Como se dice casa en náhuatl

Two metal sphere each of radius 2.0 cm, have a center-to-center separation of 3.30 m. Sphere 1 has a chrage of +1.10 10^-8 C. Sp

Leokris [45]

Answer:

A) V = -136.36 V , B) V = 4.85 10³ V , C) V = 1.62 10⁴ V

Explanation:

To calculate the potential at an external point of the spheres we use Gauss's law that the charge can be considered at the center of the sphere, therefore the potential for an external point is

V = k ∑ $q_{i} / r_{i}$

where $q_{i}$ and $r_{i}$ are the loads and the point distances.

A) We apply this equation to our case

V = k (q₁ / r₁ + q₂ / r₂)

They ask us for the potential at the midpoint of separation

r = 3.30 / 2 = 1.65 m

this distance is much greater than the radius of the spheres

let's calculate

V = 9 10⁹ (1.1 10⁻⁸ / 1.65 + (-3.6 10⁻⁸) / 1.65)

V = 9 10¹ / 1.65 (1.10 - 3.60)

V = -136.36 V

B) The potential at the surface sphere A

r₂ is the distance of sphere B above the surface of sphere A

r₂ = 3.30 -0.02 = 3.28 m

r₁ = 0.02 m

we calculate

V = 9 10⁹ (1.1 10⁻⁸ / 0.02 - 3.6 10⁻⁸ / 3.28)

V = 9 10¹ (55 - 1,098)

V = 4.85 10³ V

C) The potential on the surface of sphere B

r₂ = 0.02 m

r₁ = 3.3 -0.02 = 3.28 m

V = 9 10⁹ (1.10 10⁻⁸ / 3.28 - 3.6 10⁻⁸ / 0.02)

V = 9 10¹ (0.335 - 180)

V = 1.62 10⁴ V

6 0

3 years ago

You observe a distant galaxy. You find that a spectral line of hydrogen that is shifted from its normal location in the visible

dimulka [17.4K]

Answer:

The galaxy is moving away from us

Explanation:

Galaxy refers to a system of millions or billions of stars accompanied by gas and dust such that they are held together as a result of gravitational attraction.

When we observe a distant galaxy, we find that a spectral line of hydrogen that is shifted from its normal location in the visible part of the spectrum into the infrared part of the spectrum.

It means that the galaxy is moving away from us.

6 0

3 years ago

A circular cylinder has a diameter of 3.0 cm and a mass of 25 g. It floats in water with its long axis perpendicular to the wate

Vilka [71]

Answer:

f = 5.3 Hz

Explanation:

To solve this problem, let's find the equation that describes the process, using Newton's second law

∑ F = ma

where the acceleration is

a = $\frac{d^2 y}{dt^2 }$

B- W = m \frac{d^2 y}{dt^2 }

To solve this problem we create a change in the reference system, we place the zero at the equilibrium point

B = W

In this frame of reference, the variable y' when it is oscillating is positive and negative, therefore Newton's equation remains

B’= m $\frac{d^2 y'}{dt^2 }$

the thrust is given by the Archimedes relation

B = ρ_liquid g V_liquid

the volume is

V = π r² y'

we substitute

- ρ_liquid g π r² y’ = m \frac{d^2 y'}{dt^2 }

$\frac{d^2 y'}{dt^2} + \rho_liquid \ g \ \pi r^2/m ) y' \ =0$

this differential equation has a solution of type

y = A cos (wt + Ф)

where

w² = ρ_liquid g π r² /m

angular velocity and frequency are related

w = 2π f

we substitute

4π² f² = ρ_liquid g π r² / m

f = $\frac{1}{2\pi } \ \sqrt{ \frac{ \rho_{liquid} \ \pi r^2 \ g}{m } }$

calculate

f = $\frac{1}{2 \pi } \sqrt{ \frac{ 1000 \ \pi \ 0.03^2 \ 9.8 }{0.025} }$

f = 5.3 Hz

6 0

3 years ago

A projectile launcher has a mass of 3 kg. It fires a projectile of mass 0.08 kg horizontally at a speed of 300 m/s. a. What is t

alexdok [17]

Answer:

0

Explanation:

It’s before the projectile was fired, so nothing has happened yet.

6 0

3 years ago

The earth rotates once per day about an axis passing through the north and south poles, an axis that is perpendicular to the pla

Rama09 [41]

Please elaborate more on your question so I can help you

7 0

3 years ago

Como se dice casa en náhuatl​

Como se dice casa en náhuatl