The formation of a cell plate is a characteristic of

What is the electric potential energy of an electron at the negative end of the cable, relative to the positive end of the cable

VashaNatasha [74]

Answer:

Electric potential energy at the negative terminal: $1.92\cdot 10^{-18}J$

Explanation:

When a particle with charge $q$ travels across a potential difference $\Delta V$ , then its change in electric potential energy is

$\Delta U = q \Delta V$

In this problem, we know that:

The particle is an electron, so its charge is

$q=-1.60\cdot 10^{-19}C$

We also know that the positive terminal is at potential

$V_+=0V$

While the negative terminal is at potential

$V_-=-12 V$

Therefore, the potential difference (final minus initial) is

$\Delta V = -12-0 = -12 V$

So, the change in potential energy of the electron is

$\Delta U = (-1.6\cdot 10^{-19})(-12)=1.92\cdot 10^{-18}J$

This means that the electron when it is at the negative terminal has $1.92\cdot 10^{-18}J$ of energy more than when it is at the positive terminal.

Since the potential at the positive terminal is 0, this means that the electric potential energy of the electron at the negative end is

$1.92\cdot 10^{-18}J$

3 0

3 years ago

Two boats leave the same port at the same time, with boat A traveling north at 15 knots (nautical miles per hour) and boat B tra

Mrrafil [7]

Answer:

The chance in distance is 25 knots

Explanation:

The distance between the two particles is given by:

$s^2 = (x_A - x_B)^2+(y_A - y_B)^2$ (1)

Since A is traveling north and B is traveling east we can say that their displacement vector are perpendicular and therefore (1) transformed as:

$s^2 = x_B^2+y_A^2$ (2)

Taking the differential with respect to time:

$\displaystyle{2s\frac{ds}{dt}= 2x_B\frac{dx_B}{dt}+2y_A\frac{dy_A}{dt}}$ (3)

where $\displaystyle{\frac{dx_B}{dt}}=v_B$ and $\displaystyle{\frac{dx_A}{dt}}=v_A$ are the respective given velocities of the boats. To find $s$ and $x_B$ we make use of the given position for A, $y_A=30$ , the Pythagoras theorem and the relation between distance and velocity for a movement with constant velocity.

$\displaystyle{y_A = v_A\cdot t\rightarrow t = \frac{y_A}{v_A}=\frac{30}{15}=2 h$

with this time, we know can now calculate the distance at which B is:

$\displaystyle{x_B = v_B\cdot t= 20 \cdot 2 = 40\ nmi$

and applying Pythagoras:

$\displaystyle{s = \sqrt{x_B^2+y_A^2}=\sqrt{30^2 + 40^2}=\sqrt{2500}=50}$

Now substituting all the values in (3) and solving for $\displaystyle{\frac{ds}{dt} }$ we get:

$\displaystyle{\frac{ds}{dt} = \frac{1}{2s}(2x_B\frac{dx_B}{dt}+2y_A\frac{dy_A}{dt})}\\\displaystyle{\frac{ds}{dt} = 25 \ knots}$

4 0

3 years ago

Which could most likely describe the three surfaces?

sergiy2304 [10]

Answer:

Surface 1 is blacktop, Surface 2 is gravel, and Surface 3 is ice.

Explanation:

Hope this helps! :]

3 0

2 years ago

Read 2 more answers

A team of ecologists are studying four different ecosystems with varying levels of biodiversity. The ecologists categorize the d

Ann [662]

Answer:

B. Ecosystem B, because its high species diversity could have resulted from increased competition among its members.

Explanation:

This is because, in the ecosystem with varying level of biodiversity, Ecosystem B has medium level of species diversity found in them with High medium level of habitat diversity which causes increasing competitions among them.

4 0

3 years ago

Imagine an alternate universe where the value of the Planck constant is . In that universe, which of the following objects would

HACTEHA [7]

Question: The planck constant was not given. In this calculation, planck constant of 6.62607*10^-9 Js is used for the calculation.

Answer:

(a) A virus -------------Classical

(b) A buckyball -----Classical

(c) A mosquito ------ Quantum

(d) A turtle ------------Quantum

Explanation:

Calculating the wavelength using the formula;

λ= h/(mv)

where

λ= Wavelength

h = Planck Constant = 6.62607*10^-9 Js

m = mass in kg

v = velocity in m/s

Virus size = 280. nm = 2.80*10⁻⁷ m

a)

A Virus:

m = 9.4 x 10-17 g 9.4*10⁻²⁰ kg

v = 0.50 µm/s = 5 *10⁻⁷ m/s

h = 6.62607*10^-9 Js

Virus size = 280 nm = 2.80*10⁻⁷ m

Substituting into the formula; we have

λ= h/(mv)

λ= 6.62607*10^-9/ (9.4*10⁻²⁰* 5 *10⁻⁷)

= 6.62607*10^-9/4.7*10^-26

= 1.4*10^17 m

Classical : Wavelength is bigger than it's size

(b)

A buckyball

m = 1.2 x 10-21 g = 1.2 *10⁻²⁴ kg

V = 37 m/s

Size = 0.7 nm = 7*10⁻¹⁰ m

Substituting into the formula, we have

λ= h/(mv)

λ= 6.62607*10^-9/ ( 1.2 *10⁻²⁴* 37)

= 6.62607*10^-9/4.44*10^-23

= 1.49 *10^14 m

Classical : Wavelength is bigger than it's size

(c)

A mosquito

Mass = 1.0 mg = 1*10⁻⁶ kg

v = 1.1 m/s

Size = 6.3 mm = 6.3*10⁻³ m

Substituting into the formula, we have

λ= h/(mv)

λ= 6.62607*10^-9/ ( 1*10⁻⁶* 1.1)

= 6.62607*10^-9/1.1*10^-6

= 6.02*10^-3 m

Quantum Approach: The wavelength and the size are comparable

(d)

A turtle

Mass = 710. g = 0.71 kg

Size = 22. cm = 0.22 m

V = 2.8 cm/s. = 0.028 m/s

Substituting into the formula, we have

λ= h/(mv)

λ= 6.62607*10^-9/ ( 0.71* 0.028)

= 6.62607*10^-9/0.01988

= 3.33*10^-7 m

Quantum Approach: The wavelength and the size are comparable

8 0

3 years ago