Study the four transverse waves shown. Compare the properties of waves B, C, & D to that of wave A.

In the Bohr model of the hydrogen atom, an electron moves in a circular path around a proton. The speed of the electron is appro

blondinia [14]

In order to answer these questions, we need to know the charges on
the electron and proton, and then we need to know the electron's mass.
I'm beginning to get the creepy feeling that, in return for the generous
5 points, you also want me to go and look these up so I can use them
in calculations ... go and collect my own straw to make the bricks with,
as it were.

Ok, Rameses:

Elementary charge . . . . . 1.6 x 10⁻¹⁹ coulomb
negative on the electron
plussitive on the proton

Electron rest-mass . . . . . 9.11 x 10⁻³¹ kg

a). The force between two charges is

F = (9 x 10⁹) Q₁ Q₂ / R²

= (9 x 10⁹ m/farad) (-1.6 x 10⁻¹⁹C) (1.6 x 10⁻¹⁹C) / (5.35 x 10⁻¹¹m)²

      = ( -2.304 x 10⁻²⁸) / (5.35 x 10⁻¹¹)²

=    8.05 x 10⁻⁸ Newton .

b). Centripetal acceleration =

   v² / r .

A = (2.03 x 10⁶)² / (5.35 x 10⁻¹¹)

   =    7.7 x 10²² m/s² .

That's an enormous acceleration ... about 7.85 x 10²¹ G's !
More than enough to cause the poor electron to lose its lunch.

It would be so easy to check this work of mine ...
First I calculated the force, then I calculated the centripetal acceleration.
I didn't use either answer to find the other one, and I didn't use " F = MA "
either.

I could just take the ' F ' that I found, and the 'A' that I found, and the
electron mass that I looked up, and mash the numbers together to see
whether F = M A .

I'm going to leave that step for you.   Good luck !

4 0

3 years ago

A wedge with an inclination of angle θ rests next to a wall. A block of mass m is sliding down the plane. There is no friction b

Softa [21]

Answer:

The net force on the block F(net) = mgsinθ).

Fw =mg(cosθ)(sinθ)

Explanation:

(a)

Here, m is the mass of the block, n is the normal force, \thetaθ is the wedge angle, and Fw is the force exerted by the wall on the wedge.

Since the block sliding down, the net force on the block is along the plane of the wedge that is equal to horizontal component of weight of the block.

F(net) = mgsinθ

The net force on the block F(net) = mgsinθ).

The direction of motion of the block is along the direction of net force acting on the block. Since there is no frictional force between the wedge and block, the only force acting on the block along the direction of motion is mgsinθ.

(b)

From the free body diagram, the normal force n is equal to mgcosθ .

n=mgcosθ

The horizontal component of normal force on the block is equal to force

Fw=n*sin(θ) that exerted by the wall on the wedge.

Substitute mgcosθ for n in the above equation;

Fw =mg(cosθ)(sinθ)

Since, there is no friction between the wedge and the wall, there is component force acting on the wall to restrict the motion of the wedge on the surface and that force is arises from the horizontal component for normal force on the block.

6 0

3 years ago

PLEASE HELP!! ITS DUE AT 11:59!!

DENIUS [597]

Answer:

14.1 po

sana makatulong

3 0

3 years ago

A 1.2 L weather balloon on the ground has a temperature of 25°C and is at atmospheric pressure (1.0 atm). When it rises to an el

Irina-Kira [14]

Answer:

71.19 C

Explanation:

25C = 25 + 273 = 298 K

Applying the ideal gas equation we have

$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$

where P, V and T are the pressure, volume and temperature of the gas at 1st and 2nd stage, respectively. We can solve for the temperature and the 2nd stage:

$T_2 = T_1\frac{P_2V_2}{P_1V_1} = 298\frac{0.77*1.8}{1.2*1} = 298*1.155 = 344.19 K = 344.19 - 273 = 71.19 C$

4 0

4 years ago

What quantity of heat is needed to convert 1 kg of ice at -13 degrees C to steam at 100 degrees C?

Effectus [21]

Answer:

Heat energy needed = 3036.17 kJ

Explanation:

We have

heat of fusion of water = 334 J/g

heat of vaporization of water = 2257 J/g

specific heat of ice = 2.09 J/g·°C

specific heat of water = 4.18 J/g·°C

specific heat of steam = 2.09 J/g·°C

Here wee need to convert 1 kg ice from -13°C to vapor at 100°C

First the ice changes to -13°C from 0°C , then it changes to water, then its temperature increases from 0°C to 100°C, then it changes to steam.

Mass of water = 1000 g

Heat energy required to change ice temperature from -13°C to 0°C

H₁ = mcΔT = 1000 x 2.09 x 13 = 27.17 kJ

Heat energy required to change ice from 0°C to water at 0°C

H₂ = mL = 1000 x 334 = 334 kJ

Heat energy required to change water temperature from 0°C to 100°C

H₃ = mcΔT = 1000 x 4.18 x 100 = 418 kJ

Heat energy required to change water from 100°C to steam at 100°C

H₄ = mL = 1000 x 2257 = 2257 kJ

Total heat energy required

H = H₁ + H₂ + H₃ + H₄ = 27.17 + 334 + 418 +2257 = 3036.17 kJ

Heat energy needed = 3036.17 kJ

5 0

3 years ago