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guapka [62]
3 years ago
9

Leah watches a toy car move across a table at a constant speed. The car moves across the table in one direction and falls off th

e other side. Leah decides to try another situation. She watches the car move across the table at a constant speed, then when it reaches the other side of the table, her friend Joey pushes the car in the other direction. The car moves back to Leah.
Based on Leah's informal experiment, what forces cause the car to change direction?
A.
balanced forces from Joey pushing the car
B.
unbalanced forces from Joey pushing the car
C.
unbalanced forces from Leah watching the car
D.
balanced forces from the car moving across the table
Physics
2 answers:
Sedbober [7]3 years ago
5 0

In order to change the direction and speed, a net external force is required. A net external force is an unbalanced force which will change the direction and gives the speed in the opposite direction. Hence, its an unbalanced force from the joey that pushes the car in the other direction due to which it the car starts to move back to Leah. Without, unbalanced force there is not change in the direction of the car's motion.

Hence, option B is correct.

zalisa [80]3 years ago
5 0

Answer:

B

Explanation:

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A photoelectric effect experiment finds a stopping potential of 1.93 V when light of wavelength 200 nm is used to illuminate the
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a) Zinc (work function: 4.3 eV)

The equation for the photoelectric effect is:

E=\phi + K (1)

where

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h = Planck constant

c = speed of light

\lambda = wavelength

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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

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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

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

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

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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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