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elixir [45]
3 years ago
10

In a semiclassical model of the hydrogen atom, the electron orbits the proton at a distance of 0.053 nm. part a what is the elec

tric potential of the proton at the position of the electron? express your answer with the appropriate units.
Physics
1 answer:
Anestetic [448]3 years ago
3 0
Potential energy is energy that is found in a system, grounded on the position of objects. The Coulomb (C) is the unit of charge, and the unit of electric potential is the Volt (V), which is equivalent to (J/C) or Joule per Coulomb.So the formula for this is potential = kQ / d, plugging in the given from the questions will give us:potential = 8.99e9N·m²/C² * 1.602e-19C / 0.053e-9m = 27 V 
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Please give an answer that is coherent
Semmy [17]

Answer:

480

Explanation:

resistance equals to potential difference divide by electric current

120÷0.25

=480

6 0
3 years ago
We know that the magnitudes of the negative charge on the electron and the positive charge on the proton are equal. Suppose, how
TEA [102]

Answer:3,45 x 10^9 N

Explanation: We have considered the total charge for each coin , this is the total atoms x 29 electrons for cooper and multiplier by electron charge, the total charge for each coin is 0,464 C

Finally we use the Coulomb law,

F=k Q/ (r)^2

6 0
3 years ago
Calculate the mass (in SI units) of (a) a 160 lb human being; (b) a 1.9 lb cockatoo. Calculate the weight (in English units) of
kondaur [170]

Explanation:

1. Force applied on an object is given by :

F = W = mg

(a) A 160 lb human being, F = 160 lb

g = acceleration due to gravity, g = 32 ft/s²

m=\dfrac{F}{g}

m=\dfrac{160\ lb}{32\ ft/s^2}

m = 5 kg

(b) A 1.9 lb cockatoo, F = 1.9 lb

m=\dfrac{F}{g}

m=\dfrac{1.9\ lb}{32\ ft/s^2}

m = 0.059 kg

2. (a) A 2300 kg rhinoceros, m = 2300 kg

W=2300\ kg\times 32\ ft/s^2=73600\ lb

(b) A 22 g song sparrow, m = 22 g = 0.022 kg

W=0.022\ kg\times 32\ ft/s^2=0.704\ lb

Hence, this is the required solution.

5 0
3 years ago
A hiker begins a trip by first walking 25.0 km southeast from her car. She stops and sets up her tent for the night. On the seco
Sunny_sXe [5.5K]

Answer:

(a)

x_1=17.68km\\y_1=-17.68km\\x_2=20km\\y_2=34.64km

(b) r=41.32km

\alpha =24.23^o

Explanation:

Let us take the north direction to be the positive y-axis and the east to be positive x-axis.

First day:

25.0 km southeast, which implies 45^o south of east. The y-component will be negative and the x-component will be positive.

x_1=25cos45^o=17.68km

y_1=-25sin45^o=-17.68km

Second day:

She starts off at the stopping point of last day. This time, both the y- and x-components are positive.

x_2=40cos60^o=20km

y_2=40sin60^o=34.64km

Therefore, total displacements:

x=x_1+x_2=(17.68+20)km=37.68km

y=y_1+y_2=(-17.68+34.64)km=16.96km

Magnitude of displacements,

r=\sqrt{x^2+y^2}=41.32km

Direction,

\alpha =tan^{-1}(\frac{y}{x})=24.23^o

6 0
4 years ago
What are the signs of the velocity, force, and acceleration (components) afterthe cart is released and is moving qway from the m
SVETLANKA909090 [29]

Answer:

car is moving away its direction is negative.

the speed must also be negative.

speed this distance decreases the acceleration is negative and if the speed is increasing the acceleration is positive.

Explanation:

In the exercise they indicate that the direction to the motion sensor is positive, as they indicate that the car is moving away its direction is negative.

The speed of the car is

           v = (x₂-x₁) / t

As the positions are negative, and the car moves away the speed must also be negative.

The analysis for acceleration must be very careful if the speed this distance decreases the acceleration is negative and if the speed is increasing the acceleration is positive.

3 0
3 years ago
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