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When conducting objects (such as metals) are connected for a brief time, charge can be made to flow quite easily from one to the

sukhopar [10]

Answer:

a) charges of the opposite, b) Δq = q₄₂-q₄₁, force is repulsive

c) q = 5.27 10⁻⁵ C

Explanation:

a) In electrostatic studies it is found that charges of the same sign repel and charges of the opposite sign attract.

On the other hand, as the force between the spheres is attractive, the charges on them are of different sign

b) When the balls touch, the charges are quickly distributed between the two spheres, therefore there are two possibilities:

* if the charges were equal and as they are of the opposite sign, they are neutralized, therefore the spheres remain uncharged

The force between them is zero

* if the charges are different, a residual charge remains

Δq = q₄₂-q₄₁

that is distributed between the two spheres and the force between them is repulsive

c) For this case the charges on the two spheres is equal

q₄₁ = q₄₂ = q

the force is repulsive so the charges are of the same sign.

We can apply Coulomb's law

F = $k \frac{q_1q_2}{r^2}$

in this case

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

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

let's calculate

q = $\sqrt{ \frac{25 \ 1^2}{9 \ 10^{9}} }$ = $\sqrt{27.7778 \ 10^{-10}}$

q = 5.27 10⁻⁵ C

7 0

3 years ago

Which best describes a difference between electric current and static electricity?

NeTakaya

Perhaps one of the most useful yet taken-for-granted accomplishments of the recent centuries is the development of electric circuits. The flow of charge through wires allows us to cook our food, light our homes, air-condition our work and living space, entertain us with movies and music and even allows us to drive to work or school safely. In this unit of The Physics Classroom, we will explore the reasons for why charge flows through wires of electric circuits and the variables that affect the rate at which it flows. The means by which moving charge delivers electrical energy to appliances in order to operate them will be discussed in detail.

One of the fundamental principles that must be understood in order to grasp electric circuits pertains to the concept of how an electric field can influence charge within a circuit as it moves from one location to another. The concept of electric field was first introduced in the unit on Static Electricity. In that unit, electric force was described as a non-contact force. A charged balloon can have an attractive effect upon an oppositely charged balloon even when they are not in contact. The electric force acts over the distance separating the two objects. Electric force is an action-at-a-distance force.

Action-at-a-distance forces are sometimes referred to as field forces. The concept of a field force is utilized by scientists to explain this rather unusual force phenomenon that occurs in the absence of physical contact. The space surrounding a charged object is affected by the presence of the charge; an electric field is established in that space. A charged object creates an electric field - an alteration of the space or field in the region that surrounds it. Other charges in that field would feel the unusual alteration of the space. Whether a charged object enters that space or not, the electric field exists. Space is altered by the presence of a charged object; other objects in that space experience the strange and mysterious qualities of the space. As another charged object enters the space and moves deeper and deeper into the

6 0

4 years ago

Please help me - i beg u

oksano4ka [1.4K]

Answer:

516526.863 m

Explanation:

From the question given above, the following data were obtained:

Acceleration due to gravity (g) = 2.24 m/s²

Mass (M) = 8.96×10²¹ Kg

Gravitational constant (G) = 6.67×10¯¹¹Nm²/Kg²

Radius (r) =?

The radius of the planet can be obtained as follow:

g = GM/r²

2.24 = 6.67×10¯¹¹ × 8.96×10²¹ / r²

2.24 = 5.97632×10¹¹ / r²

Cross multiply

2.24 × r² = 5.97632×10¹¹

Divide both side by 2.24

r² = 5.97632×10¹¹ / 2.24

r² = 2.668×10¹¹

Take the square root of both side

r = √2.668×10¹¹

r = 516526.863 m

Thus, the radius of the planet is 516526.863 m

8 0

3 years ago

A 1.0 liter pot is filled with water at sea level and is brought to a boil. the same pot is filled with water at 5,000 feet abov

kykrilka [37]

The second pot will boil faster

8 0

4 years ago

Gold, which has a density of 19.32 g/cm3, is the most ductile metal and can be pressed into a thin leaf or drawn out into a long

NNADVOKAT [17]

Answer:

a) 578.0 cm²

b) 25.18 km

Explanation:

We're given the density and mass, so first calculate the volume.

D = M / V

V = M / D

V = (6.740 g) / (19.32 g/cm³)

V = 0.3489 cm³

a) The volume of any uniform flat shape (prism) is the area of the base times the thickness.

V = Ah

A = V / h

A = (0.3489 cm³) / (6.036×10⁻⁴ cm)

A = 578.0 cm²

b) The volume of a cylinder is pi times the square of the radius times the length.

V = πr²h

h = V / (πr²)

h = (0.3489 cm³) / (π (2.100×10⁻⁴ cm)²)

h = 2.518×10⁶ cm

h = 25.18 km

3 0

3 years ago