Answer:
the work done by the field is positive and the potential energy of the electron field system decreases
Explanation:
This exercise asks to find the work and the potential energy of an electron in an electric field.
Work is defined by
W = F .d = F d cos θ
the electric force is
F_e = q E
W = q E d cos θ
since the charge of the electron is negative the force is in the opposite direction to the electric field
W = - e E d
we select the direction to the right is positive, point i is to the left of point f,
therefore the work moving from point i to point F has two possibilities
* The electric field lines go from i to f point , so that point i is on the side of the positive charges, so the electron approaches them, This movement is opposite to that indicated
* the field line reaches point i, this implies that the charges are negative, so the electrioc field is then negativeand the electron charge is negative too. The electron moves away from this point, this is in accordance with the indicated movement
In the latter case the electric field lines go from f to i point, therefore the Work is positive
Now let's examine the potential energy
ΔU = - q E .d
so we see that this definition is related to work,
ΔU = -W
Therefore, as the work is positive, the power energy must decrease
When reviewing the different answers, the correct ones are:
the work done by the field is positive and the potential energy of the electron field system decreases
The work done by the electron while moving from point [tex]i[/tex] to point [tex]f[/tex] in the direction of uniform electric field is negative and the potential energy of the electron increases.
An electron moves from point i to point f, in the direction of a uniform electric field, then the potential energy of the electron can be calculated s given below.
[tex]\Delta V=-qEd[/tex]
Where [tex]\Delta V[/tex] is the potential energy, [tex]E[/tex] is the electric field, [tex]q[/tex] is the charge and [tex]d[/tex] is the displacement of the electron.
The work done by the electron in the uniform electric field can be calculated as,
[tex]W = F\times d \times cos\theta[/tex]
Where [tex]W[/tex]is the work done by electron, [tex]F[/tex] is the electric force, [tex]d[/tex] is the displacement of the electron and for uniform electric field, the value of [tex]\theta[/tex] is zero.
Hence [tex]W=F\times d\times 1\\W=F \times d[/tex]
Electric force [tex]F = q E[/tex]
By substituting the value of electric force on the above formula,
[tex]W = qEd[/tex]
Hence, the relation between the work done the electron in an uniform electric field and potential energy of the electron can be given below.
[tex]W = -\Delta V[/tex]
The work done by the electron is negative and the potential energy of the electron increases.
For more information, follow the link given below.
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Question 18 of 25
Which type of reaction is shown in this energy diagram?
Energy
Products
Activation
Energy
Reoctants
to
ti
Time
A. Endothermic, because the products are lower in energy
B. Exothermic, because the reactants are lower in energy
C. Endothermic, because the reactants are lower in energy
D. Exothermic, because the products are lower in energy
Answer:
Endothermic, because the reactants are lower in energy (C)
Explanation:
From the graph, you can see the energy of the products is higher than the energy of the reactants. If you recall that when the enthalpy change Eproducts is gretater than Ereactants, the reaction is said to be endothermic.
5.0 L/s water flows through a horizontal pipe that narrows smoothly from 10.0 cm diameter to 5.0 cm diameter. A pressure gauge in the narrow section reads 50 kPa. What is the reading of the pressure gauge in the wide section
Solution :
The volume rate of flow is given by : R = 5.0 L/s
[tex]$ = 5.0 \times 10^{-3} \ m^3/s$[/tex]
The radius of the pipe, [tex]$r_1= 5 \times 10^{-2} \ m$[/tex]
∴ [tex]$ 5.0 \times 10^{-3} = \pi (2.5 \times 10^{-2})^2 v_1$[/tex]
then, [tex]$v_1 = \frac{5.0 \times 10^{-3}}{(3.14)(5 \times 10^{-2})^2}$[/tex]
= 0.637 meter per second
Then the speed of the water at wider section,
[tex]$R=A_1v_1$[/tex]
Similarly, the speed of water at narrow pipe.
The radius of the [tex]$r_2 = 2.5 \times 10^{-2}$[/tex] m
[tex]$5.0 \times 10^{-3} = \pi (2.5 \times 10^{-2})^2 v_1$[/tex]
then, [tex]$v_2 = \frac{5.0 \times 10^{-3}}{(3.14)(2.5 \times 10^{-2})^2}$[/tex]
= 2.55 meter per sec
Now from Bernoulli's theorem,
[tex]$P_1 + \frac{1}{2} \rho v_1^2 =P_2 + \frac{1}{2} \rho v_2^2 $[/tex]
[tex]$P_1 = P_2 + \frac{1}{2} \rho (v_2^2 - v_1^2)$[/tex]
[tex]$= 50 \kPa + (0.5)(10^3)[(2.55)^2-(0.637)^2]$[/tex]
= 50 kPa + 3.05 kPa
= 53.05 kPa
or 53000 Pa
This question involves the concepts of Bernoulli's Theorem and Volumetric Flowrate.
The pressure reading in the wide section is "53.05 KPa".
First, we will use the volumetric flow rate to find the velocities of the water at wide and narrow sections.
[tex]V = A_1v_1[/tex]
where,
V = Volumetric Flow Rate = 5 L/s = 5 x 10⁻³ m³/s
r₁ = radius of narrow section = 5 cm/2 = 2.5 cm = 0.025 m
A₁ = Area of narrow section = πr₁² = π(0.025 m)²
v₁ = velocity at narrow section = ?
Therefore,
[tex]5\ x\ 10^{-3}\ m^3=[\pi(0.025\ m)^2](v_1)\\\\v_1=\frac{5\ x\ 10^{-3}\ m^3}{\pi (0.025\ m)^2}\\\\v_1=2.55\ m/s\\[/tex]
Similarly,
[tex]V = A_2v_2[/tex]
where,
V = Volumetric Flow Rate = 5 L/s = 5 x 10⁻³ m³/s
r₂ = radius of wide section = 10 cm/2 = 5 cm = 0.05 m
A₂ = Area of wide section = πr₁² = π(0.05 m)²
v₂ = velocity at wide section = ?
Therefore,
[tex]5\ x\ 10^{-3}\ m^3=[\pi(0.05\ m)^2](v_2)\\\\v_2=\frac{5\ x\ 10^{-3}\ m^3}{\pi (0.05\ m)^2}\\\\v_2=0.64\ m/s\\[/tex]
Now, we will use Bernoulli's Theorem to find out the pressure wide section.
[tex]P_1 + \frac{1}{2}\rho v_1^2=P_2 + \frac{1}{2}\rho v_2^2[/tex]
where,
[tex]\rho[/tex] = density of water = 1000 kg/m³
P₁ = pressure in narrow section = 50 KPa = 50000 Pa
P₂ = pressure in wide section = ?
Therefore,
[tex]50000\ Pa + \frac{1}{2}(1000\ kg/m^3)(2.55\ m/s)^2=P_2 + \frac{1}{2}(1000\ kg/m^3)(0.64\ m/s)^2[/tex]
P₂ = 50000 Pa + 3251.25 Pa - 204.8 Pa
P₂ = 53046.45 Pa = 53.05 KPa
Learn more about Bernoulli's Theorem here:
https://brainly.com/question/13098748?referrer=searchResults
The attached picture shows Bernoulli's Theorem.
Did I hear correctly that the speed of light is different in deep space observation?
Answer:
Astronomers can learn about the elements in stars and galaxies by decoding the information in their spectral lines. There is a complicating factor in learning how to decode the message of starlight, however. If a star is moving toward or away from us, its lines will be in a slightly different place in the spectrum from where they would be in a star at rest. And most objects in the universe do have some motion relative to the Sun.
what is the mathematical definition of momentum? what is a more conceptual or descriptive definition of momentum?
Answer:
Momentum can be defined as "mass in motion." All objects have mass; so if an object is moving, then it has momentum - it has its mass in motion.
Explanation:
