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3 years ago

5

Suppose a photon with an energy of 1.60 eV strikes a piece of metal. If the electron that it hits loses 0.800 eV leaving the met

al, what is the kinetic energy of the electron once it has broken free of the metal surface?

1 answer:

JulsSmile [24]3 years ago

7 0

To solve this problem it is necessary to apply the concepts related to the change of Energy in photons and the conservation of energy.

From the theory we could consider that the energy change is subject to

$\Delta E = E_0 -W_f$

Where

$E_0 =$ Initial Energy

$W_f =$ Energy loses

Replacing we have that

$\Delta E = 1.6-0.8$

$\Delta E = 0.8eV$

Therefore the Kinetic energy of the electron once it has broken free of the metal surface is 0.8eV

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A satellite circles the Earth in an orbit whose radius is twice the Earth’s radius. The Earth’s mass is 5.98 x 1024 kg, and its

gavmur [86]

Hello!

Recall the period of an orbit is how long it takes the satellite to make a complete orbit around the earth. Essentially, this is the same as 'time' in the distance = speed * time equation. For an orbit, we can define these quantities:

$d = 2\pi r$ ← The circumference of the orbit

speed = orbital speed, we will solve for this later

time = period

Therefore:

$T = \frac{2\pi r}{v}$

Where 'r' is the orbital radius of the satellite.

First, let's solve for 'v' assuming a uniform orbit using the equation:
$v = \sqrt{\frac{Gm}{r}}$

G = Gravitational Constant (6.67 × 10⁻¹¹ Nm²/kg²)

m = mass of the earth (5.98 × 10²⁴ kg)

r = radius of orbit (1.276 × 10⁷ m)

Plug in the givens:
$v = \sqrt{\frac{(6.67*10^{-11})(5.98*10^{24})}{(1.276*10^7)}} = 5590.983 m/s$

Now, we can solve for the period:

$T = \frac{2\pi (1.276*10^7)}{5590.983} =\boxed{ 14339.776 s}$

7 0

2 years ago

The stopcock connecting a 2.14 L bulb containing oxygen gas at a pressure of 8.19 atm, and a 9.84 L bulb containing krypton gas

marshall27 [118]

Answer : The final pressure of the system in atm is, 3.64 atm

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

$P\propto \frac{1}{V}$

or,

$P_1V_1+P_2V_2=P_fV_f$

where,

$P_1$ = first pressure = 8.19 atm

$P_2$ = second pressure = 2.65 atm

$V_1$ = first volume = 2.14 L

$V_2$ = second volume = 9.84 L

$P_f$ = final pressure = ?

$V_f$ = final volume = 2.14 L + 9.84 L = 11.98 L

Now put all the given values in the above equation, we get:

$8.19atm\times 2.14L+2.65atm\times 9.84L=P_f\times 11.98L$

$P_f=3.64atm$

Therefore, the final pressure of the system in atm is, 3.64 atm

4 0

3 years ago

You have four fixed-volume containers at STP . Container A has 0.5 mol of gas in 11.2 L. Container B has 2 mol of gas in 22.4 L.

Colt1911 [192]

Answer:

Container A and C

Explanation:

ideal gas equation gives P=nRT/V

so at constant Temperature and pressure, P=n/T

Container A and C after dividing number of moles and Volume, are found to be the same=0.0446

8 0

3 years ago

If we keep on applying force on a material object, can it gain speed of light?

Xelga [282]

Answer:

As an object moves faster, its mass increases. ... Because masses approach infinity with increasing speed, it is impossible to accelerate a material object to (or past) the speed of light. To do so would require an infinite force.

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3 years ago

A hockey puck moving at 0.4600 m/s collides with another puck that was at rest. The pucks have equal mass. The first puck is def

Sladkaya [172]

Answer:

Speed =0.283m/ s

Direction = 47.86°

Explanation:

Since it is a two dimensional momentum question with pucks having the same mass, we derive the momentum in xy plane

MU1 =MU2cos38 + MV2cos y ...x plane

0 = MU2sin38 - MV2sin y .....y plane

Where M= mass of puck, U1 = initial velocity of puck 1=0.46, U2 = final velocity of puck 1 =0.34, V2 = final velocity of puck 2, y= angular direction of puck2

Substitute into equation above

.46 = .34cos38 + V2cos y ...equ1

.34sin38 = V2sin y...equ2

.19=V2cos Y...x

.21=V2sin Y ...y

From x

V2 =0.19/cost

Sub V2 into y

0.21 = 0.19(Sin y/cos y)

1.1052 = tan y

y = 47.86°

Sub Y in to x plane equ

.19 = V2 cos 47.86°

V2=0.283m/s

7 0

3 years ago