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3 years ago

Match the following moves to the correct total distance and displacement of the moves.

1 answer:

5 0

A. Move 2 m east and then 12 m east; displacement is 14 m east and the distance is 14 m
B. Move 10 m east and then 12 m west, the displacement is 2 m west and the distance is 22 m.
C. Move 8 m west and then 16 m east; the displacement is 8 m east and the distance is 24 m
D. Move 12 m west and then 8 m east; the displacement is 4 m and the distance is 20 m

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If the second harmonic of a certain string is 42 Hz, what is the fundamental frequency of the string?

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3 years ago

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3 years ago

How long should a spring be stretched for it to store 45 J of energy? The force constant of the spring is

ELEN [110]

Answer:

x = 0.4 m

Explanation:

When a spring is stretched from its equilibrium position. Some energy is stored in the spring. This energy is called the elastic potential energy of the spring. The formula used to calculate the magnitude of this stored energy is given as follows:

P.E = (1/2)kx²

where,

P.E = Elastic Potential Energy Stored in the spring = 45 J

k = Spring Constant = 540 N/m

x = amount of stretching = ?

Therefore,

45 J = (1/2)(540 N/m)x²

x² = (45 J)(2)/(540 N/m)

x = √(0.167 m²)

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3 years ago

A baseball of mass m = 0.31 kg is spun vertically on a massless string of length L = 0.51m. The string can only support a tensio

natulia [17]

Given data:

* The mass of the baseball is 0.31 kg.

* The length of the string is 0.51 m.

* The maximum tension in the string is 7.5 N.

Solution:

The centripetal force acting on the ball at the top of the loop is,

$\begin{gathered} T+mg=\frac{mv^2}{L}_{} \\ v^2=\frac{L(T+mg)}{m} \\ v=\sqrt[]{\frac{L(T+mg)}{m}} \end{gathered}$

For the maximum velocity of the ball at the top of the vertical circular motion,

$v_{\max }=\sqrt[]{\frac{L(T_{\max }+mg)}{m}}$

where g is the acceleration due to gravity,

Substituting the known values,

$\begin{gathered} v_{\max }=\sqrt[]{\frac{0.51(7.5_{}+0.31\times9.8)}{0.31}} \\ v_{\max }=\sqrt[]{\frac{0.51(10.538)}{0.31}} \\ v_{\max }=\sqrt[]{17.34} \\ v_{\max }=4.16\text{ m/s} \end{gathered}$

Thus, the maximum speed of the ball at the top of the vertical circular motion is 4.16 meters per second.

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1 year ago