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Ilya [14]
3 years ago
12

If I hit myself, and it hurts, does that make me weak or strong? A. Yes B. No

Physics
2 answers:
cestrela7 [59]3 years ago
8 0

Answer:

Yes

Explanation:

I've done plenty of research by hitting myself repeatedly. It does hurt after awhile.

Bumek [7]3 years ago
4 0

Yes but no here's why if you hit yourself you mean to inflict pain, so your brain transmits the pain to where you hit yourself, so I would think it would be answer b.

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How does kinetic energy affect the stopping distance of a vehicle traveling at 30 mph compared to the same vehicle traveling at
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Both cars start from the same point. Which describes the motion shown?
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3 years ago
Read 2 more answers
(a) Calculate the magnitude of the gravitational force exerted by Mars on a 80 kg human standing on the surface of Mars. (The ma
andrew11 [14]

Answer:

a) F=1.044\times 10^9\ N

b)F'=1.044\times 10^9\ N

c) F_p=1.0672\times10^{-7}\ N

d) Treat the humans as though they were points or uniform-density spheres.

Explanation:

Given:

  • mass of Mars, M=6.4\times 10^{23}\ kg
  • radius of the Mars, r=3.4\times 10^{6}\ m
  • mass of human, m=80\ kg

a)

Gravitation force exerted by the Mars on the human body:

F=G.\frac{M.m}{r^2}

where:

G=6.67 \times 10^{-11}\ m^3.kg^{-1}.s^{-2} = gravitational constant

F=6.67\times10^{-11}\times \frac{6.4\times 10^{23}\times 80}{(3.4\times 10^{6})^2}

F=1.044\times 10^9\ N

b)

The magnitude of the gravitational force exerted by the human on Mars is equal to the force by the Mars on human.

F'=F

F'=1.044\times 10^9\ N

c)

When a similar person of the same mass is standing at a distance of 4 meters:

F_p=6.67\times10^{-11}\times \frac{80\times 80}{4}

F_p=1.0672\times10^{-7}\ N

d)

The gravitational constant is a universal value and it remains constant in the Universe and does not depends on the size of the mass.

  • Yes, we have to treat Mars as spherically symmetric so that its center of mass is at its geometric center.
  • Yes, we also have to ignore the effect of sun, but as asked in the question we have to calculate the gravitational force only due to one body on another specific body which does not brings sun into picture of the consideration.
4 0
3 years ago
In the ground state of hydrogen, according to the Bohr model, an electron orbits 5.3 x 10-11 m from the nucleus. It undergoes a
Readme [11.4K]

Answer:

Explanation:

Given

radius of electron(r)=5.3\times 10^{-11} m

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we know

a_c=\frac{v^2}{r}

v=\sqrt{r\times a_c}

v=\sqrt{5.3\times 10^{-11}\times 9\times 10^{22}}

v=\sqrt{47.7\times 10^11}

v=21.84\times 10^5 m/s

(b)For n=10

r=100\times 5.3\times 10^{-11} m\approx 5.3\times 10^{-9} m

a_c=10^4\times 9\times 10^{22} m/s^2

a_c=9\times 10^{26} m/s^2

v=\sqrt{r\times a_c}

v=\sqrt{9\times 10^{26}\times 5.3\times 10^{-9}}

v=21.84\times 10^8 m/s

8 0
3 years ago
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