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Mrac [35]
3 years ago
7

An electron has an initial speed of

Physics
1 answer:
Inessa05 [86]3 years ago
7 0

Initial velocity, u= 4.18E^5 m/s

Final velocity, v= 5.19E^5 m/s

a = 3E^14 m/s²

from,

a = <u>(</u><u>v</u><u> </u><u>-</u><u> </u><u>u</u><u>)</u><u> </u>

t

t = <u>(</u><u>v</u><u> </u><u>-</u><u> </u><u>u</u><u>)</u><u> </u>

a

t = <u>(</u><u>5</u><u>.</u><u>1</u><u>9</u><u>E</u><u>^</u><u>5</u><u> </u><u>-</u><u> </u><u>4</u><u>.</u><u>1</u><u>8</u><u>E</u><u>^</u><u>5</u><u>)</u>

3E^14

t = 3.37 × 10^-10s

side note; the E represents 10 power.

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Help me yall plz..........
patriot [66]

Answer:

mark me brainlyest and ill answer

Explanation:

6 0
3 years ago
Liquid water at 120 kPa enters a 7-kW pump where its pressure is raised to 5.6 MPa. If the elevation difference between the exit
ziro4ka [17]

Answer:

The answer is given below

Explanation:

Things provided in the statement:

Pressure <em>P1</em> = 120 kPa and <em>P2</em> = 5.6 MP or 5600 kPa

Power, <em>W</em> = 7 kW

Elevation difference = ∆z = 10 m

Mass of flow = m˙

So potential energy changes may be significant

Specific volume of water V= 0.001 m³/kg

Now putting the values in the formula

Power, <em>W </em>= m˙  x V (<em>P1 - P2</em>) + m˙ x g x ∆z

             7 = m˙  x 0.001 (5600 - 120 ) + m˙ x 9.8 x 10 x (1 kJ/kg/ 1000 m^2/s^2)

             7 = m˙ x 5.48 + m˙ x 0.098

             7 = m ˙x 5.38

            m˙ = 7/5.38

          So mass flow m˙  =  1.301 kJ/s

8 0
3 years ago
Why is it more difficult to lean over and push a heavy box across the floor than it is to attach a rope and pull the box at the
melisa1 [442]

Answer:

Case 1: <u>Pushing</u> Diagram 1

Leaning over and Pushing the heavy box from the floor, the push will be divided in to two parts, one is horizontal that can help the box move, and one is vertically downwards, which increases the downward force of the heavy object (an addition to the gravity) and thus increases friction, making it very hard to push.  When you push at certain angle, you are exhibiting two forces as shown in diagram 1.

  1. Horizontal force acting along the plane.
  2. Vertical force downward perpendicular to the surface.

Case 2: <u>Pulling</u> Diagram 2

Pulling on a rope similar object at the same angle, the pull can be divided into two parts, one is horizontal that can help the box move, and one is vertically upwards, which decreases the downwards force of the box (a subtraction in the gravity) and thus decreases friction, making it very easy to pull. When you pull at a certain angle, you are exhibiting two forces as shown in diagram 2.

  1. Horizontal force acting along the plane.
  2. Vertical force upward perpendicular to the surface.

So, in the case of pushing, it adds an extra weight on the object, which results in difficulty to push that object at the same angle.  In case of pulling, the upward perpendicular force, it tries to lift the  object upward and divided the weight partially. Thus making it easier to move the object at same angle.

8 0
3 years ago
A(n) 19000 kg freight car is rolling along a track at 1.5 m/s. Calculate the time needed for a force of 2600 N to stop the car.
garik1379 [7]

Answer: 10.96secs

Explanation:

According to newton's second law

Force = mass × acceleration

F = ma

F = m(v-u)/t

Cross multiplying we have

Ft = m(v-u)

t = m(v-u)/F

Given F = 2600N m = 19,000kg v = 1.5m/s u = 0m/s

Substituting this values in the formula for the time we have

t = 19000(1.5-0)/2600

t = 28,500/2600

t= 10.96secs

7 0
3 years ago
15) In which of the following cases would sound reach each ear out of phase? A. You are standing directly in front of the sound
Sonja [21]

In one of the greatest coincidences to arise on Brainly in quite some time, Choice-C is the correct choice for BOTH #15 and #16 .

Choice-C is the only situation in which the source is a different distance from each ear.

7 0
3 years ago
Read 2 more answers
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