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MrMuchimi
3 years ago
13

Order the layers, with the oldest at the bottom and most recent at the top.

Physics
1 answer:
givi [52]3 years ago
7 0

Answer:

adbce i think

Explanation:

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What is the SPEED of a ball that travels 120 m in 6s?
AVprozaik [17]

Answer:

20m/second

Explanation:

The reason the answer is 20m/second is because to find the speed of the ball  in this question you have to divide the distance over the time giving you the result of 20m/second

7 0
3 years ago
The work function for tungsten metal is 4.52eV a. What is the cutoff (threshold) wavelength for tungsten? b. What is the maximum
Tanya [424]

Answer: a) 274.34 nm; b) 1.74 eV c) 1.74 V

Explanation: In order to solve this problem we have to consider the energy balance for the photoelectric effect on tungsten:

h*ν = Ek+W ; where h is the Planck constant, ek the kinetic energy of electrons and W the work funcion of the metal catode.

In order to calculate the cutoff wavelength we have to consider that Ek=0

in this case  h*ν=W

(h*c)/λ=4.52 eV

λ= (h*c)/4.52 eV

λ= (1240 eV*nm)/(4.52 eV)=274.34 nm

From this h*ν = Ek+W;  we can calculate the kinetic energy for a radiation wavelength of 198 nm

then we have

(h*c)/(λ)-W= Ek

Ek=(1240 eV*nm)/(198 nm)-4.52 eV=1.74 eV

Finally, if we want to stop these electrons we have to applied a stop potental equal to 1.74 V . At this potential the photo-current drop to zero. This potential is lower to the catode, so this  acts to slow down the ejected electrons from the catode.

5 0
3 years ago
Anna Litical analyzes the force between a planet and its moon, varying the mass of
Travka [436]

Answer:

Trial 1 is the largest, trial 3 is the smallest

Explanation:

Given:

<em>Trial 1</em>

M₁ = 6·10²² kg

d₁ = 3 500 km = 3.5·10⁶ м

<em>Trial  2</em>

M₂ = 6·10²² kg

d₂ = 7 000 km = 7·10⁶ м

<em>Trial  3</em>

M₃ = 3·10²² kg

d₃ = 7 000 km = 7·10⁶ м

___________

F - ?

Gravitational force:

F₁ = G·m·M₁ / d₁² = m·6.67·10⁻¹¹·6·10²² / (3.5·10⁶)² = 0.37·m  (N)

F₂ = G·m·M₂ / d₂² = m·6.67·10⁻¹¹·6·10²² / (7·10⁶)² = 0.08·m  (N)

F₃ = G·m·M₃ / d₃² = m·6.67·10⁻¹¹·3·10²² / (7·10⁶)² = 0.04·m  (N)

Trial 1 is the largest, trial 3 is the smallest

5 0
1 year ago
What displacement in cm would occur with a 75 N/m spring if you placed a 300 N weight on the spring?
ollegr [7]

Surface tension=75N/m

Weight=300N

\\ \bull\tt\longmapsto Surface\:Tension=\dfrac{Weight}{Displacement}

\\ \bull\tt\longmapsto Displacement=\dfrac{Weight}{Surface\:Tension}

\\ \bull\tt\longmapsto Displacement=\dfrac{300}{75}

\\ \bull\tt\longmapsto Displacement=4m

\\ \bull\tt\longmapsto Displacement=400cm

6 0
3 years ago
A 100-n force has a horizontal component of 80 n and a vertical component of 60 n. the force is applied to a box which rests on
iragen [17]
<span>160 Joules

   For this problem, we can ignore the vertical component of the applied force and focus on only the horizontal component of 80 N and since work is defined as force over distance, let's multiply the force by the distance: 80 N * 2.0 m = 160 Nm = 160 kg*m^2/s^2 = 160 Joules.

   So the cart has a final kinetic energy of 160 Joules.</span>
6 0
3 years ago
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