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Yanka [14]
4 years ago
14

An undamped 2.89-kg horizontal spring oscillator has a spring constant of 24.3 n/m. while oscillating, it is found to have a spe

ed of 3.89 m/s as it passes through its equilibrium position. what is its amplitude of oscillation?
Physics
1 answer:
ololo11 [35]4 years ago
4 0
The speed of the system when it passes through the equilibrium position is v=3.89 m/s, and this corresponds to the maximum speed of the oscillator.

The angular frequency of the oscillator is given by
\omega= \sqrt{ \frac{k}{m} }
where k is the spring constant, and m the mass. Substituting the data, we find
\omega= \sqrt{ \frac{24.3 N/m}{2.89 kg} }=2.9 s^{-1}

and since the amplitude of the oscillation, A, is related to the maximum speed by
v_{max}=\omega A
we can re-arrange this equation to calculate A:
A= \frac{v_{max}}{\omega}= \frac{3.89 m/s}{2.9 s^{-1}}=1.34 m
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What happens if a mid-ocean ridge occurs on land
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Answer:

Despite being such prominent feature on our planet, much of the mid-ocean ridge system remains a mystery. While we have mapped about half of the global mid-ocean ridge in high resolution, less than one percent of the mid-ocean ridge has been explored in detail using submersibles or remotely operated vehicles. so therefore we do not have enough information about them to know what will happen

Explanation:

A mid-ocean ridge or mid-oceanic ridge is an underwater mountain range, formed by plate tectonics. This uplifting of the ocean floor occurs when convection currents rise in the mantle beneath the oceanic crust and create magma where two tectonic plates meet at a divergent boundary. Mid-ocean ridges occur along divergent plate boundaries, where new ocean floor is created as the Earth’s tectonic plates spread apart. As the plates separate, molten rock rises to the seafloor, producing enormous volcanic eruptions of basalt. The speed of spreading affects the shape of a ridge  slower spreading rates result in steep, irregular topography while faster spreading rates produce much wider profiles and more gentle slopes.

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3 years ago
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harkovskaia [24]

Answer:

Socratic app

Explanation:

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7 0
3 years ago
a ball on a string makes 30.0 revolutions in 14.4s, in a circle of radius 0.340m. what is its period.(unit=s)
STALIN [3.7K]

Answer:

0.480 seconds

Explanation:

The period is the time for 1 revolution.  Writing a proportion:

14.4 s / 30.0 rev = t / 1 rev

t = 0.480 s

The period is 0.480 seconds.

7 0
3 years ago
On flat ground, a 70-kg person requires about 300 W of metabolic power to walk at a steady pace of 5.0 km/h (1.4 m/s). Using the
Darina [25.2K]

Answer:

C 350W

Explanation:

Given power output to walk on a flat ground to be 300W, h = 0.05x, v = 1.4m/s

m = 70kg and g =9.8m/s².

x = horizontal distance covered

Total energy used = potential energy used in climbing and the energy used in a walking the horizontal distance.

E = mgh + 300t

Where t is the time taken to cover the distance

x = vt and h = 0.05vt

So

E = mg×0.05×vt + 300t

Substituting respective values

E = 70×9.8×0.05×1.4t +300t = 348t

P = E/t = 348W ≈ 350W.

4 0
4 years ago
A cannon is fired straight up into the air. If the cannon ball comes back down to the launch point in 5 seconds, what was the ma
Nana76 [90]

Answer:

30.63 m

Explanation:

From the question given above, the following data were obtained:

Total time (T) spent by the ball in air = 5 s

Maximum height (h) =.?

Next, we shall determine the time taken to reach the maximum height. This can be obtained as follow:

Total time (T) spent by the ball in air = 5 s

Time (t) taken to reach the maximum height =.?

T = 2t

5 = 2t

Divide both side by 2

t = 5/2

t = 2.5 s

Thus, the time (t) taken to reach the maximum height is 2.5 s

Finally, we shall determine the maximum height reached by the ball as follow:

Time (t) taken to reach the maximum height = 2.5 s

Acceleration due to gravity (g) = 9.8 m/s²

Maximum height (h) =.?

h = ½gt²

h = ½ × 9.8 × 2.5²

h = 4.9 × 6.25

h = 30.625 ≈ 30.63 m

Therefore, the maximum height reached by the cannon ball is 30.63 m

3 0
4 years ago
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