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2 years ago

Charged beads are placed at the corners of a square in the various configurations shown in

Physics

1 answer:

Akimi4 [234]2 years ago

8 0

Answer:

The consecutive charge configuration has a more intense field than alternating

Explanation:

In each corner we place a different account there are only two different settings, see attached.

In the case of alternating charging (+ - + -) see diagram 1, the electric field in the center is canceled in pairs, resulting in a zero field

In the case of consecutive loads (+ + - -) in this case we have a result between the two charges, therefore the total field is

E = 2 k q / ra2 a cos 45

The consecutive charge configuration has a more intense field than alternating

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Particle q1 has a positive 6 µC charge. Particle q2 has a positive 2 µC charge. They are located 0.1 meters apart.

m_a_m_a [10]

Answer:

F = 10.788 N

Explanation:

Given that,

Charge 1, $q_1=6\ \mu C=6\times 10^{-6}\ C$

Charge 2, $q_2=2\ \mu C=2\times 10^{-6}\ C$

Distance between charges, d = 0.1 m

We know that there is a force between charges. It is called electrostatic force. It is given by :

$F=\dfrac{kq_1q_2}{d^2 }\\\\F=\dfrac{8.99\times 10^9\times 6\times 10^{-6}\times 2\times 10^{-6}}{(0.1)^2 }\\\\F=10.788\ N$

So, the force applied between charges is 10.788 N.

3 0

2 years ago

Read 2 more answers

where again p is the phonon momentum, E is the photon energy and c is the speed of light. When you divide the photon energy foun

creativ13 [48]

Answer:

Yes

Explanation:

p = momentum of photon

E = energy of photon

c = velocity of light

Units of p = kg m /s

Units of E = kg m^2 / s^2

Units of E / p = {kg m^2 / s^2} / {kg m /s} = m/s

It is the unit of speed, so by the division of energy to the momentum, we get the speed. yes it is correct.

8 0

1 year ago

If it were possible to remove gravity and friction, think about what would happen to a football if it were tossed into the air.

elena-14-01-66 [18.8K]

Ignoring fluid resistance, football will <span>maintain a constant speed until other forces accelerate the football.</span>

6 0

2 years ago

Read 2 more answers

Match the words in the left-hand column to the appropriate blank in the sentences in the right-hand column. use each word only o

shepuryov [24]

Answer:

An annular Solar Eclipse

Explanation:

Solar eclipse is an event that occurs naturally on Earth when the moon in its orbit is positioned between the Earth and the Sun.Solar Eclipse can be total ,partial or annular.In the total solar eclipse, the moon completely covers the sun where as in the annular solar eclipse the moon covers the center of the Sun leaving outer edges of the Sun to be visible forming the<em> ring of fire.</em>In partial solar eclipse the Earth moves through the lunar penumbra as the moon moves between Earth and Sun.The moon blocks only some parts of the solar disk.Annular solar eclipse happens during new moon and the moon is at its farthest position from the Earth called Apogee.

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2 years ago

Air at 3 104 kg/s and 27 C enters a rectangular duct that is 1m long and 4mm 16 mm on a side. A uniform heat flux of 600 W/m2 is

ad-work [718]

Answer:

$T_{out}=27.0000077$ ºC

Explanation:

First, let's write the energy balance over the duct:

$H_{out}=H_{in}+Q$

It says that the energy that goes out from the duct (which is in enthalpy of the mass flow) must be equals to the energy that enters in the same way plus the heat that is added to the air. Decompose the enthalpies to the mass flow and specific enthalpies:

$m*h_{out}=m*h_{in}+Q\\m*(h_{out}-h_{in})=Q$

The enthalpy change can be calculated as Cp multiplied by the difference of temperature because it is supposed that the pressure drop is not significant.

$m*Cp(T_{out}-T_{in})=Q$

So, let's isolate $T_{out}$ :

$T_{out}-T_{in}=\frac{Q}{m*Cp}\\T_{out}=T_{in}+\frac{Q}{m*Cp}$

The Cp of the air at 27ºC is 1007 $\frac{J}{kgK}$ (Taken from Keenan, Chao, Keyes, “Gas Tables”, Wiley, 1985.); and the only two unknown are $T_{out}$ and Q.

Q can be found knowing that the heat flux is 600W/m2, which is a rate of heat to transfer area; so if we know the transfer area, we could know the heat added.

The heat transfer area is the inner surface area of the duct, which can be found as the perimeter of the cross section multiplied by the length of the duct:

Perimeter:

$P=2*H+2*A=2*0.004m+2*0.016m=0.04m$