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3 years ago

When developing a question for a scientific inquiry, the question will ideally

2 answers:

7 0

When developing a question for a scientific inquiry, the question will ideally :

- be measurable
- Will have various variables that could be used for further testing
- not just simply ask why

hope this helps

Ierofanga [76]3 years ago

3 0

Answer: the answer is all of the above

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Two small identical conducting spheres are placed with their centers 0.41 m apart. One is given a charge of 12 ✕ 10−9 C, the oth

nataly862011 [7]

(a) $-1.48\cdot 10^{-5}N$

The electrostatic force exerted between the two sphere is given by:

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1, q2 are the charges on the two spheres

r is the separation between the centres of the two spheres

In this problem,

$q_1 = 12\cdot 10^{-9} C\\q_2 = -23\cdot 10^{-9} C\\r = 0.41 m$

Substituting these values into the equation, we find the force

$F=(9\cdot 10^9 Nm^2 C^{-2} )\frac{(12\cdot 10^{-9}C)(-23\cdot 10^{-9} C)}{(0.41 m)^2}=-1.48\cdot 10^{-5}N$

And the negative sign means the force is attractive, since the two spheres have charges of opposite sign.

(b) $+1.62\cdot 10^{-6}N$

The total net charge over the two sphere is:

$Q=q_1 +q_2 = 12\cdot 10^{-9}C+(-23\cdot 10^{-9}C)=-11\cdot 10^{-9} C$

When the two spheres are connected, the charge distribute equally over the two spheres (since they are identical, they have same capacitance), so each sphere will have a charge of

$q=\frac{Q}{2}=\frac{-11\cdot 10^{-9}C}{2}=-5.5\cdot 10^{-9}C$

So the electrostatic force between the two spheres will now be

$F=k\frac{q^2}{r^2}$

And substituting numbers, we find

$F=(9\cdot 10^9 Nm^2 C^{-2} )\frac{(-5.5\cdot 10^{-9} C)^2}{(0.41 m)^2}=+1.62\cdot 10^{-6}N$

and the positive sign means the force is repulsive, since the two spheres have same sign charges.

7 0

3 years ago

a major difference radio waves, visible light, and gamma rays is the _________ of the photons, which results in different photon

Aleks04 [339]

There is a spectrum of electromagnetic radiation with variable wavelengths and frequency, which in turn imparts different characteristics. ... X-rays and gamma rays have the same nature as visible light, radiant heat, and radio waves; however, they have shorter wavelengths and consequently a larger photon energy.

3 0

3 years ago

Read 2 more answers

what is the energy now stored if the capacitor was disconnected from the potential source before the separation of the plates wa

diamong [38]

Answer:

Final energy = Uf = initial energy × d₂/d₁

Explanation:

Energy is the ability to do work.

capacitor is an electronic device that store charges

where

V is the potential difference

d is the distance of seperation between the two plates

ε₀ is the dielectric constant of the material used in seperating the two plates, e.g., paper, mica, glass etc.

A = cross sectional area

U =¹/₂CV²

C =ε₀A/d

C × d=ε₀A=constant

C₂d₂=C₁d₁

C₂=C₁d₁/d₂

charge will 'q' remains same in the capacitor, if the capacitor was disconnected from the electric potential source (v) before the separation of the plates was replaced

Energy=U =(1/2)q²/C

U₂C₂ = U₁C₁

U₂ =U₁C₁ /C₂

U₂ =U₁d₂/d₁

Final energy = Uf = initial energy × d₂/d₁

7 0

3 years ago

Can some answer 7 a and b please

pav-90 [236]

Answer:

a.After $15$ second Mr Comer's speed $=$ $1.7 m/s$

b.Distance travelled by Mr.Comer in $15$ seconds $=$ $18.75\ m$

Explanation:

a. Lets recall our first equation of motion $V_{f} =V_{i} + at$

Now we know that $V_{i} = 0.8m/s$ , $a = 0.06\ m/s^2$ and

$t = 15 \ sec$

Plugging the values we have.

$V_{f} =V_{i} + at$

$V_{f} =0.8 + 0.06 \times 15$

$V_{f} =0.8+ 0.9$

$V_{f} =1.7 m/s$

Then Mr.Comer's speed after $15$ sec $=$ $1.7ms^-1$

Lets find the distance and recall our third equation of motion.

$V_{f} ^2-V{i}^2 = 2as$

So $s =$ distance covered.

Dividing both sides with 2a we have.

$\frac{V_{f} ^2-V{i}^2}{2a} = 2as$

Plugging the values.

$\frac{(1.7)^2-(0.8)^2}{2\times0.06} = 2as$

$s= 18.75\ m$

So Mr.Comer will travel a distance of $s= 18.75\ m$ .

4 0

3 years ago

PLEASE HELP ASAP!! Nitrogen, Helium, and Neon would all be gases at room temperature; Why is this?

kvasek [131]

-- The boiling points of the first group are all at temperatures
that are way lower than a comfortable room.
. . . . . Nitrogen . . . -320° F
. . . . . Helium . . . . -452° F
. . . . . Neon . . . . . -411° F

The freezing points of the second group are all at temperatures
that are way higher than a comfortable room.
. . . . . Lithium. . . . . . 357° F
. . . . . Sodium . . . . . 208° F
. . . . . Potassium . . . 146 °F

7 0

3 years ago