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3 years ago

A change from one phase to another involves

Physics

2 answers:

Oksi-84 [34.3K]3 years ago

7 0

A transition
that may be wrong because i dont know the other choices

Mashutka [201]3 years ago

6 0

I think it is c because my mom told me its not

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Two 0.50 g spheres are charged equally and placed 2.5 cm apart. When released, they begin to accelerate at 170 m/s^2 .What is th

vitfil [10]

Answer:

$q=7.65*10^{-8}C$

Explanation:

Using Newton's second law, we calculate the magnitude of the electric force between the spheres:

$F=ma\\F=0.5*10^{-3}kg(170\frac{m}{s^2})\\F=0.085N$

The magnitude of the charge in both spheres is the same. So, we calculate the charge, using Coulomb's law:

$F=\frac{kq^2}{d^2}\\q=\sqrt\frac{Fd^2}{k}\\q=\sqrt\frac{(0.085N)(2.5*10^{-2}m)^2}{8.99*10^9\frac{N\cdot m^2}{C^2}}\\q=7.65*10^{-8}C$

8 0

3 years ago

A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 4.444 N

Dmitriy789 [7]

Answer:

The value of third charge is 0.8μC.

Explanation:

Given that.

Magnitude of net force=4.444 N

According to figure,

Suppose, First charge = 2.4 μC

Second charge = 6.2 μC

Distance r₁ = 9.8 cm

Distance r₂ = 2.1 cm

We need to calculate the value of r

Using Pythagorean theorem

$r=\sqrt{(r_{1})^2+(r_{2})^2}$

Put the value into the formula

$r=\sqrt{(9.8)^2+(2.1)^2}$

$r=10.02\ cm$

We need to calculate the force

Using formula of force

$F_{12}=\dfrac{kq_{1}q_{2}}{(r)^2}$

Force F₁₂,

$F_{12}=\dfrac{9\times10^{9}\times2.4\times10^{-6}\times6.2\times10^{-6}}{(10.02\times10^{-2})^2}$

$F_{12}=13.33\ N$

$F_{21}=-13.33\ N$

Force F₂₃,

$F_{23}=\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(10.02)^2}$

We need to calculate the value of third charge

$F_{net}=F_{21}+F_{23}$

$4.444=-13.33+\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(5.01)^2}$

$q_{3}=\dfrac{(4.444+13.33)\times(5.01\times10^{-2})^2}{9\times10^{9}\times6.2\times10^{-6}}$

$q_{3}=7.99\times10^{-7}\ C$

$q_{3}=0.8\times10^{-6}\ C$

Hence, The value of third charge is 0.8μC.

4 0

3 years ago

At locations A and B, the electric potential has the values VA = 1.83 V and VB = 5.17 V, respectively. A proton released from re

densk [106]

Answer:

a. It starts at point B.

vp = 2.53*10⁴ m/s

a. it starts at point A.

ve= 1.08*10⁶ m/s

Explanation:

a) As the proton is a positive charge, when released from rest, it will be accelerated due to the potential difference, from the higher potential to the lower one, so it is at the point B when released.

Once released, as the total energy must be conserved, the increase in kinetic energy must be equal (in magnitude) to the change in the electric potential energy, as follows:

ΔK + ΔUe = 0 ⇒ ΔK = -ΔUe =- (e*ΔV)

⇒ $-( e* (VA-VB) ) = \frac{1}{2}*mp*v^{2}$

where e= elementary charge= 1.6*10⁻¹⁹ C, VA = 1.83 V, VB= 5.17V, and mp= mass of proton = 1.67*10⁻²⁷ kg.

Replacing by these values, and solving for v, we have:

$v = \sqrt{\frac{2*1.6e-19C*3.34 V}{1.67e-27kg} } = 2.53e4 m/s$

⇒ vp = 2.53*10⁴ m/s

b) If, instead of a proton, the charge realeased from rest, had been an electron, a few things would change:

First, as the electrons carry negative charges, they move from the lower potentials to the higher ones, which means that it would have started at point A.

Second, as its charge is (-e) the change in electric potential energy had been negative also:

ΔUe = -e*ΔV = -e* (VB-VA)

In order to find the speed of the electron when it is just passing point B, we can apply the conservation of energy principle as for the proton, as follows:

$-( (-e)* (VB-VA) ) = \frac{1}{2}*me*v^{2}$

where e= elementary charge= 1.6*10⁻¹⁹ C, VA = 1.83 V, VB= 5.17V, and me= mass of electron = 9.1*10⁻³¹ kg.

Replacing by these values, and solving for v, we have:

$v = \sqrt{\frac{2*1.6e-19C*3.34 V}{9.1e-31kg} } = 1.08e6 m/s$

⇒ ve = 1.08*10⁶ m/s

4 0

3 years ago

What forces are acting upon a coasting car

Ket [755]

Answer:

f grav and f norm

Explanation:

5 0

3 years ago

Are Constellations based on the magnitude of stars.

Nikolay [14]

Constellations are based off of many Greek and Roman fables. Many of their gods and beliefs are pictured in the stars, which is where we get most of our constellations. Hope this helps!

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3 years ago

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