What would the position of arrows on a target need to be to illustrate measurements that are neither accurate nor precise

Complete question:

An electron is a subatomic particle (m = 9.11 x 10-31 kg) that is subject to electric forces. An electron moving in the +x direction accelerates from an initial velocity of +6.18 x 105 m/s to a final velocity of 2.59 x 106 m/s while traveling a distance of 0.0708 m. The electron's acceleration is due to two electric forces parallel to the x axis: F₁ = 8.87 x 10-17 N, and , which points in the -x direction. Find the magnitudes of (a) the net force acting on the electron and (b) the electric force F₂.

Answer:

(a) The net force of the electron, ∑F = 4.07 x 10⁻¹⁷ N

(b) the electric force, F₂ = 4.8 x 10⁻¹⁷ N

Explanation:

Given;

initial velocity of the electron, = +6.18 x 10⁵ m/s

final velocity of the electron, = 2.59 x 10⁶ m/s

the distance traveled by the electron, d = 0.0708 m

The first electric force,

(a) The net force of the electron is given as;

∑F = F₁ - F₂ = ma

where;

a is the acceleration of the electron

∑F = ma = (9.11 x 10⁻³¹ kg)(4.468 x 10¹³)

∑F = 4.07 x 10⁻¹⁷ N

(b) the electric force, F₂ is given as;

∑F = F₁ - F₂

F₂ = F₁ - ∑F

F₂ = 8.87 x 10⁻¹⁷ - 4.07 x 10⁻¹⁷

F₂ = 4.8 x 10⁻¹⁷ N

Final answer:

The problem involves calculating the acceleration of an electron, then using Newton's second law to find the net force on the electron. This is used to find the magnitude of a second electric force acting on the electron.

Explanation:

First, we can calculate the acceleration of the electron using the formula a = Δv/Δt, where 'a' is acceleration, 'Δv' is the change in velocity, and 'Δt' is the change in time. In this case, Δv = vf - vi = 2.59 x 106 m/s - 6.18 x 105 m/s = 1.972 x 106 m/s. The time taken by the electron to travel 0.0708 m can be found using the equation d = vi t + 0.5 a t₂. We use these values to get Δt which we use to find 'a'.

Next, let's use Newton's second law F = ma to find the net force acting on the electron. The only forces acting on the electron are electric forces, and we know one them is 8.87 x 10-17 N. If we designate this known force as F₁ then the total force F total = F₁ + F₂ where F₂ is the unknown electric force.

Finally, we can find F₂ = F total - F₁. This gives the magnitude of the second electric force.

Learn more about Physics - Electric Forces here:

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Answer:

Good conductor of heat

Explanation:

Because metals are shiny, ductile, malleable, sonorous, good conductors of heat and electricity and have high melting and boiling points

Answer: mine is different so im sorry im here for points

Explanation:

Write the velocity vectors in component form.

• initial velocity:

v₁ = 4 m/s at 45º N of E

v₁ = (4 m/s) (cos(45º) i + sin(45º) j)

v₁ ≈ (2.83 m/s) i + (2.83 m/s) j

• final velocity:

v₂ = 4 m/s at 10º N of E

v₂ = (4 m/s) (cos(10º) i + sin(10º) j)

v₂ ≈ (3.94 m/s) i + (0.695 m/s) j

The average acceleration over this 3-second interval is then

a = (v₂ - v₁) / (3 s)

a ≈ (0.370 m/s²) + (-0.711 m/s²)

with magnitude

||a|| = √[(0.370 m/s²)² + (-0.711 m/s²)²] ≈ 0.802 m/s²

and direction θ such that

tan(θ) = (-0.711 m/s²) / (0.370 m/s²) ≈ -1.92

→   θ ≈ -62.5º

which corresponds to an angle of about 62.5º S of E, or 27.5º E of S. To use the notation in the question, you could say it's E 62.5º S or S 27.5º E.

Answer:

The maximum wavelength of light that could liberate electrons from the aluminum metal is 303.7 nm

Explanation:

Given;

wavelength of the UV light, λ = 248 nm = 248 x 10⁻⁹ m

maximum kinetic energy of the ejected electron, K.E = 0.92 eV

let the work function of the aluminum metal = Ф

Apply photoelectric equation:

E = K.E + Ф

Where;

Ф is the minimum energy needed to eject electron the aluminum metal

E is the energy of the incident light

The energy of the incident light is calculated as follows;

The work function of the aluminum metal is calculated as;

Ф = E - K.E

Ф = 8.02 x 10⁻¹⁹  -  (0.92 x 1.602 x 10⁻¹⁹)

Ф =  8.02 x 10⁻¹⁹ J   -  1.474 x 10⁻¹⁹ J

Ф = 6.546 x 10⁻¹⁹ J

The maximum wavelength of light that could liberate electrons from the aluminum metal is calculated as;

To solve this problem it is necessary to apply the concepts related to the conservation of energy, through the balance between the work done and its respective transformation from the gravitational potential energy.

Mathematically the conservation of these two energies can be given through

Where,

W = Work

Final gravitational Potential energy

Initial gravitational Potential energy

When the spacecraft of mass m is on the surface of the earth then the energy possessed by it

Where

M = mass of earth

m = Mass of spacecraft

R = Radius of earth

Let the spacecraft is now in an orbit whose attitude is then the energy possessed by the spacecraft is

Work needed to put it in orbit is the difference between the above two

Therefore the work required to launch a spacecraft from the surface of the Eart andplace it ina circularlow earth orbit is

Answer:

-30=5(x+1) is -7

Explanation:

distribute flip subtract 5 from both sides divide both sides by 5