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slavikrds [6]
3 years ago
11

Ana walks from 4 m to 200 cm. Which of the following statements is true about

Physics
1 answer:
Gemiola [76]3 years ago
3 0
Distance=2m

because 200cm = 2m
so 4m-2m=2m
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1 (3 points)
trapecia [35]

Answer:

300

Explanation:

15x20

5 0
3 years ago
The maximum wavelength that an electromagnetic wave can have and still eject electrons from a metal surface is 542 nm. What is t
Veseljchak [2.6K]

Answer:

2.295 eV

Explanation:

maximum wavelength, λ = 542 nm = 542 x 10^-9 m

The work function of the metal is defined as the minimum amount of energy falling on the metal so that the photo electrons just ejects the surface of metal.

W_{o}=\frac{hc}{\lambda }

where, h is the Plank's constant and c be the speed of light

h = 6.634 x 10^-34 Js

c = 3 x 10^8 m/s

W_{o}=\frac{6.634\times 10^{-34}\times 3\times 10^{8}}{542\times10^{-9} }

W_{o}=3.67\times 10^{-19} J=\frac{3.67\times 10^{-19}}{1.6\times 10^{-19}} eV

Wo = 2.295 eV

Thus, the work function of this metal is 2.295 eV.

5 0
3 years ago
What is the efficient cause if acceleration​
vladimir1956 [14]

Answer:

acceleration= velocity ÷ time

Explanation:

the question is outrageous

4 0
3 years ago
Read 2 more answers
In a container of negligible mass, 0.400 kg of ice at an initial temperature of -29.0 ∘C is mixed with a mass m of water that ha
Pie

Answer:

1 kg

Explanation:

The container has negligible mass and no heat is loss to the surrounding.

Mass of ice = 0.4kg, initial temperature of ice = -29oC, final temperature of the mixture = 26oC, mass of water (m2) = ?kg, initial temperature of water = 80oC, c ( specific heat capacity of water ) = 4200J/kg.K, Lf = heat of fusion of water = 3.36 × 10^5 J/kg

Using the formula:

Quantity of heat gain by ice = Quantity of heat loss by water

Quantity of heat gain by ice = mass of ice × heat of fusion of ice + mass of water × specific heat capacity of water = (0.4 × 3.36 × 10^ 5) + (0.4 × 4200 × (26- (-29) = 13.44 × 10^4 + 9.24 × 10^ 4 = 22.68 × 10^4 J

Quantity of heat loss by water = m2cΔT

Quantity of heat loss by water = m2 ×4200× (80 - 26) = m(226800)

since heat gain = heat loss

22.68 × 10^4 = 226800 m2

divide both side by 226800

226800 / 226800 = m2

m2 = 1 kg

5 0
3 years ago
What are the two factors that affect the frictional force between two surfaces​
Talja [164]

Static friction and normal force? I would Google to double check if I'm right.

6 0
3 years ago
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