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yaroslaw [1]
2 years ago
7

An electron is fired horizontally at 2.5 x 10 m/s between two horizontal parallel plates 7.5 cm long, as shown in the diagram. T

he magnitude of the electric field is 130 N/C. The plate separation is great enough to allow the electron to escape. Edge effects and gravitation are negligible. Find the velocity of the electron as it escapes from between the plates.
Physics
1 answer:
DochEvi [55]2 years ago
8 0

Answer:

the answer to your question is 2.5

Explanation:

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A 190 mH inductor is connected to an emf given by
Bas_tet [7]

(a) The reactance of the inductor is 25.46 ohms.

(b) The expression for the current through the inductor is I(t) = (6.32 A) sin(134t)

<h3>Ractance of the inductor</h3>

The reactance of the inductor is calculated as follows;

XL = ωL

where;

  • ω is angular frequency
  • L is 190 mH

v(t) = (161 V) sin(134t)

v(t) =  V sin(ωt)

The reactance of the inductor is calculated as follows;

XL = (134) x (190 x 10⁻³)

XL = 25.46 ohms

<h3>Peak current in the circuit</h3>

I₀ = V₀/XL

I₀ = (161) / (25.46)

I₀ = 6.32 A

<h3>Expression for the current through the inductor</h3>

I(t) = (6.32 A) sin(134t)

Learn more about inductance here: brainly.com/question/16765199

7 0
1 year ago
Problem 2: (15 pts) A 10-m high cylindrical container is half-filled at the bottom with water of density =1000 kg/m3 while the t
Step2247 [10]

Answer:

\Delta p = 90.7 kPa

Explanation:

specific gravity of oil is = \frac{\rho_{oil}}{\rho_w}

\rho_{oil} = 0.85*1000 = 850 kg/m3

we know that

change in pressure  for oil is given as

\Delta p = \rho gh

here density and h is for oil

\Delta p = 850*5 *9.81 = 41,692.5 kPa

change in pressure  for WATER is given as

\Delta p = \rho gh

here density is for water and h is for water

\Delta p = 1000*5 *9.81 = 49,050 kPa

pressure change due to both is given as

\Delta p = 41692.3 + 49050 = 90742.5 N/m2

\Delta p = 90.7 kPa

8 0
3 years ago
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Determine the minimum applied force p required to move wedge a to the right. the spring is compressed a distance of 175 mm. negl
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<span>b. The coefficient of static friction for all contacting surfaces is μs=0.35. neglect friction at the rollers.

</span>
6 0
3 years ago
If the velocity versus time graph of an object is a horizontal line, the object is
FromTheMoon [43]
A. Moving with constant non-zero speed
3 0
3 years ago
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A 4 kg textbook sits on a desk. It is pushed horizontally with a 50 N applied force against a 15 N frictional force.
GarryVolchara [31]

a) See free-body diagram in attachment

b) The book is stationary in the vertical direction

c) The net horizontal force is 35 N in the forward direction

d) The net force on the book is 35 N in the forward horizontal direction

e) The acceleration is 8.75 m/s^2 in the forward direction

Explanation:

a)

The free-body diagram of a body represents all the forces acting on the body using arrows, where the length of each arrow is proportional to the magnitude of the force and points in the same direction.

From the diagram of this book, we see there are 4 forces acting on the book:

- The applied force, F = 50 N, pushing forward in the horizontal direction

- The frictional force, F_f = 15 N, pulling backward in the horizontal direction (the frictional force always acts in the direction opposite to the motion)

- The weight of the book, W=mg, where m is the mass of the book and g=9.8 m/s^2 is the acceleration of gravity, acting downward. We can calculate its magnitude using the mass of the book, m = 4 kg:

W=(4)(9.8)=39.2 N

- The normal reaction exerted by the desk on the book, N, acting upward, and balancing the weight of the book

b)

The book is in equilibrium in the vertical direction, therefore there is no motion.

In fact, the magnitude of the normal reaction (N) exerted by the desk on the book is exactly equal to the weight of the book (W), so the equation of motion along the vertical direction is

N-W=ma

where a is the acceleration; however, since N = W, this becomes

a=0

And since the book is initially at rest on the desk, this means that there is no motion.

c)

We said there are two forces acting in the horizontal direction:

- The applied force, F = 50 N, forward

- The frictional force, F_f = 15 N, backward

Since they act along the same line, we can calculate their resultant as

\sum F = F - F_f = 50 - 15 = 35 N

and therefore the net force is 35 N in the forward direction.

d)

The net force is obtained as the resultant  of the net forces in the horizontal and vertical direction. However, we have:

- The net force in the horizontal direction is 35 N

- The net force in the vertical direction is zero, because the weight is balanced by the normal reaction

Therefore, this means that the total net force acting on the book is just the net force acting on the horizontal direction, so 35 N forward.

e)

The acceleration of the book can be calculated by using Newton's second law:

\sum F = ma

where

\sum F is the net force

m is the mass

a is the acceleration

Here we have:

\sum F = 35 N (in the forward direction)

m = 4 kg

Therefore, the acceleration is

a=\frac{\sum F}{m}=\frac{35}{4}=8.75 m/s^2 (forward)

Learn more about forces, weight and Newton's second law:

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#LearnwithBrainly

8 0
3 years ago
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