
A car drives at a constant speed around a banked circular track with a diameter of 136 m . The motion of the car can be describe
galina1969 [7]

Answer:

speed = 44.9m/s

x = 35.5 m,  y = 58.0m

Explanation:

A car on a circular track with constant angular velocity ω can be described by the equation of position r:

\overrightarrow {r(t)} = Rsin(\omega t)\hat{i} + Rcos(\omega t)\hat{j}

The velocity v is given by:

\overrightarrow {v(t)} = \overrightarrow{\frac{dr}{dt}}= \omega Rcos(\omega t)\hat{i} - \omega Rsin(\omega t)\hat{j}

The acceleration a:

\overrightarrow {a(t)} = \overrightarrow{\frac{dv}{dt}}= -\omega^2 Rsin(\omega t)\hat{i} - \omega^2 Rcos(\omega t)\hat{j}

From the given values we get two equations:

-\omega^2 Rsin(\omega t)=-15.4\\-\omega^2 Rcos(\omega t)=-25.4

We also know:

\overrightarrow {a(t)} = -\omega^2 Rsin(\omega t)\hat{i} - \omega^2 Rcos(\omega t)\hat{j}=-\omega^2\overrightarrow{r(t)}

The magnitude of the acceleration a is:

a=\sqrt{(-15.4)^2+(-25.4)^2}=29.7

The magnitude of position r is:

r=R=68m

Plugging in to the equation for a(t):

\overrightarrow {a(t)} =-\omega^2\overrightarrow{r(t)}

and solving for ω:

|\omega|=0.66

Now solve for time t:

\frac{sin(0.66t)}{cos(0.66t)}=tan(0.66t)=\frac{15.4}{25.4}\\t=0.83

Using the calculated values to compute v(t):

\overrightarrow {v(t)}= \omega Rcos(\omega t)\hat{i} - \omega Rsin(\omega t)\hat{j}\\\overrightarrow {v(t)}=44.88cos(0.55)\hai{i}-44.88sin(0.55)\hat{j}\\\overrightarrow {v(t)}=38.3\hat{i}-23.5\hat{j}

The speed of the car is:

\sqrt{38.3^2 + (-23.5)^2} = 44.9

The position r:

\overrightarrow {r(t)} = Rsin(\omega t)\hat{i} + Rcos(\omega t)\hat{j}\\\overrightarrow {r(t)} = 68sin(0.55)\hat{i} + 68cos(0.55)\hat{j}\\\overrightarrow {r(t)} = 35.5{i} + 58.0\hat{j}

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Answer:

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