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3 years ago

6

The pivot point of a lever is called the fulcrum. please select the best answer from the choices provided t f user: the thread a

round a screw is a lever. please select the best answer from the choices provided t f

2 answers:

kkurt [141]3 years ago

4 0

It is TRUE. The pivot point of a lever is called the fulcrum .

Andreas93 [3]3 years ago

3 0

Answer:

true

Explanation:

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There is 4000 balls in the earth of the world why is that so hard for you

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1 year ago

As you may well know, placing metal objects inside a microwave oven can generate sparks. Two of your friends are arguing over th

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Answer:

$5.04\cdot 10^8 A$

Explanation:

The work function of the metal corresponds to the minimum energy needed to extract a photoelectron from the metal. In this case, it is:

$\phi = 3.950\cdot 10^{-19}J$

So, the energy of the incoming photon hitting on the metal must be at least equal to this value.

The energy of a photon is given by

$E=\frac{hc}{\lambda}$

where

h is the Planck's constant

c is the speed of light

$\lambda$ is the wavelength of the photon

Using $E=\phi$ and solving for $\lambda$ , we find the maximum wavelength of the radiation that will eject electrons from the metal:

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3 years ago

A lab cart with a mass of 15 kg is moving with constant velocity, v, along a straight horizontal track. A student drops a 2 kg m

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The equation $15v_{i} + 2*0 = (15 + 2)v_{f}$ (option 3) represents the horizontal momentum of a 15 kg lab cart moving with a constant velocity, v, and that continues moving after a 2 kg object is dropped into it.

The horizontal momentum is given by:

$p_{i} = p_{f}$

$m_{1}v_{1}_{i} + m_{2}v_{2}_{i} = m_{1}v_{1}_{f} + m_{2}v_{2}_{f}$

Where:

m₁: is the mass of the lab cart = 15 kg

m₂: is the <em>mass </em>of the object dropped = 2 kg

$v_{1}_{i}$ : is the initial velocity of the<em> lab cart </em>

$v_{2}_{i}$ : is the <em>initial velocit</em>y of the <em>object </em>= 0 (it is dropped)

$v_{1}_{f}$ : is the final velocity of the<em> lab cart </em>

$v_{2}_{f}$ : is the <em>final velocity</em> of the <em>object </em>

Then, the horizontal momentum is:

$15v_{1}_{i} + 2*0 = 15v_{1}_{f} + 2v_{2}_{f}$

When the object is dropped into the lab cart, the final velocity of the lab cart and the object <u>will be the same</u>, so:

$15v_{1}_{i} + 2*0 = v_{f}(15 + 2)$

Therefore, the equation $15v_{i} + 2*0 = (15 + 2)v_{f}$ represents the horizontal momentum (option 3).

Learn more about linear momentum here:

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I hope it helps you!

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2 years ago

Please help, thanks!

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3 years ago

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