un automóvil de 70Cv de potencia se mueve con velocidad de 90km/h constante, ¿cual es la fuerza desarrollada?

Two parallel-plate capacitors, identical except that one has twice the plate separation of the other, are charged by the same vo

erik [133]

Answer:

The capacitor having less distance of separation has a stronger electric field.

Explanation:

The capacitors are identical and only difference between them is that one has twice the plate separation of the other. Therefore, capacitance of the given capacitors C1 and C2 is,

C1= Aε/d and C2=Aε/2d

The charges Q1 and Q2 on the capacitors of capacitance C1 and C2 respectively, is then given by the equation,

Q1=VC1

Q1=VAε/d

Q2=VC2

Q2=VAε/2d

Therefore, the surface charge density σ1 and σ2 for the capacitors is,

σ1=Q1/A

σ1=VAε/(d*A)

σ1=Vε/d

Similarly,

σ2=Q2/A

σ2=Vε/2d

The electric field between the plates is directly proportional to the surface charge density. And so electric field is inversely proportional to the distance of separation. Therefore the capacitor whose distance of separation is less has a stronger electric field.

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3 years ago

We are sending a 30 Mbit file from source host A to destination host B. All links in the path between source and destination hav

Dmitrij [34]

Answer:

$t=1.5\times 10^{-4}\ s$

Explanation:

Given:

file size to be transmitted, $D=30\ Mb$

transmission rate of data, $\dot D=10\ Mb.s^{-1}$

propagation speed, $v=2\times 10^8\ m.s^{-1}$

distance of data transfer, $s=10000\ km=10^4\ m$

<u>Now the delay in data transfer from source to destination for each 10 Mb:</u>

$t'=\frac{s}{v}$

$t'=\frac{10^4}{2\times 10^8}$

$t'=5\times 10^{-5}\ s$

<u>Now this time is taken for each 10 Mb of data transfer and we have 30 Mb to transfer:</u>

So,

$t=3\times t'$

$t=3\times 5\times 10^{-5}$

$t=1.5\times 10^{-4}\ s$

3 0

4 years ago

A block with mass m =6.4 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.28 m.

Zanzabum

Answer

given,

mass of block (m)= 6.4 Kg

spring is stretched to distance, x = 0.28 m

initial velocity = 5.1 m/s

a) computing weight of spring

k x = m g

$k = \dfrac{mg}{x}$

$k = \dfrac{6.4 \times 9.8}{0.28}$

k = 224 N/m

b) $f = \dfrac{\omega}{2\pi}$

$\omega = \sqrt{\dfrac{k}{m}}= \sqrt{\dfrac{224}{6.4}} = 5.92 \ rad/s$

$f = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

$f = \dfrac{1}{2\pi}\sqrt{\dfrac{224}{6.4}}$

$f =0.94\ Hz$

c) $v_b = -v cos \omega t$

$v_b = -5.1 \times cos (5.92 \times 0.42)$

$v_b = 4.04\ m/s$

d) $a_{max} = v \omega$

$a_{max} = 4.04 \times 5.92$

$a_{max} =23.94\ m/s^2$

e) $Y =- A sin (\omega t)$

$A = \dfrac{v}{\omega}$

$A = \dfrac{4.04}{5.92}$

A = 0.682 m

$Y =- 0.682 \times sin (5.92 \times 0.42)$

$Y =- 0.42$

Force = $m \omega^2 |Y|$

= $6.4 \times 5.92^2\times 0.42$

F = 94.20 N

4 0

4 years ago

Which of these chemical equations demonstrates the law of conservation of mass as it applies to the combustion of ethane? A. 2 C

WINSTONCH [101]

A. 2 C₂ H₆ + 7 O₂ → 4 C O₂ + 6 H₂ O

according to law of conservation of mass , the total mass of reactants side must be same as the total mass of product side. so we need to check if each atom in the equation has same number on both side of the equation or not.

in this equation , we have

4 atoms of carbon left and 4 atoms of carbon on right

12 atoms of hydrogen on left and 12 atoms of hydrogen on right

14 atoms of oxygen on left and 14 atoms of oxygen on right

3 0

4 years ago

An example of an anaerobic exercise is what?

san4es73 [151]

Answer:

1st: Theatre History

4th Quarter

Upcoming

Due today

Syllabus

Due Sunday1st: Theatre History

4th Quarter

Upcoming

Due today

Syllabus

Due Sunday1st: Theatre History

4th Quarter

Upcoming

Due today

Syllabus

Due Sunday

Explanation:

1st: Theatre History

4th Quarter

Upcoming

Due today

Syllabus

Due Sunday

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3 years ago