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ankoles [38]
3 years ago
10

Calculate the tangential speed of a yo-yo twirled at the end of a 4 meter long string at 2 revolutions per second.

Physics
1 answer:
Oliga [24]3 years ago
8 0

Answer: 50 m/s

Explanation: speed v = 2· pi·n·r = 2· 3.14· 2 s^-1· 4 m

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How quickly do muscles become fatigued?​
agasfer [191]

Answer:

Fatigue is usually defined as the reversible decline of performance during activity, and most recovery occurs within the first hour. However, there is also a slowly reversible component that can take several days to reverse (155). Muscle injury also causes a decline in performance that reverses only very slowly.

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2 years ago
How does the input distance of a third-class lever compare to the output distance​
Alexandra [31]

Answer:

A first-class lever: fulcrum is between input and output force; second-class lever: output force is between input force and fulcrum; third-class lever: input force is between fulcrum and output force

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2 years ago
Conversion of one form of energy into another form is ​
Artist 52 [7]

Answer: energy transformation

5 0
2 years ago
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A particle leaves the origin with a speed of 3 106 m/s at 38 degrees to the positive x axis. It moves in a uniform electric fiel
Salsk061 [2.6K]

Answer:

If the particle is an electron E_y = 3.311 * 10^3 N/C

If the particle is a proton, E_y = 6.08 * 10^6 N/C

Explanation:

Initial speed at the origin, u = 3 * 10^6 m/s

\theta = 38^0 to +ve x-axis

The particle crosses the x-axis at , x = 1.5 cm = 0.015 m

The particle can either be an electron or a proton:

Mass of an electron, m_e = 9.1 * 10^{-31} kg

Mass of a proton, m_p = 1.67 * 10^{-27} kg

The electric field intensity along the positive y axis E_y, can be given by the formula:

E_y = \frac{2 m u^2 sin \theta cos \theta}{qx} \\

If the particle is an electron:

E_y = \frac{2 m_e u^2 sin \theta cos \theta}{qx} \\

E_y = \frac{2 * 9.1 * 10^{-31} * (3*10^6)^2 *(sin38)( cos38)}{1.6*10^{-19} * 0.015} \\

E_y = 3311.13 N/C\\E_y = 3.311 * 10^3 N/C

If the particle is a proton:

E_y = \frac{2 m_p u^2 sin \theta cos \theta}{qx} \\

E_y = \frac{2 * 1.67 * 10^{-27} * (3*10^6)^2 *(sin38)( cos38)}{1.6*10^{-19} * 0.015} \\

E_y = 6.08 * 10^6 N/C

8 0
3 years ago
Convert 14 mg to kg ....<br>​
____ [38]
<h2>♨ANSWER♥</h2>

14mg = 14 × 10^-3 g

= 14 × 10^-3 / 10^3 kg

= 14 × 10^-6 kg

= 0.014 kg

<u>☆</u><u>.</u><u>.</u><u>.</u><u>hope this helps</u><u>.</u><u>.</u><u>.</u><u>☆</u>

_♡_<em>mashi</em>_♡_

7 0
1 year ago
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