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GalinKa [24]
3 years ago
7

In the oscillating spring ball system, what is true about the energy of the ball when it is located at its amplitude? Select the

correct answer A. The kinetic energy is at its maximum value and the potential energy is zero. B. The potential energy is at its maximum value and the kinetic energy is zero. C. The total energy is zero. D. Both the kinetic and potential energy are at their maximum our swer values
Physics
1 answer:
Nitella [24]3 years ago
8 0

When the maximum amplitude is reached, the body is also positioned in the maximum position which it can reach. At that point, the maximum potential energy stored in the system will therefore be taken.

PE = \frac{1}{2} kh_{max} ^2

However, when it reaches the maximum amplitude it is also in a change of direction of the velocity vector, reaching its minimum point in magnitude, thereby generating the kinetic energy to become zero.

KE = \frac{1}{2} mv^2

Therefore of the options shown, the correct answer would be B.

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The curvature of the helix r​(t)equals(a cosine t )iplus(a sine t )jplusbt k​ (a,bgreater than or equals​0) is kappaequalsStartF
4vir4ik [10]

Answer:

\kappa = \frac{1}{2 b}

Explanation:

The equation for kappa ( κ) is

\kappa = \frac{a}{a^2 + b^2}

we can find the maximum of kappa for a given value of b using derivation.

As b is fixed, we can use kappa as a function of a

\kappa (a) = \frac{a}{a^2 + b^2}

Now, the conditions to find a maximum at a_0 are:

\frac{d \kappa(a)}{da} \left | _{a=a_0} = 0

\frac{d^2\kappa(a)}{da^2}  \left | _{a=a_0} < 0

Taking the first derivative:

\frac{d}{da} \kappa = \frac{d}{da}  (\frac{a}{a^2 + b^2})

\frac{d}{da} \kappa = \frac{1}{a^2 + b^2} \frac{d}{da}(a)+ a * \frac{d}{da}  (\frac{1}{a^2 + b^2} )

\frac{d}{da} \kappa = \frac{1}{a^2 + b^2} * 1 + a * (-1)  (\frac{1}{(a^2 + b^2)^2} ) \frac{d}{da}  (a^2+b^2)

\frac{d}{da} \kappa = \frac{1}{a^2 + b^2} * 1 - a  (\frac{1}{(a^2 + b^2)^2} ) (2* a)

\frac{d}{da} \kappa = \frac{1}{a^2 + b^2} * 1 -  2 a^2  (\frac{1}{(a^2 + b^2)^2} )

\frac{d}{da} \kappa = \frac{a^2+b^2}{(a^2 + b^2)^2}  -  2 a^2  (\frac{1}{(a^2 + b^2)^2} )

\frac{d}{da} \kappa = \frac{1}{(a^2 + b^2)^2} (a^2+b^2 -  2 a^2)

\frac{d}{da} \kappa = \frac{b^2 -  a^2}{(a^2 + b^2)^2}

This clearly will be zero when

a^2 = b^2

as both are greater (or equal) than zero, this implies

a=b

The second derivative is

\frac{d^2}{da^2} \kappa = \frac{d}{da} (\frac{b^2 -  a^2}{(a^2 + b^2)^2} )

\frac{d^2}{da^2} \kappa = \frac{1}{(a^2 + b^2)^2} \frac{d}{da} ( b^2 -  a^2 ) + (b^2 -  a^2) \frac{d}{da} ( \frac{1}{(a^2 + b^2)^2}  )

\frac{d^2}{da^2} \kappa = \frac{1}{(a^2 + b^2)^2} ( -2  a ) + (b^2 -  a^2) (-2) ( \frac{1}{(a^2 + b^2)^3}  ) (2a)

\frac{d^2}{da^2} \kappa = \frac{-2  a}{(a^2 + b^2)^2} + (b^2 -  a^2) (-2) ( \frac{1}{(a^2 + b^2)^3}  ) (2a)

We dcan skip solving the equation noting that, if a=b, then

b^2 -  a^2 = 0

at this point, this give us only the first term

\frac{d^2}{da^2} \kappa = \frac{- 2  a}{(a^2 + a^2)^2}

if a is greater than zero, this means that the second derivative is negative, and the point is a minimum

the value of kappa is

\kappa = \frac{b}{b^2 + b^2}

\kappa = \frac{b}{2* b^2}

\kappa = \frac{1}{2 b}

3 0
3 years ago
The Moon and Earth rotate about their common center of mass, which is located about RcM 4700 km from the center of Earth. (This
erica [24]

To solve this problem it is necessary to apply the concepts related to gravity as an expression of a celestial body, as well as the use of concepts such as centripetal acceleration, angular velocity and period.

PART A) The expression to find the acceleration of the earth due to the gravity of another celestial body as the Moon is given by the equation

g = \frac{GM}{(d-R_{CM})^2}

Where,

G = Gravitational Universal Constant

d = Distance

M = Mass

R_{CM} = Radius earth center of mass

PART B) Using the same expression previously defined we can find the acceleration of the moon on the earth like this,

g = \frac{GM}{(d-R_{CM})^2}

g = \frac{(6.67*10^{-11})(7.35*10^{22})}{(3.84*10^8-4700*10^3)^2}

g = 3.4*10^{-5}m/s^2

PART C) Centripetal acceleration can be found throughout the period and angular velocity, that is

\omega = \frac{2\pi}{T}

At the same time we have that centripetal acceleration is given as

a_c = \omega^2 r

Replacing

a_c = (\frac{2\pi}{T})^2 r

a_c = (\frac{2\pi}{26.3d(\frac{86400s}{1days})})^2 (4700*10^3m)

a_c = 3.34*10^{-5}m/s^2

3 0
3 years ago
Neon has 10 protons and 10 electrons. How many additional electrons would it take to fill the outer (second) electron shell?
Yakvenalex [24]
None because Neon is already stable
8 0
3 years ago
When 1 kg of water and 1 kg of wood absorb the same amount of heat, the change in temperature of the wood is greater than the ch
vampirchik [111]

Your Answer Will Be C

8 0
4 years ago
Early psychologists were predominately all of the following except __________. A. white B. male C. European D. wealthy
yulyashka [42]

Answer: D. wealthy

Explanation: on the e2020 test its right

7 0
4 years ago
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