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3 years ago

To review relationships among electric potential, electric potential energy, and force on a test charge

Physics

1 answer:

adelina 88 [10]3 years ago

5 0

Answer:

positive, negative, positive

Explanation:

Electric field lines are the imaginary lines that is the path followed by an isolated unit positive charged particle in an electric field.

The direction of electric field at that point is determined by drawing the tangent at that point on the electric field line.

Part A

The electric field lines always begin at positive charges and end at negative charges.

One could also say that the lines we use to represent an electric field indicate the direction in which a positive test charge would initially move when released from rest.

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is plugged into the outlet of a 120−V circuit that has a 20−A circuit breaker. You plug an electric hair dryer into the same out

WINSTONCH [101]

Answer:

Given the point of maximun electric power of the hair dryer 1500 (w); The circuit breaker won´t trip at all

Explanation:

The most simplyfied relation between power, voltage and current is:

Electric power (in watts) =Voltage (in volts) * current (in ampers)

In the case of P= 1500 (w) and V= 120 (v) we have:

I = 1500/120 (a) = 12,65 (a)

This value is far away of 20 (a) (the nominal trip current

8 0

3 years ago

Which list places the layers of the sun in the correct order from outermost to innermost?

Valentin [98]

Answer - corona, chromosphere, photosphere

4 0

3 years ago

A loop of wire is carrying current of 2 A . The radius of the loop is 0.4 m. What is the magnetic field at a distance 0.09 m alo

HACTEHA [7]

Answer:

$B=2.91\ \mu T$

Explanation:

Given that,

The current in the loop, I = 2 A

The radius of the loop, r = 0.4 m

We need to find the magnetic field at a distance 0.09 m along the axis and above the center of the loop. The formula for the magnetic field at some distance is given as follows :

$B=\dfrac{\mu_o}{4\pi }\dfrac{2\pi r^2 I}{(r^2+d^2)^{3/2}}$

Put all the values,

$B=10^{-7}\times \dfrac{2\pi \times 0.4^2 \times 2}{(0.4^2+0.09^2)^{3/2}}\\\\=2.91\times 10^{-6}\ T\\\\or\\\\B=2.91\ \mu T$

So, the required magnetic field is equal to $2.91\ \mu T$ .

3 0

3 years ago

Express 0.00000000062 kg in scientific notation. 6.2 x 1010kg 6.2 x 10-10kg 0.62 x 10-9kg

PilotLPTM [1.2K]

I believe the correct answer from the choices listed above is the second option. The scientific notation of the measurement 0.00000000062 kg would be 6.2 x 10^-10 kg. Scientific notation is used to express too large and too small values of numbers. Hope this helps. Have a nice day.

7 0

3 years ago

Read 2 more answers

An electron moving to the left at 0.8c collides with a photon moving to the right. After the collision, the electron is moving t

SVETLANKA909090 [29]

Answer:

Wavelength = 2.91 x 10⁻¹² m, Energy = 6.8 x 10⁻¹⁴

Explanation:

In order to show that a free electron can’t completely absorb a photon, the equation for relativistic energy and momentum will be needed, along the equation for the energy and momentum of a photon. The conservation of energy and momentum will also be used.

E = y(u) mc²

Here c is the speed of light in vacuum and y(u) is the Lorentz factor

y(u) = 1/√[1-(u/c)²], where u is the velocity of the particle

The relativistic momentum p of an object of mass m and velocity u is given by

p = y(u)mu

Here y(u) being the Lorentz factor

The energy E of a photon of wavelength λ is

E = hc/λ, where h is the Planck’s constant 6.6 x 10⁻³⁴ J.s and c being the speed of light in vacuum 3 x 108m/s

The momentum p of a photon of wavelenght λ is,

P = h/λ

If the electron is moving, it will start the interaction with some momentum and energy already. Momentum of the electron and photon in the initial and final state is

p(pi) + p(ei) = p(pf) + p(ef), equation 1, where p refers to momentum and the e and p in the brackets refer to proton and electron respectively

The momentum of the photon in the initial state is,

p(pi) = h/λ(i)

The momentum of the electron in the initial state is,

p(ei) = y(i)mu(i)

The momentum of the electron in the final state is

p(ef) = y(f)mu(f)

Since the electron starts off going in the negative direction, that momentum will be negative, along with the photon’s momentum after the collision

Rearranging the equation 1 , we get

p(pi) – p(ei) = -p(pf) +p(ef)

Substitute h/λ(i) for p(pi) , h/λ(f) for p(pf) , y(i)mu(i) for p(ei), y(f)mu(f) for p(ef) in the equation 1 and solve

h/λ(i) – y(i)mu(i) = -h/λ(f) – y(f)mu(f), equation 2

Next write out the energy conservation equation and expand it

E(pi) + E(ei) = E(pf) + E(ei)

Kinetic energy of the electron and photon in the initial state is

E(p) + E(ei) = E(ef), equation 3

The energy of the electron in the initial state is

E(pi) = hc/λ(i)

The energy of the electron in the final state is

E(pf) = hc/λ(f)

Energy of the photon in the initial state is

E(ei) = y(i)mc2, where y(i) is the frequency of the photon int the initial state

Energy of the electron in the final state is

E(ef) = y(f)mc2

Substitute hc/λ(i) for E(pi), hc/λ(f) for E(pf), y(i)mc² for E(ei) and y(f)mc² for E(ef) in equation 3

Hc/λ(i) + y(i)mc² = hc/λ(f) + y(f)mc², equation 4

Solve the equation for h/λ(f)

h/λ(i) + y(i)mc = h/λ(f) + y(f)mc

h/λ(f) = h/lmda(i) + (y(i) – y(f)c)m

Substitute h/λ(i) + (y(i) – y(f)c)m for h/λ(f) in equation 2 and solve

h/λ(i) -y(i)mu(i) = -h/λ(f) + y(f)mu(f)

h/λ(i) -y(i)mu(i) = -h/λ(i) + (y(f) – y(i))mc + y(f)mu(f)

Rearrange to get all λ(i) terms on one side, we get

2h/λ(i) = m[y(i)u(i) +y(f)u(f) + (y(f) – y(i)c)]

λ(i) = 2h/[m{y(i)u(i) + y(f)u(f) + (y(f) – y(i))c}]

λ(i) = 2h/[m.c{y(i)(u(i)/c) + y(f)(u(f)/c) + (y(f) – y(i))}]

Calculate the Lorentz factor using u(i) = 0.8c for y(i) and u(i) = 0.6c for y(f)

y(i) = 1/[√[1 – (0.8c/c)²] = 5/3

y(f) = 1/√[1 – (0.6c/c)²] = 1.25

Substitute 6.63 x 10⁻³⁴ J.s for h, 0.511eV/c2 = 9.11 x 10⁻³¹ kg for m, 5/3 for y(i), 0.8c for u(i), 1.25 for y(f), 0.6c for u(f), and 3 x 10⁸ m/s for c in the equation derived for λ(i)

λ(i) = 2h/[m.c{y(i)(u(i)/c) + y(f)(u(f)/c) + (y(f) – y(i))}]

λ(i) = 2(6.63 x 10-34)/[(9.11 x 10-31)(3 x 108){(5/3)(0.8) + (1.25)(0.6) + ((1.25) – (5/3))}]

λ(i) = 2.91 x 10⁻¹² m

So, the initial wavelength of the photon was 2.91 x 10-12 m

Energy of the incoming photon is

E(pi) = hc/λ(i)

E(pi) = (6.63 x 10⁻³⁴)(3 x 10⁸)/(2.911 x 10⁻¹²) = 6.833 x 10⁻¹⁴ = 6.8 x 10⁻¹⁴

So the energy of the photon is 6.8 x 10⁻¹⁴ J

6 0

3 years ago