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Firlakuza [10]
3 years ago
5

Light of frequency 2.5 x 1015 Hz illuminates a

Physics
1 answer:
Nikitich [7]3 years ago
3 0

Answer:

h f = W + KE

Input energy equals work function plus KE of emitted electron

W = 6.63E-34 * 2.5E15 - 6.3 * 1.6E-19

W = 6.63 * 2.5 * 10^-19 - 10.1 * E-19 ev     (1ev = 1.6E-19 J)

W = (16.6 - 10.1)E-19 = 6.5E-19 J

h f = 6.5E-19 J        for electrons to be emitted with zero KE

f = 6.5E-19 / 6.63E-34 = .98E-15 / sec = 9.8E-14 / sec  (threshold)

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What process changes a liquid to a solid?
kipiarov [429]

Answer:

D. Freezing?

Explanation:

Get water, put it in the freezer, turns into ice after a few hours.

5 0
3 years ago
for a body moving in a straight line / its distance and displacement can be same. justify with an example?
Tema [17]
The distance of an object is the total distance it travels, while the displacement of the object is how far away the object is from the starting point.

Because the body is moving in a STRAIGHT line, that means it does not change directions, therefore when the body gets to the destination, the total distance will be the same as the displacement. If the body were to change directions, then the magnitude of the vectors will need to be added up and calculated.

For example, let's say you are walking to your friends house directly across the street from your house. All you need to do is walk in a strsight line from the front of your house and you will get to your friends house. The distance and displacement will be the same.

Now if your friend lived right across the street, but 5 houses down, and you cross the street directly from your house then turn in another direction and walk straight, then the distance in this case will quite likely be greater than the displacement because the displacement is the distance from your house to your friends house when measured diagonally.
6 0
3 years ago
A charge Q is placed on the x axis at x = +4.0 m. A second charge q is located at the origin. If Q = +75 nC and q = −8.0 nC, wha
Stells [14]

Answer:

23.1 N/C

Explanation:

OP = 3 m , OQ = 4 m

PQ = \sqrt{4^{2}+3^{2}}=5 m

q = - 8 nC, Q = 75 nC

Electric field at P due to the charge Q is

E_{1}=\frac{KQ}{PQ^{2}}=\frac{9\times 10^{9}\times 75\times 10^{-9}}{25}=27 N/C

Electric field at P due to the charge q is

E_{2}=\frac{Kq}{PO^{2}}=\frac{9\times 10^{9}\times 8\times 10^{-9}}{9}=8 N/C

According to the diagram, tanθ = 3/4

Resolve the components of E1 along x axis and along y axis.

So, Electric field along X axis, Ex = - E1 Cos θ

Ex = - 27 x 4 / 5 = - 21.6 N/C

Electric field along y axis, Ey = E1 Sinθ - E2

Ey = 27 x 3 /5 - 8 = 8.2 N/C

The resultant electric field at P is given by

E=\sqrt{E_{x}^{2}+E_{y}^{2}}=\sqrt{(-21.6)^{2}+(8.2)^{2}}=23.1 N/C

3 0
3 years ago
What are the things on top of the titanic called?
KengaRu [80]
Simple answer: they are like large chimneys for the fuel's excess gas.
6 0
3 years ago
A 1600N force is applied to a 85kg mass what is the acceleration of the mass?
Fantom [35]
Newton 3rd law:
F=ma
1600=85*a
1600/85=a
a=18.82 m/s^2
4 0
3 years ago
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