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3 years ago

A bus starts to move from rest. If it is accelerated by 0.8m/s2, calculate the velocity and distance traveled after 8 s.

Physics

1 answer:

Colt1911 [192]3 years ago

6 0

We are given:_______________________________________________

Initial velocity (u) = 0 m/s

acceleration (a) = 0.8 m/s²

time interval (t) = 8 seconds

final velocity (v) = v m/s

distance travelled (s) = s m

Solving for the final velocity:___________________________________

We know that:

v = u + at [first equation of motion]

v = 0 + (0.8)(8) [replacing the variables]

v = 6.4 m/s

Hence, the final velocity is 6.4 m/s

Solving for the distance travelled:_______________________________

We know that:

s = ut + 1/2(at²) [second equation of motion]

s = (0)(8) + 1/2(0.8)(8)(8) [replacing the variables]

s = 1/2(6.4*8)

s = 25.6 m

Hence, the distance travelled is 25.6 m

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here the charge density of metal plate is given as

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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.

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3 years ago

15 points! An atomic nucleus initially moving at 420 m/s emits an alpha particle in the direction of its velocity, and the remai

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where:

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$v_2 = 350 m/s$ is the final velocity of the nucleus

Solving for $v_1$ , we find the final velocity of the alpha particle:

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