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Darya [45]
3 years ago
15

A skier starts from rest and accelerates down a slope 2 at 23 m/s^2. How much time required for the skier to reach a speed of 9.

3 m/a? NEED AN ANSWER show work
​
Physics
1 answer:
gayaneshka [121]3 years ago
6 0

The skier's speed at time <em>t</em> is

<em>v</em> = (23 m/s²) <em>t</em>

To reach a speed of 9.3 m/s, the skier would need

9.3 m/s = (23 m/s²) <em>t</em>

<em>t</em> = (9.3 m/s) / (23 m/s²)

<em>t</em> ≈ 0.404 s

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Two objects moving with the same speed may not have the same velocity. Why?​
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Changing the length of an air column will alter its
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We can model the motion of a dragonfly's wing as simple harmonic motion. The total distance between the upper and lower limits o
PtichkaEL [24]

Answer:

The maximum speed of the wing tip is 1.25 m/s

Explanation:

In simple harmonic motion, the body experiences a restoring force whose magnitude is directly proportional to the distance from its mean position and its direction is towards the mean position.

The speed in the case of simple harmonic motion is given by the equation :

v = Aωsinωt

Here A is maximum displacement, t is time and ω is the angular frequency of oscillation.

For maximum speed, sinωt = 1 . So, the above equation becomes:

v = Aω

But, ω = 2πf , here f is the frequency of oscillation. So,

v = A x 2πf      .....(1)

In this problem,

Frequency of oscillation, f = 40 Hz

Total distance between the upper and lower limits of motion of wing tip is equal to the twice of maximum displacement of the wing tip from the equilibrium position, i.e. ,

2 x A = 1 cm

A = 0.5 cm = 0.5 x 10⁻² m

Substitute the value of f and A in equation (1).

v=2\pi\times40\times0.5\times10^{-2}

v = 1.25 m/s

7 0
3 years ago
A horizontal platform in the shape of a circular disk rotates on a frictionless bearing about a vertical axle through the center
uranmaximum [27]

Answer:

4.36 rad/s

Explanation:

Radius of platform r = 2.97 m

rotational inertia I = 358 kg·m^2

Initial angular speed w = 1.96 rad/s

Mass of student m = 69.5 kg

Rotational inertia of student at the rim = mr^2 = 69.5 x 2.97^2 = 613.05 kg.m^2

Therefore initial rotational momentum of system = w( Ip + Is)

= 1.96 x (358 + 613.05)

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When she walks to a radius of 1.06 m

I = mr^2 = 69.5 x 1.06^2 = 78.09 kg·m^2

Rotational momentuem of system = w(358 + 78.09) = 436.09w

Due to conservation of momentum, we equate both momenta

436.09w = 1903.258

w = 4.36 rad/s

7 0
3 years ago
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