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amm1812
8 days ago
7

Mary starts from her house, walks 80 meters south, and stops to chat with her aunt on the sidewalk. After chatting for a few min

utes, she walks 125 meters to the west to meet a friend and then walks 45 meters north with her friend to a café for lunch. Overall, she takes 10 minutes to do all of this. What is Mary's average speed? A) 8 m/s B) 1.25 m/s C) 25 m/s D) 4.5 E) 0.42
Physics
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Consider an electron with charge −e and mass m orbiting in a circle around a hydrogen nucleus (a single proton) with charge +e.
kicyunya [3294]

Result:

v=\sqrt{k\frac{e^2}{m_e r}}, 2.18\cdot 10^6 m/s

Explanation:

The electromagnetic attraction between the electron and the proton in the nucleus is equivalent to the centripetal force:

k\frac{(e)(e)}{r^2}=m_e \frac{v^2}{r}

where

k represents the Coulomb constant

e denotes the charge of the electron

e denotes the charge of the proton in the nucleus

r signifies the distance from the electron to the nucleus

v indicates the velocity of the electron

is the mass of the electron

Rearranging for v, we determine

v=\sqrt{k\frac{e^2}{m_e r}}

Inside a hydrogen atom, the distance separating the electron from the nucleus is roughly

r=5.3\cdot 10^{-11}m

while the mass of the electron is

m_e = 9.11\cdot 10^{-31}kg

and the charge is

e=1.6\cdot 10^{-19} C

By plugging in the values into the formula, we achieve

v=\sqrt{(9\cdot 10^9 m/s) \frac{(1.6\cdot 10^{-19} C)^2}{(9.11\cdot 10^{-31} kg)(5.3\cdot 10^{-11} m)}}=2.18\cdot 10^6 m/s

7 0
2 months ago
The δe of a system that releases 12.4 j of heat and does 4.2 j of work on the surroundings is __________ j.
Yuliya22 [3333]

For this issue, the answer is clarified as the system takes in energy (+). The surroundings contribute 84 KJ of work. Whenever a system is receiving work from its surroundings, the value will be positive. Therefore, it sums to 12.4 KJ + 4.2 = 16.6 KJ.

5 0
3 months ago
The newly formed xenon nucleus is left in an excited state. Thus, when it decays to a state of lower energy a gamma ray is emitt
ValentinkaMS [3465]

Answer:3.87*10^-4

Explanation:

To determine the mass reduction, delta mass Xe, of the xenon nucleus due to its decay, we first use the provided wavelength of the gamma radiation to calculate its frequency via c = freq*wavelength.

From C=f*lambda we set up: 3*10^8=f*3.44*10^-12.

Solving gives frequency F=0.87*10^20 Hz.

Next, we calculate the emitted energy using the equation E=hf, which translates to E=f*Planck's constant.

Thus, E=0.87*10^20*6.62*10^-34, resulting in E=575.94*10^(-16).

This energy is then converted from joules to MeV.

Utilizing the formula E=mc^2, with c^2 = 931.5 MeV/u, enables us to find the reduction in mass, yielding

3.87*10^-4 u.

6 0
3 months ago
The air temperature over a lake decreases linearly with height after sunset, since air cools faster than water. The mean molar m
Yuliya22 [3333]

Answer: t = 0.878s

Explanation: A note for you,

since the temperature decreases in a straight line, you can expect the movement speed to also behave linearly. However, this isn't exactly true (referring to the formula). Alternatively, utilize the interpolation principle: (x/v_surface + x/v_top)/2 = t.

While the answer may not match exactly, it should be a close approximation. You can use this formula, thus avoiding large distance calculations.

7 0
4 months ago
Read 2 more answers
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