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Evgesh-ka [11]
3 years ago
5

What is the mathematical expression for the electric ptential difference between points A and B when a charged particle is movin

g a displacement (s) from A to B in an external electric?
Physics
1 answer:
ratelena [41]3 years ago
4 0

Answer:

chang in V=\int\limits^B_A {E} \, ds

Explanation:

Since, as we know, the potential difference 'ΔV' is the difference of between the Potential energy per unit charge U/qo at one point 'B' to Potential energy per unit charge at other point 'A'. It so happens when a test charge 'qo' moves from point A to B, the potential difference becomes the change of potential energy of the system, i.e.

Potential Difference =\frac{change in U}{qo}=\int\limits^B_A {E} \, ds

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Explanation: The first one
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3 years ago
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A pair of eyeglass frames is made of epoxy plastic. At room temperature (20.0°C), the frames have circular lens holes 2.50 cm in
nignag [31]

Answer:

T₂ = 114 °C

Explanation:

Area or superficial expansion: can be defined as an increase in area, per unit area per degree rise in temperature. It unit is (1/K) or (1/°C).

β = ΔA/(A₁ΔT) ............................................. equation 1

β = 2α ......................................................... equation 2

Area of circle = πr².................................... equation 3

Where β = Area expansivity, α = linear expansivity, ΔA = increase in area = (A₂ - A₁), ΔT = change in temperature, A₁ = initial area, A₂ = Final area r = radius,

from the question, The coefficient of linear expansion for epoxy = 1.3 × 10⁻⁴ °C⁻¹,  

∴ Coefficient of area expansion for epoxy = 2 × 1.3 × 10⁻⁴ =

β = 2.6 × 10⁻⁴ °C⁻¹,

Using equation 3 to calculate for area, and taking (π = 3.143)

r₁ = 2.5 cm ∴ A₁ = πr₁² = 3.143 × 2.5² =19.64 cm².

r₂ = 2.53 cm ∴ A₂ = πr₂² = 3.143 × 2.53² =20.12 cm².

ΔA = A₂ - A₁ = 20.12 - 19.64 = 0.48 cm².

Making ΔT the subject of the relation in equation 1.

ΔT = ΔA/(βA₁) ...................................... equation 4

Substituting the values above into equation 4,

ΔT = 0.48/(2.6 × 10⁻⁴ × 19.64)

ΔT = 0.48/(51.064 × 10⁻⁴)

ΔT =( 0.48/51.064) × 10000

ΔT = 0.0094 × 10000 = 94 °C

But, ΔT = T₂ -T₁,

Then, T₂ = ΔT  + T₁

Where T₁ = 20 °C,  ΔT =94 °C

∴ T₂  = 94 + 20 = 114 °C.

Therefore, temperature at which the frame must be heated if the the lenses 2.53 cm are to be inserted in them = 114  °C

3 0
4 years ago
How much work is done if you push a 200 N box across a floor with a force of 50 N for a distance of 20m
vovikov84 [41]
<h2>Answer: 1000 J</h2>

The Work W done by a Force F refers to the release of potential energy from a body that is moved by the application of that force to overcome a resistance along a path.  

It should be noted that it is a scalar magnitude, and its unit in the International System of Units is the Joule (like energy). Therefore, 1 Joule is the work done by a force of 1 Newton when moving an object, in the direction of the force, along 1 meter:  

1J=(1N)(1m)=Nm  

Now, when the applied force is constant and the direction of the force and the direction of the movement are parallel, the equation to calculate it is:  

W=(F)(d) (1)

When they are not parallel, both directions form an angle, let's call it \alpha. In that case the expression to calculate the Work is:  

W=Fdcos{\alpha} (2)

For example, in order to push the 200 N box across the floor, you have to apply a force along the distance d to overcome the resistance of the weight of the box (its 200 N).  

In this case both <u>(the force and the distance in the path) are parallel</u>, so the work W performed is the product of the force exerted to push the box F=50N by the distance traveled d. as shown in equation (1).

Hence:

W=(50N)(20m)  

W=1000Nm=1000J >>>>This is the work  

8 0
3 years ago
The pilot of an airplane notes that the compass indicates a heading due west. the airplane's speed relative to the air is 130 km
Alja [10]
I got 21 once I answer that question
3 0
3 years ago
(1 point) An object moves along a straight track from the point (−5,4,3)(−5,4,3) to the point (−2,13,−6)(−2,13,−6). The only for
Pavel [41]

Answer:

-15Nm

Explanation:

Work is said to be done when a force causes an object to change its position.

Workdone = Force × Distance

Given the force (−2i+2j+3k)N

Since the body moves from point (−5,4,3) to the point (−2,13,−6), we will take the difference of both points to get the distance covered.

The difference will be final point (-2,13,-6) minus initial point (-5,4,3) i.e (-2,13,-6) - (-5,4,3) = (3,9,-9)

This point can be represented vectorially as (3i+9j-9k)meters

Work done = (−2i+2j+3k)×(3i+9j-9k)

Note that i.i = j.j = k.k = 1, the multiplication of different components will give "zero"

Work done = (-2×3)+(2×9)+(3×-9)

Work done = -6+18-27

Work done= -15Nm

6 0
3 years ago
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