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VladimirAG [237]
2 years ago
11

Which is an indication that a man might be choking

Physics
2 answers:
Maksim231197 [3]2 years ago
5 0

Answer:

if they are struggling to breathe, attempting to cough (like coughing something up) or asking for help whilst holding their stomach, neck

nalin [4]2 years ago
4 0

Answer:

If they put their hands near their neck, if they're not breathing properly, if they aren't coughing.

Explanation:

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What happens with the plates at this Juan de fuca
djverab [1.8K]

Answer:

Subduction of the Juan de Fuca plate causes melting and magma generation in the mantle which rises to the surface to create the Cascade volcanoes.

Explanation:

8 0
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Why won't a very bright beam of red light impart more energy to an ejected electron than a feeble beam of violet light?
bearhunter [10]
This is related to the energy carried by photons of light the energy of each photon is proportional to the frequency of the light since red light has a lower frequency then violet light and photons of red light carry less energy than the photons of violet light as a result the red protons eject electrons that have less energy than the ejected electrons by Violet photons
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What are some good biotechnology topics?
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2 years ago
A point charge q is located at the center of a spherical shell of radius a that has a charge −q uniformly distributed on its sur
muminat

Answer:

a) E = 0

b) E =  \dfrac{k_e \cdot q}{ r^2 }

Explanation:

The electric field for all points outside the spherical shell is given as follows;

a) \phi_E = \oint E \cdot  dA =  \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}

From which we have;

E \cdot  A =  \dfrac{{\Sigma Q}}{\varepsilon _{0}} = \dfrac{+q + (-q)}{\varepsilon _{0}}  = \dfrac{0}{\varepsilon _{0}} = 0

E = 0/A = 0

E = 0

b) \phi_E = \oint E \cdot  dA =  \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}

E \cdot  A  = \dfrac{+q }{\varepsilon _{0}}

E  = \dfrac{+q }{\varepsilon _{0} \cdot A} = \dfrac{+q }{\varepsilon _{0} \cdot 4 \cdot \pi \cdot r^2}

By Gauss theorem, we have;

E\oint dS =  \dfrac{q}{\varepsilon _{0}}

Therefore, we get;

E \cdot (4 \cdot \pi \cdot r^2) =  \dfrac{q}{\varepsilon _{0}}

The electrical field outside the spherical shell

E =  \dfrac{q}{\varepsilon _{0} \cdot (4 \cdot \pi \cdot r^2) }= \dfrac{q}{4 \cdot \pi \cdot \varepsilon _{0} \cdot r^2 }=  \dfrac{q}{(4 \cdot \pi \cdot \varepsilon _{0} )\cdot r^2 }

k_e=  \dfrac{1}{(4 \cdot \pi \cdot \varepsilon _{0} ) }

Therefore, we have;

E =  \dfrac{k_e \cdot q}{ r^2 }

5 0
2 years ago
Using the image provided, which moon phase is next?
ella [17]

Answer:

Waning Crescent

5 0
2 years ago
Read 2 more answers
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