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3 years ago

What was the acceleration of the 2 kg cart?

Physics

2 answers:

Setler79 [48]3 years ago

7 0

It was <em>(1/2) (Net force on the cart) m/s²) </em>.

inysia [295]3 years ago

6 0

Answer:

it is b

Explanation:

girl just go with it i say that it is right so just put b

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o illustrate the work-energy concept, consider the case of a stone falling from xi to xf under the influence of gravity. Using t

Phantasy [73]

Answer: force of gravity on the body due to height difference above the earth's surface

Explanation: as you increase the height of a body above ground, you do work against gravity in moving it from a point on the earth's surface to that point. So a body falling has a stored up gravito-potential energy which acts on it downward due to its mass, accelerating it downwards

Answer b): kinetic energy of the body

Explanation: the downward force produces an acceleration of magnitude 9.81m/s2 downwards which means an increasing velocity. This increasing velocity means the kinetic energy of the body is increasing (kinetic energy is proportional to velocity of the body squared)

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3 years ago

A photoelectric effect experiment finds a stopping potential of 1.93 V when light of wavelength 200 nm is used to illuminate the

GenaCL600 [577]

a) Zinc (work function: 4.3 eV)

The equation for the photoelectric effect is:

$E=\phi + K$ (1)

where

$E=\frac{hc}{\lambda}$ is the energy of the incident photon, with

h = Planck constant

c = speed of light

$\lambda$ = wavelength

$\phi$ = work function of the metal

K = maximum kinetic energy of the photoelectrons emitted

The stopping potential (V) is the potential needed to stop the photoelectrons with maximum kinetic energy: so, the corresponding electric potential energy must be equal to the maximum kinetic energy,

$eV=K$

So we can rewrite (1) as

$E=\phi + eV$

where we have:

$\lambda=200 nm = 2\cdot 10^{-7} m$

V = 1.93 V

e is the electron charge

First of all, let's find the energy of the incident photon:

$E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{2\cdot 10^{-7}m}=9.95\cdot 10^{-19} J$

Converting into electronvolts,

$E=\frac{9.95\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=6.22 eV$

And now we can solve eq.(1) to find the work function of the metal:

$\phi = E-eV=6.22 eV-1.93 eV=4.29 eV$

so, the metal is most likely zinc, which has a work function of 4.3 eV.

b) The stopping potential is still 1.93 V

Explanation:

The intensity of the incident light is proportional to the number of photons hitting the surface of the metal. However, the energy of the photons depends only on their frequency, so it does not depend on the intensity of the light. This means that the term E in eq.(1) does not change.

Moreover, the work function of the metal is also constant, since it depends only on the properties of the material: so $\phi$ is also constant in the equation. As a result, the term (eV) must also be constant, and therefore V, the stopping potential, is constant as well.

6 0

4 years ago

A solid cylinder with a mass of 2.72 kg and a radius of 0.083 m starts from rest at a height of 4.20 m and rolls down a 88.7 ◦ s

Bingel [31]

Explanation:

According to the law of conservation of energy ,

Potential energy = kinetic energy

$mgh = \frac{1}{2} \times mv^{2} + \frac{1}{2} \times I \times \omega^{2}$