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2 years ago

1.25 is closer to 1.04 or not ? plz heelp plz

Physics

2 answers:

sweet [91]2 years ago

8 0

Answer:

no it's not

Explanation:

it's closer to 1.30 than 1.04

Aloiza [94]2 years ago

3 0

No 1.10 is between both of those numbers

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A 53-N force is needed to keep a 50.0-kg box sliding across a flat surface at a constant velocity. What is the coefficient of ki

earnstyle [38]

The weight of the box is (mass) x (gravity) = (50 kg) x (9.8m/s²) = 490 newtons.

If the box is sliding at constant speed, and not speeding up or slowing down,
that means that the horizontal forces on it add up to zero.

Since you're pushing on it with 53N in that direction, friction must be pulling
on it with 53N in the other direction.

The 53N of friction is (the weight) x (the coefficient of kinetic friction).

53N = (490N) x (coefficient).

Divide each side by 490N : Coefficient = (53N) / (490N) = 0.1082 .

Rounded to the nearest hundredth, that's 0.11 . (choice 'd')

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3 years ago

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A penny rides on top of a piston as it undergoes vertical simple harmonic motion with an amplitude of 4.0cm. If the frequency is

aniked [119]

1) At the moment of being at the top, the piston will not only tend to push the penny up but will also descend at a faster rate at which the penny can reach in 'free fall', in that short distance. Therefore, at the highest point, the penny will lose contact with the piston. Therefore the correct answer is C.

2) To solve this problem we will apply the equations related to the simple harmonic movement, hence we have that the acceleration can be defined as

$a = -\omega^2 A$

Where,

a = Acceleration

A = Amplitude

$\omega$ = Angular velocity

From a reference system in which the downward acceleration is negative due to the force of gravity we will have to

$a = -g$

$-\omega^2 A = -g$

$\omega = \sqrt{\frac{g}{A}}$

From the definition of frequency and angular velocity we have to

$\omega = 2\pi f$

$f = \frac{1}{2\pi} \sqrt{\frac{g}{A}}$

$f = \frac{1}{2\pi} \sqrt{\frac{9.8}{4*10^{-2}}}$

$f = 2.5Hz$

Therefore the maximum frequency for which the penny just barely remains in place for the full cycle is 2.5Hz

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3 years ago

A gymnast is swinging on a high bar. The distance between his waist and the bar is 0.905 m, as the drawing shows. At the top of

Citrus2011 [14]

Answer:

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u = Initial velocity

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$h_f$ = Final height

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