Pls tell the answer

Please Help ASAP I Need Help with question 5 I got the answer wrong this is my second and final attempt for the unit test

victus00 [196]

The answer is Electrical

8 0

3 years ago

Read 2 more answers

An initially stationary electron is accelerated by a uniform 640 N/C Electric Field. a) Calculate the kinetic energy of the elec

bulgar [2K]

Answer:

(a) 1.298 * 10^(-4) J

(b) 5.82 * 10^6 m/s

Explanation:

Parameters given:

Electric field, E = 640 N/C

Distance traveled by electron, r = 15 cm = 0.15 m

Mass of electron, m = 9.11 * 10^(-31) kg

Electric charge of electron, q = 1.602 * 10^(-19) C

(a) The kinetic energy of the electron in terms of Electric field is given as:

K = (q² * E² * r²) / 2m

Therefore, Kinetic energy, K, is:

K = [(1.602 * 10^(-19))² * 640² * 0.15²] / [2 * 9.11 * 10^(-31)]

K = {23651.981 * 10^(-38)} / [18.22 * 10^(-31)]

K = 1298.13 * 10^(-7) J = 1.298 * 10^(-4) J

(b) To find the final velocity of the electron, we have to first find the acceleration of the electron. This can be gotten by using the equations of force.

Force is generally given as:

F = ma

Electric force is given as:

F = qE

Therefore, equating both, we have:

ma = qE

a = (qE) / m

a = (1.602 * 10^(-19) * 640) / (9.11 * 10^(-31))

a = 112.54 * 10^(12) m/s² = 1.13 * 10^(14) m/s²

Using one of the equations of motion, we have that:

v² = u² + 2as

Since the electron started from rest, u = 0 m/s

Therefore:

v² = 2 * 1.13 * 10^(14) * 0.15

v² = 3.39 * 10^(13)

v = 5.82 * 10^6 m/s

The velocity of the electron after moving a distance of 15 cm is 5.82 * 10^6 m/s.

8 0

3 years ago

1. Take the speed at which the body moves as 10 m/s 2. Calculate the distance travelled by the body every second 3. Take the tim

exis [7]

Answer:

a)this graph is also a line b) in both cases we have a uniform movement

Explanation:

In this exercise we have a uniform movement

v = d / t

d = v t

in the table we give some values to make the graph

t (s) d (m)

1 10

2 20

3 30

In the attached we can see the graph that is a straight line

we have another vehicle at v = 50 me / S

t (s) d (m)

1 50

2 100

3 150

this graph is also a line

b) in both cases we have a uniform movement

3 0

4 years ago

A wheel rotates about a fixed axis with a constant angular acceleration of 4.0 rad/s2. The diameter of the wheel is 40 cm. What

Ray Of Light [21]

Answer:

v = 0.42m/s

Explanation:

In order o calculate the linear speed of the point at the border of the wheel, you first take into account that the total acceleration of such a point is given by:

$a_{total}^2=a_r^2+a_t^2$ (1)

atotal: total acceleration = 1.2m/s^2

ar: radial acceleration of the wheel

at: tangential acceleration

The tangential acceleration is also given by:

$a_t=r\alpha$ (2)

r: radius of the wheel = (40cm/2 )= 20cm = 0.2m

α: angular acceleration = 4.0rad/s^2

You replace the expression (2) into the expression (1) and solve for the radial acceleration:

$a_{total}^2=a_r^2+(r\alpha)^2\\\\a_r=\sqrt{(a_{total})^2-(r\alpha)^2}\\\\a_r=\sqrt{(1.2m/s^2)^2-((0.2m)(4.0rad/s^2))^2}=0.894\frac{m}{s^2}$

Next, you use the following formula for the radial acceleration and solve for the linear speed:

$a_r=\frac{v^2}{r}\\\\v=\sqrt{ra_r}=\sqrt{(0.2m)(0.894m/s^2)}=0.42\frac{m}{s}$

The linear speed of the point at the border of the wheel is 0.42m/s

5 0

3 years ago

An electron moving in the direction of the +x-axis enters a magnetic field. If the electron experiences a magnetic deflection in

algol13

Answer:

- z axis

Explanation:

The magnetic force is perpendicular to its speed and the direction of the magnetic field, its equation being

F = q v x B

Where the bold indicates vectors, q is a positive charge,

The way to use this equation is with the rule of the right hand, the thumb points in the direction of the velocity, the other fingers extended in the direction of the magnetic field and the palm is the direction of the force for a positive charge.

Let's apply this to our case the speed is on the x axis and the force in ex. negative Y axis, the other fingers point perpendicular to the sheet, but since the charge is negative the magnetic field enters the sheet, this would be the negative z axis

8 0

4 years ago

Pls tell the answer​

Pls tell the answer