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2 years ago

A student charges a balloon and then brings it near a metal sphere hanging from the

Physics

2 answers:

Neko [114]2 years ago

8 0

Answer: B

Explanation:

The balloon will pick up stray electrons, so it will become negatively charged. Since it is attracted to the metal sphere, the sphere must be positively charged.

Hunter-Best [27]2 years ago

3 0

Answer:

Explanation:

the balloon has a negative charge and the metal sphere has a positive charge

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You have two small spheres, each with a mass of 2.40 grams, separated by a distance of 10.0 cm. You remove the same number of el

nordsb [41]

Answer:

q = 2.066* 10⁻¹³ C.

n = 1,291,250 electrons.

Explanation:

If the gravitational attraction is equal to their electrical repulsion, we can write the following equation:

$F_{g} = F_{c} (1)$

where Fg is the gravitational attraction, that can be written as follows according Newton's Universal Law of Gravitation:

$F_{g} = G*\frac{m_{1}*m_{2}}{r_{12}^{2}} (2)$

Fc, due to it is the electrical repulsion between both charged spheres, must obey Coulomb's Law (assuming we can treat both spheres as point charges), as follows:

$F_{c} = k*\frac{q_{1}*q_{2}}{r_{12}^{2}} (3)$

since m₁ = m₂ = 0.0024 kg, and r₁₂ = 0.1m, G and k universal constants, and q₁ = q₂ = Q, we can replace the values in (2) and (3), so we can rewrite (1) as follows:

$G*\frac{(0.0024kg)^{2}}{r_{12}^{2}} = k*\frac{Q^{2}}{r_{12}^{2}} (4)$

Since obviously the distance is the same on both sides, we can cancel them out, and solve (4) for Q² first, as follows:

$Q^{2} = \frac{6.67e-11*(0.0024kg)^{2}}{9e9Nm2/C2} = 4.27*e-26 C2 (5)$

Since both charges are the same, the charge on each sphere is just the square root of (5):
Q = 2.066* 10⁻¹³ C.

Assuming that both spheres were electrically neutral before being charged, the negative charge removed must be equal to the positive charge on the spheres.
Now, since each electron carries an elementary charge equal to -1.6*10⁻¹⁹ C, in order to get the number of electrons removed from each sphere, we need to divide the charge removed from each sphere (the outcome of part 1) with negative sign) by the elementary charge, as follows:
$n_{e} =\frac{-2.066e-13C}{-1.6e-19C} = 1,291,250 electrons. (6)$

4 0

3 years ago

the picture shows six panels (a, b, c, d, e, and f), each displaying two snapshots of a box moving on a rough flor. forces may b

Agata [3.3K]

K = 1/2mv^2 of kinetic energy. The change in the object's kinetic energy is equal to the net work performed on it.

<h3>What causes the kinetic energy to change?</h3>

Equations. Mass and the square of the velocity are directly related to translational kinetic energy. The difference between the end and starting kinetic energies is known as change in kinetic energy.

<h3>In solar panels, is there kinetic energy?</h3>

employing semiconductor-cell-based panels. technique that uses solar thermal systems to store solar energy. This heat is used directly or transformed into concentrated solar power, or the sum of the potential energy and kinetic energy of an object or system, and electricity.

Learn more about kinetic energy here:

brainly.com/question/26472013

#SPJ4

8 0

1 year ago

An ocean thermal energy conversion system is being proposed for electric power generation. Such a system is based on the standar

defon

Answer:

Explanation:

Dear Student, this question is incomplete, and to attempt this question, we have attached the complete copy of the question in the image below. Please, Kindly refer to it when going through the solution to the question.

To objective is to find the:

(i) required heat exchanger area.

(ii) flow rate to be maintained in the evaporator.

Given that:

water temperature = 300 K

At a reasonable depth, the water is cold and its temperature = 280 K

The power output W = 2 MW

Efficiency $\zeta$ = 3%

where;

$\zeta = \dfrac{W_{out}}{Q_{supplied }}$

$Q_{supplied } = \dfrac{2}{0.03} \ MW$

$Q_{supplied } = 66.66 \ MW$

However, from the evaporator, the heat transfer Q can be determined by using the formula:

Q = UA(L MTD)

where;

$LMTD = \dfrac{\Delta T_1 - \Delta T_2}{In (\dfrac{\Delta T_1}{\Delta T_2} )}$

Also;

$\Delta T_1 = T_{h_{in}}- T_{c_{out}} \\ \\ \Delta T_1 = 300 -290 \\ \\ \Delta T_1 = 10 \ K$

$\Delta T_2 = T_{h_{in}}- T_{c_{out}} \\ \\ \Delta T_2 = 292 -290 \\ \\ \Delta T_2 = 2\ K$

$LMTD = \dfrac{10 -2}{In (\dfrac{10}{2} )}$

$LMTD = \dfrac{8}{In (5)}$

LMTD = 4.97

Thus, the required heat exchanger area A is calculated by using the formula:

$Q_H = UA (LMTD)$

where;

U = overall heat coefficient given as 1200 W/m².K

$66.667 \times 10^6 = 1200 \times A \times 4.97 \\ \\ A= \dfrac{66.667 \times 10^6}{1200 \times 4.97} \\ \\ \mathbf{A = 11178.236 \ m^2}$

The mass flow rate:

$Q_{H} = mC_p(T_{in} -T_{out} ) \\ \\ 66.667 \times 10^6= m \times 4.18 (300 -292) \\ \\ m = \dfrac{ 66.667 \times 10^6}{4.18 \times 8} \\ \\ \mathbf{m = 1993630.383 \ kg/s}$

3 0

3 years ago

How do iquantum tunel through a wall and how long will it take me?

Anon25 [30]

Answer:

The phenomenon known as "tunneling" is one of the best-known predictions of quantum physics, because it so dramatically confounds our classical intuition for how objects ought to behave. If you create a narrow region of space that a particle would have to have a relatively high energy to enter, classical reasoning tells us that low-energy particles heading toward that region should reflect off the boundary with 100% probability. Instead, there is a tiny chance of finding those particles on the far side of the region, with no loss of energy. It's as if they simply evaded the "barrier" region by making a "tunnel" through it.

Explanation:

5 0

2 years ago

The force that probably propels the movement of the earth s gigantic rock plates is

il63 [147K]

The correct answer is:

c. convection.

The heating of magma and the continuous cycle of evolution of the magma creating a convection current is the reason for the evolution of Earths tectonic plates.

Explanation:

Tectonic plates are ready to move because the Earth's lithosphere has higher strength than the underlying asthenosphere. Lateral density changes in the mantle appear in convection. Plate movement is believed to be driven by a succession of the motion of the seafloor apart from the extended ridge (due to variations in topography and density of the crust.

4 0

2 years ago