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Anettt [7]
3 years ago
9

Identical isolated conducting spheres 1 and 2 have equal charges and are separated by a distance that is large compared with the

ir diameters. The magnitude of the electrostatic force acting on sphere 2 due to sphere 1 is F = 0.42 N. Suppose now that a third identical sphere 3, having an insulating handle and initially neutral, is touched first to sphere , then to sphere 2, and finally removed. What is the magnitude of the electrostatic force F' that now acts on sphere 2?
Physics
1 answer:
scoundrel [369]3 years ago
3 0

Answer:

F = 0.1575 N

Explanation:

When the third sphere touches the first sphere, the charge is distributed between both spheres, then now the first sphere has only half of his original charge.

In this moment then

Sphere one has a charge = Q/2

Sphere three has a charge = Q/2

Now when the third sphere touches the second sphere again the charge is distributed in a manner that both sphere has the same charge.

How the total charge is Q = Q/2 + Q = 3/2Q, when the spheres are separated each one has 3/4Q

Sphere two has a charge = 3/4Q

Sphere three has a charge = 3/4Q

The electrostatic force that acts on sphere 2 due to sphere 1 is:

F = \frac{kQ_{1}Q_{2} }{r^{2} }

F= \frac{K(Q/2)(3Q/4)}{r^{2} }

how \frac{KQ^{2} }{r^{2} } = 0.42

Then

F = \frac{0.42*3}{8}

F = 0.1575 N

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A container of gas is at a pressure of 1.3x10^5 Pa and a volume of 6 m^3. How much work is done by the gas if it expands at a co
Ivenika [448]
In thermodynamics, work of a system at constant pressure conditions is equal to the product of the pressure and the change in volume. It is expressed as follows:

W = P(V2 - V1)
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Hope this answers the question. Have a nice day.
5 0
3 years ago
What is the temperature of a 3.72 mm cube (e=0.288) that radiates 56.6 W?
blsea [12.9K]

Answer:

The temperature is 2541.799 K

Explanation:

The formula for black body radiation is given by the relation;

Q = eσAT⁴

Where:

Q = Rate of heat transfer 56.6

σ = Stefan-Boltzman constant = 5.67 × 10⁻⁸ W/(m²·k⁴)

A = Surface area of the cube = 6×(3.72 mm)² = 8.3 × 10⁻⁵ m²

e = emissivity = 0.288

T = Temperature

Therefore, we have;

T⁴ = Q/(e×σ×A) = 56.6/(5.67 × 10⁻⁸ × 8.3 × 10⁻⁵ × 0.288) = 4.174 × 10¹⁴ K⁴

T  =  2541.799 K

The temperature = 2541.799 K.

7 0
3 years ago
A block of mass m=9.0 kg and speed V and is behind a block of mass M= 27 kg and speed of .50 m/s. The surface is frictionless, a
sammy [17]

Answer:

2.06 m/s

Explanation:

From the law of conservation of linear momentum, the sum of momentum before and after collision are equal. Considering this case where we have frictionless surface, no momentum is lost in the process.

Momentum before collision

Momentum is given by p=mv where m and v represent mass. The initial sum of momentum will be 9v+(27*0.5)=9v+13.5

Momentum after collision

The momentum after collision will be given by (9+27)*0.9=32.4

Relating the two then 9v+13.5=32.4

9v=18.5

V=2.055555555555555555555555555555555555555 m/s

Rounded off, v is approximately 2.06 m/s

5 0
3 years ago
A 50.0-kg block is being pulled up a 16.0° slope by a force of 250 n that is parallel to the slope. the coefficient of kinetic
Drupady [299]
When dealing with multiple forces acting on a body, it is advisable to draw a free-body diagram like that shown in the picture. There are four forces acting on the box: weight (W) pointing straight down, normal force perpendicular to the slope denoted as Fn, force used to push the box upwards along the slope and the frictional force acting opposite to the direction of motion of the box denoted as Ff. Frictional force is equal to coefficient of kinetic friction (μk) multiplied with Fn.

∑Fy = Fn - mgcos30° = 0
           Fn = (50)(9.81)(cos 16) = 471.5 N

When in motion, the net force is equal to mass times acceleration according to Newton's 2nd Law of Motion:

Fnet = F - μk*Fn - mgsin30° = ma
          250 - (0.2)(471.5 N) - (50)(sin 16°) = (50)(a)
          a = 2.84 m/s²

8 0
3 years ago
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