The flow of electrons from one atom to another is called

The melting of a glacier is an example of the interactions among which of Earth’s spheres?

Anuta_ua [19.1K]

Atmosphere, geosphere, cryosphere would be the correct answer I think

4 0

3 years ago

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

If an object has zero acceleration, does it have to have zero velocity?

adelina 88 [10]

Answer:

Yes, the velocity would also be zero.

Explanation:

Acceleration is the change in velocity over time, therefore, there has to be a change in velocity for something to accelerate. which means without acceleration, the object has no velocity.

3 0

3 years ago

after a circuit has been turned off, so it is important to make sure they are discharged before you touch them. Suppose a 120 mF

Snezhnost [94]

Answer:

The time is $110.16\times10^{-3}\ sec$

Explanation:

Given that,

Capacitor = 120 μF

Voltage = 150 V

Resistance = 1.8 kΩ

Current = 50 mA

We need to calculate the discharge current

Using formula of discharge current

$i_{0}=\dfrac{V_{0}}{R}$

Put the value into the formula

$i_{0}=\dfrac{150}{1.8\times10^{3}}$

$i_{0}=83.3\times10^{-3}\ A$

We need to calculate the time

Using formula of current

$i=\dfrac{V_{0}}{R}e^{\frac{-t}{RC}}$

Put the value into the formula

$50=\dfrac{150}{1.8\times10^{3}}e^{\frac{-t}{RC}}$

$\dfrac{50}{83.3}=e^{\frac{-t}{RC}}$

$\dfrac{-t}{RC}=ln(0.600)$

$t=0.51\times1.8\times10^{3}\times120\times10^{-6}$

$t=110.16\times10^{-3}\ sec$

Hence, The time is $110.16\times10^{-3}\ sec$

4 0

3 years ago

Assuming a vertical trajectory with no drag, derive the applicable form of the rocket equation for this application

VARVARA [1.3K]

Answer:

The vertical trajectory is governed by Ordinary Differential Equation.

Time derivatives of each state variables.

d(d)/dt = v, d(m)/dt = -d(m-fuel)/dt, d(v)/dt = F/m.

Where V is velocity positive upwards, t is time, m is mass, m-fuel is fuel mass, F is Total force, positive upwards.

Therefore,

F = -mg - D + T, If V is positive and

F = -mg + D - T, If T is negative.

D is drag and the questions gave it as zero.

Explanation:

The two sign cases in derivative equations above are required because F is defined positive up, so the drag D and thrust T can subtract or add to F depending in the sign of V . In contrast, the gravity force contribution mg is always negative. In general, F will be some function of time, and may also depend on the characteristics of the particular rocket. For example, the T component of F will become zero after all the fuel is expended, after which point the rocket will be ballistic, with only the gravity force and the aerodynamic drag force being p

8 0

4 years ago