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3 years ago

The fulcrum is the

Physics

1 answer:

olga_2 [115]3 years ago

6 0

Answer:

point of support on which a lever rotates.

Explanation:

The fulcrum is the point of support on which a lever rotates. Fulcrum is a pivotal part of simple machines.

The fulcrum provides the platform for a lever to torque.

The force that opposes motion by the applied force is termed the frictional force.
Friction is a force that opposes motion.
The stored energy of an object is its potential energy.
The potential energy is the energy due to the position of a body.
The distance an object moves when doing work is termed its displacement.

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What is the work function of cadmium metal if light with = 9.85×1014hz is necessary to eject electrons?

Alika [10]

We use the Planck’s formula:

E = hv

where,

E = energy, h = planck’s constant = 6.6x10^-34 J s, v = frequency in Hz (Hz = 1 / s)

Subsituting the values to find for E:

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3 years ago

The United States and South Korean soccer teams are playing in the first round of the World Cup. An American kicks the ball, giv

Aleks04 [339]

Answer:v=2.82 m/s

Explanation:

Given

initial velocity of ball(u)=3.6 m/s

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3 years ago

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irinina [24]

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4 years ago

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A cylinder of mass mm is free to slide in a vertical tube. The kinetic friction force between the cylinder and the walls of the

Zinaida [17]

Answer:

$y = \frac{-f +/- \sqrt{f^{2} +2kmg}}{k}$

Explanation:

Let y₀ be the initial position of the cylinder when the spring is attached and y its position when it is momentarily at rest.From work-kinetic energy principles, The work done by the spring force + work done by friction + work done by gravity = kinetic energy change of the cylinder

work done by the spring force = ¹/₂k(y₀² - y²)

work done by friction = - f(y - y₀)

work done by gravity = mg(y - y₀)

kinetic energy change of the cylinder = ¹/₂m(v₁² - v₀²)

So ¹/₂k(y₀² - y²) - f(y - y₀) + mg(y - y₀) = ¹/₂m(v₁² - v₀²)

Since the cylinder starts at rest, v₀ = 0. Also, when it is momentarily at rest, v₁ = 0

¹/₂k(y₀² - y²) - f(y - y₀) + mg(y - y₀) = ¹/₂m(0² - 0²)

¹/₂k(y₀² - y²) - f(y - y₀) + mg(y - y₀) = 0

¹/₂ky₀² + fy₀ - mgy₀ -¹/₂ky² - fy + mgy = 0

¹/₂ky₀² + fy₀ - mgy₀ = ¹/₂ky² + fy - mgy

Let y₀ = 0, then the left hand side of the equation equals zero. So,

0 = ¹/₂ky² + fy - mgy

¹/₂ky² + fy - mgy = 0

Using the quadratic formula

$y = \frac{-f +/- \sqrt{f^{2} - 4 X\frac{k}{2} X -mg}}{2 X \frac{k}{2} }\\ y = \frac{-f +/- \sqrt{f^{2} +2kmg}}{k}$

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3 years ago