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3 years ago

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Three equal negative point charges are placed at three of the corners of a square of side d. What is the magnitude of the net el

ectric field at the center of the square? (k=1/4πϵ0=8.99×109N⋅m2/C2 )
Express your answer in terms of the variables d, q, and appropriate constants.

2 answers:

Rina8888 [55]3 years ago

8 0

<span>this may help you
As far as the field goes, the two charges opposite each other cancel!
So E = kQ / d² = k * Q / (d/√2)² = 2*k*Q / d² ◄
and since k = 8.99e9N·m²/C²,
E = 1.789e10N·m²/C² * Q / d² </span>

Vaselesa [24]3 years ago

7 0

Answer:

E = 2kq /d^2

Explanation:

Let d = side of the square

Distance between the center and corner of the square = a

a = √(d^2 + d^2) / 2

a = √2d^2 / 2

a = (√2 * d) /2

a = d /√2

Let the magnitude of an electric field = E

E = kq/a^2

Put a = d/√2

E = kq / (d/√2)^2

E = kq / d^2 / 2

E = 2kq / d^2

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A particle moves along a straight path through a displacement d = 2.5i + cj while a force F = 8.5i + -8.5j acts on it. The displ

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Answer:

Explanation:

Work is defined as the scalar product of force and distance

W=F•d

Given that

F = 8.5i + -8.5j. +×-=-

F=8.5i-8.5j

d = 2.5i + cj

If the work in the practice is zero, then W=0

therefore,

W=F•ds

0=F•ds

0=(8.5i -8.5j)•(2.5i + cj)

Note that

i.i=j.j=k.k=1

i.j=j.i=k.i=i.k=j.k=k.j=0

So applying this

0=(8.5i -8.5j)•(2.5i + cj)

0= (8.5×2.5i.i + 8.5×ci.j -8.5×2.5j.i-8.5×cj.j)

0=21.25-8.5c

Therefore,

8.5c=21.25

c=21.25/8.5

c=2.5

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3 years ago

How much work is done when an engine generates 400 Watts of power in 25 seconds?

IceJOKER [234]

Answer:

10000 J or 10 KJ

Explanation:

power = workdone/time taken

400 = workdone/25

workdone = 400 * 25

=10000 J

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How fast is Barry Allen ( The Flash) in season 5 of the Flash?

Burka [1]

Answer: The Flash, Allen's top speed is Mach 3.3, or 2,532 miles per hour.

Explanation:

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3 years ago

A solution containing 10.0 g of an unknown liquid and 90.0 g water has a freezing point of -3.33 °C. Given Kf = 1.86 °C/m for wa

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Answer:

62.06 g/mol

Explanation:

We are given that a solution containing 10 g of an unknown liquid and 90 g

Given mass of solute = $W_B$ =10 g

Given mass of solvent= $W_A$ =90 g

$k_f=1.86^{\circ}C/m$

Freezing point of solution =-3.33 $^{\circ}$ C

Freezing point of solvent = $0^{\circ}$ C

Change in freezing point =Depression in freezing point

=Freezing point of solvent - freezing point of solution=0+3.33= $3.33^{\circ}$

$\Delta T_f=\frac{W_B\times K_f\times 1000}{W_A\times M_B}$

$M_B=\frac{10\times 1.86\times 1000}{3.33\times 90}$

$M_B=62.06 g/mol$

Hence, molar mass of unknown liquid is 62.06g/mol.

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3 years ago

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Based on the following equation, answer the questions below. ρ = (2γϕ + ψ)/rg where ρ [=] moles per cubic foot [mol/ft3] γ [=] j

AlekseyPX

1) Fundamental units of $\Psi$ are $[\frac{mol}{m\cdot s^2}]$

2) Fundamental units of $\Phi$ are $[\frac{mol}{m^3}]$

Explanation:

The equation for the variable $\rho$ is

$\rho =\frac{2\gamma \Phi+\Psi}{rg}$

where we have:

$\rho$ measured in $[\frac{mol}{ft^3}]$

$\gamma$ measured in $[\frac{J}{kg}]$

$r$ measured in $[in]$

$g$ measured in $[\frac{m}{s^2}]$

We can re-write the equation as

$\rho rg = 2\gamma \Phi + \Psi$

And we notice that the units of the term on the left must be equal to the units of the term on the right.

This means that:

1) First of all, $\Psi$ must have the same units of $\rho r g$ . So,

$[\rho r g]=[\frac{mol}{ft^3}][in][\frac{m}{s^2}]$

However, both ft (feet) and in (inches) are not fundamental dimensions: this means that they can be expressed as meters. Therefore, the fundamental units of $\Psi$ are

$[\Psi]=[\frac{mol}{m^3}][m][\frac{m}{s^2}]=[\frac{mol}{m\cdot s^2}]$

2)

The term $2\gamma \Phi$ must have the same units of $\Psi$ in order to be added to it. Therefore,

$[\gamma \Phi] = [\frac{mol}{m\cdot s^2}]$

We also know that the units of $\gamma$ are $[\frac{J}{kg}]$ , therefore

$[\frac{J}{kg}][\Phi]= [\frac{mol}{m\cdot s^2}]$

And so, the fundamental units of $\Phi$ are

$[\Phi]= [\frac{mol\cdot kg}{J\cdot m\cdot s^2}]$

However, the Joules can be written as

$[J]=[kg][\frac{m^2}{s^2}]$

Therefore

$[\Phi]= [\frac{mol\cdot kg}{(kg \frac{m^2}{s^2})\cdot m\cdot s^2}]=[\Phi]= [\frac{mol}{m^3}]$

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7 0

3 years ago