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larisa86 [58]
3 years ago
14

¿Dónde se produce la mayor parte del dióxido de carbono? ¿Por qué?​

Physics
1 answer:
levacccp [35]3 years ago
6 0

Answer:

hi how are you doing today Jasmine

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You are standing 1 meter from a squawking parrot. If you move to a distance three meters away, the sound
spayn [35]
Use the inverse square law, thus if you move a distance of 3m away, the sound intensity decrease by 1/3^2= 1/9
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Analyze how friction can be both a positive and negative aspect on our lives
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Positive - Friction allows us to create heat in a desperate situation, like being lost in the woods. If we didn't have friction, we would probably freeze to death.
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8 0
3 years ago
Two charges are located in the x–y plane. If q1 = -2.90 nC and is located at x = 0.00 m, y = 0.840 m and the second charge has m
Lunna [17]

Answer:

Epx= - 21.4N/C

Epy= 19.84N/C

Explanation:

Electric field theory

The electric field at a point P due to a point charge is calculated as follows:

E= k*q/r²

E= Electric field in N/C

q = charge in Newtons (N)

k= electric constant in N*m²/C²

r= distance from load q to point P in meters (m)

Equivalences

1nC= 10⁻⁹C

known data

q₁=-2.9nC=-2.9 *10⁻⁹C

q₂=5nC=5  *10⁻⁹C

r₁=0.840m

r_{2} =\sqrt{1^{2} +0.8^{2} } =\sqrt{1.64}

sin\beta =\frac{0.8}{\sqrt{1.64} } =0.6246

cos\beta =\frac{1}{\sqrt{1.64} } =0.7808

Calculation of the electric field at point P due to q1

Ep₁x=0

Ep_{1y} =\frac{k*q_{1} }{r_{1}^{2}  } =\frac{8.99*10^{9}*2.9*10^{-9}  }{0.84^{2} } =36.95\frac{N}{C}

Calculation of the electric field at point P due to q2

Ep_{2x} =-\frac{k*q_{2} *cos\beta }{r_{2}^{2}  } =-\frac{8.99*10^{9}*5*10^{-9} *0.7808 }{(\sqrt{1.64})^{2}  } =-21.4\frac{N}{C}

Ep_{2y} =-\frac{k*q_{2} *sin\beta }{r_{2}^{2}  } =-\frac{8.99*10^{9}*5*10^{-9} *0.6242 }{(\sqrt{1.64})^{2}  } =-17.11\frac{N}{C}

Calculation of the electric field at point P(0,0) due to q1 and q2

Epx= Ep₁x+ Ep₂x==0 - 21.4N/C =- 21.4N/C

Epy= Ep₁y+ Ep₂y=36.95 N/C-17.11N =19.84N/C

7 0
3 years ago
Help me please I can't get the final step​
inna [77]

Answer:

\displaystyle m=\frac{2}{3},\ n=\frac{4}{3}

Explanation:

<u>Dimensional Analysis</u>

It's given the relation between quantities A, B, and C as follows:

\displaystyle A=\frac{3}{2}B^mC^n

and the dimensions of each variable is:

A=L^2T^2

B=LT^{-1}

C=LT^2

Substituting the dimensions into the relation (the coefficient is not important in dimension analysis):

\displaystyle L^2T^2=\left(LT^{-1}\right)^m\left(LT^2\right)^n

Operating:

L^2T^2=\left(L^mT^{-m}\right)\left(L^nT^{2n}\right)

L^2T^2=L^{m+m}T^{-m+2n}

Equating the exponents:

m+n=2

-m+2n=2

Adding both equations:

3n=4

Solving:

n=4/3

m=2-4/3=2/3

Answer:

\mathbf{\displaystyle m=\frac{2}{3},\ n=\frac{4}{3}}

6 0
3 years ago
Help me please !!!!!
bezimeni [28]

Answer:

confounding cause they had exposure to many programmes

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3 years ago
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