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Scorpion4ik [409]
2 years ago
8

Explain your answers to 9a and 9b in terms of Newton's laws of motion.

Physics
1 answer:
kogti [31]2 years ago
4 0

Answer:i One way to solve the quadratic equation x2  = 9 is to subtract 9 from both sides to get one side equal to 0: x2  – 9 = 0. The expression on the left can be factored:

Explanation:

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Identify the relationship between kinetic energy and gravitational potential energy for the cyclist at each position
sasho [114]

Answer:

follows are the responses to the given question:

Explanation:

The kinetic energy of every object moving.  PK=m \times v^2 any entity lifted against the strength of gravity stores elastic potential of gravity.

PE = height \times mass  \times  g

Its total power of an independent device stays constant underneath the Mass conservation on Energy and it is, the kinetic energy plus potential energy is just like a fixed (KE+PE=Standard). So, if KE improves, it is valid that PE declines.

If the PE is now at least then KE has been at the highest. It is also valid that KE is reduced as PE is increased as well as the maximum PE, the minimum KE.

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The upper part of the mantle and the crust are called the__________
swat32

Answer: Lithosphere

Explanation:

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yanalaym [24]
Uh.. what's the question..?
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3 years ago
Each blade of a fan has a radius of 11 inches. If the fan’s rate of turn is 1440o /sec, find the following. (a) The angular spee
notka56 [123]

Answer:

a) 24.43 radians per second

b) 268.73 inches per second

Explanation:

a) The angular speed of the fan on Celsius degrees/second is 1400, so we should convert that value to radians using the fact that 2π rad = 360 °C:

\omega = 1400\frac{C}{s}=1400\frac{C}{s}*\frac{2\pi\,rad}{360\,C}

\omega = 1400\frac{C}{s}=24.43\frac{rad}{s}

b) Linear speed on a point of the blade is related with angular speed of the fan by the equation

v=\omega r

with v linear speed, ω angular speed and r the radius of the blades. So:

v=(24.43\frac{rad}{s})(11 in)

Radians isn't really a unity; it is dimensionless so we can put it or not. So:

v=268.73\frac{in}{s}

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Low-frequency red light is used in dark rooms because it does not expose the film. This is because the red light has a low
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energy. It will not disrupt the picture developing process by overexposing too much light on the film.

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