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scoundrel [369]
3 years ago
14

Two large, flat, horizontally oriented plates are parallel to each other, a distance d apart. Half way between the two plates th

e electric field has magnitude E. If the separation of the plates is reduced to d/2 what is the magnitude of the electric field half way between the plates?
a. 0
b. E/2
c. 2E
d. 4E
e. E
Physics
1 answer:
melisa1 [442]3 years ago
8 0

Answer:

E

Explanation:

From the equation describing the Electric field strength of a charge

E=F/q

We can deduce that the electric field strength is dependent on the force arcing on the magnitude of the charge.

As the force increases the the electric field increases, also the electric field strength decreases with increasing charge.

Hence the distance has no effect on the electric field strength. The magnitude remain as E

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Mr Smith is working in a muddy garden. When he picks up a paving stone his feet sink deeper into the mud. Explain why his feet s
Anni [7]

because he is carrying more mass and as the ground is muddy his feet goes in due to the pull of gravity

3 0
3 years ago
1- A train travels 92 kilometers in 3 hours, and then 55 kilometers in 3 hours. What is its average speed?
Hunter-Best [27]
<h3>Answer:</h3>
  1. 24.5 km/h
  2. 4 17/27 m/s
  3. 11/3 m/s²
<h3>Explanation:</h3>

1. The average speed is the ratio of total distance to total time:

... speed = distance/time = (92 km +55 km)/(3 h +3h) = (147 km)/(6 h)

... = 24.5 km/h

2. speed = distance/time = (125 m)/(27 s) = 4 17/27 m/s

3. a = ∆v/∆t = (15 m/s -4 m/s)/(3 s) = 11/3 m/s²

8 0
3 years ago
The mass of a cannon ball is 10 kg. If the speed of the cannon ball is 50 m/s in
Maurinko [17]

Answer:

Explanation:

Kinetic Energy = 0.5(Mass)(Velocity2)

Kinetic energy= 0.5 × 10kg × (50m/s)2

Kinetic Energy = 5kg × 2500m/s

Kinetic energy = 125000 J ( Ans)

7 0
3 years ago
The design speed of a multilane highway is 60 mi/hr. What is the minimum stopping sight distance that should be provided on the
kicyunya [14]

Answer:

Part a: When the road is level, the minimum stopping sight distance is 563.36 ft.

Part b: When the road has a maximum grade of 4%, the minimum stopping sight distance is 528.19 ft.

Explanation:

Part a

When Road is Level

The stopping sight distance is given as

SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}

Here

  • SSD is the stopping sight distance which is to be calculated.
  • u is the speed which is given as 60 mi/hr
  • t is the perception-reaction time given as 2.5 sec.
  • a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
  • G is the grade of the road, which is this case is 0 as the road is level

Substituting values

                              SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0)}\\SSD=220.5 +342.86 ft\\SSD=563.36 ft

So the minimum stopping sight distance is 563.36 ft.

Part b

When Road has a maximum grade of 4%

The stopping sight distance is given as

SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}

Here

  • SSD is the stopping sight distance which is to be calculated.
  • u is the speed which is given as 60 mi/hr
  • t is the perception-reaction time given as 2.5 sec.
  • a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
  • G is the grade of the road, which is given as 4% now this can be either downgrade or upgrade

For upgrade of 4%, Substituting values

                              SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35+0.04)}\\SSD=220.5 +307.69 ft\\SSD=528.19 ft

<em>So the minimum stopping sight distance for a road with 4% upgrade is 528.19 ft.</em>

For downgrade of 4%, Substituting values

                              SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0.04)}\\SSD=220.5 +387.09 ft\\SSD=607.59ft

<em>So the minimum stopping sight distance for a road with 4% downgrade is 607.59 ft.</em>

As the minimum distance is required for the 4% grade road, so the solution is 528.19 ft.

3 0
3 years ago
Question 8 of 10
Marrrta [24]

Answer:

C

Explanation:

If the theory were to be proved you you need to repeat the experiment over and over again so that way you can prove that it is true wuth the same results.

3 0
3 years ago
Read 2 more answers
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