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julia-pushkina [17]
3 years ago
9

you find out that it takes 2 sec. for the swing to complete one cycle. what is the swing`s period and frequency?

Physics
1 answer:
Gnesinka [82]3 years ago
6 0

The time it takes "to complete one cycle" is a perfect definition for <em>period</em>.
So the <em>2 sec</em> that you measured is the swing's period.

Frequency is just   1/period  (the 'reciprocal' of the period).
For this swing, the frequency is   1/(2sec)  =  0.5 per second = <em>0.5 Hz</em>.

( "Hertz" means "per second")


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3 years ago
Block B (of mass m) is initially at rest. Block A (of mass 3m) travels toward B with an initial speed v0 (vee nought) and collid
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Answer:

The maximum height reached by the two blocks is approximately 0.1147959 × v₀²

Explanation:

The mass of block B = m

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The initial velocity of block B, v₂ = 0 m/s

The initial velocity of block A, v₁ = v₀

The amount of friction between the blocks and the surface = Negligible friction

By the Law of conservation of linear momentum, we have;

Total initial momentum = Total final momentum

3·m·v₁ + m·v₂ = (3·m + m)·v₃ = 4·m·v₃

Plugging in the values for the velocities gives;

3·m × v₀ + m × 0 = (3·m + m)·v₃ = 4·m·v₃

∴ 3·m × v₀ =  4·m·v₃

\therefore v_3 = \dfrac{3}{4} \times v_0 = 0.75 \times v_0

The kinetic energy, K.E. of the combined blocks after the collision is given as follows;

K.E. = 1/2 × mass × v²

\therefore K.E. = \dfrac{1}{2} \times 4\cdot m \times \left (\dfrac{3}{4} \cdot v_0 \right )^2 = \dfrac{9}{8} \cdot m\cdot v_0^2

The potential energy, P.E., gained by the two blocks at maximum height = The kinetic energy, K.E., of the two blocks before moving vertically upwards

The potential energy, P.E. = m·g·h

Where;

m = The mass of the object at the given height

g = The acceleration due to gravity

h = The height at which the object of mass, 'm', is located

Therefore, for h = The maximum height reached by the two blocks, we have;

P.E. = K.E.

m \cdot g \cdot h =  \dfrac{9}{8} \cdot m\cdot v_0^2

h = \dfrac{\dfrac{9}{8} \cdot m\cdot v_0^2}{m \cdot g }  = \dfrac{9}{8} \cdot \dfrac{ v_0^2}{ g }  =  \dfrac{9}{8} \cdot \dfrac{ v_0^2}{ 9.8} = \dfrac{42}{392} \cdot  v_0^2 \approx 0.1147959   \cdot  v_0^2

The maximum height reached by the two blocks, h ≈ 0.1147959·v₀².

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2 years ago
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