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vovikov84 [41]
2 years ago
5

If the axes of the two cylinders are parallel, but displaced from each other by a distance d, determine the resulting electric f

ield in the region R>R3, where the radial distance R is measured from the metal cylinder's axis. Assume d<(R2−R1). Express your answer in terms of the variables ρE, R1, R2, R3, d, R, and appropriate constants.
Physics
1 answer:
Novay_Z [31]2 years ago
6 0

Answer:

E =  ρ ( R1²) / 2 ∈o R

Explanation:

Given data

two cylinders are parallel

distance = d

radial distance = R

d < (R2−R1)

to find out

Express answer in terms of the variables ρE, R1, R2, R3, d, R, and constants

solution

we have two parallel cylinders

so area is 2 \pi R × l

and we apply here gauss law that is

EA = Q(enclosed) / ∈o   ......1

so first we find  Q(enclosed) = ρ Volume

Q(enclosed) = ρ ( \pi R1² × l )

so put all value in equation 1

we get

EA = Q(enclosed) / ∈o

E(2 \pi R × l)  = ρ ( \pi R1² × l ) / ∈o

so

E =  ρ ( R1²) / 2 ∈o R

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sergiy2304 [10]

After 6 seconds, the car will surpass the cyclist.

<h3><u>Explanation:</u></h3>

The speed of the cyclist = 6 m/s.

Let after time t sec, the car will overtake the cyclist.

So, distance covered by the cyclist in t sec = 6t m

Initial velocity of the car is 0 m/s, because the car is just starting.

Acceleration of the car =2 m/s^2.

Final velocity of the car =6 m/s.

So to cover the distance 6t, the time required by the car = \frac{1}{2} \times a \times t^2 = \frac{1}{2} \times 2 \times t^2

6t = \frac{1}{2} \times 2 \times t^2\\6t = t^2

t =6 sec

So, after 6 seconds, the car will surpass the cycle.  

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2 years ago
A box of mass 5.0 kg is accelerated from rest by a force across a floor at a rate of 2.0 m/s2 for 7.0 s. find the net work done
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W=Fd. Force is not given so we solve for it. F=ma, m=5kg, a=2m/s^2, F=10N. Distance is not given so we solve for it, x=.5a(t^2)=.5(2)(7x7)=49m. F=10N, d=49m, W=490J.
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2 years ago
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Notice that in each conversion factor the numerator equals the denominator when units are taken into account. A common error in
navik [9.2K]

Answer:

he factor for the temporal part 1.296 107 s² = h²

 m / s² = 12960 km / h²

Explanation:

This is a unit conversion exercise.

In the unit conversion, the size of the object is not changed, only the value with respect to which it is measured is changed, for this reason in the conversion the amount that is in parentheses must be worth one.

In this case, it is requested to convert a measure km/h²

Unfortunately, it is not clearly indicated what measure it is, but the most used unit in physics is   m / s² , which is a measure of acceleration. Let's cut this down

the factor for the distance is 1000 m = 1 km

the factor for time is 3600 s = 1 h

let's make the conversion

        m / s² (1km / 1000 m) (3600 s / 1h)²

note that as time is squared the conversion factor is also squared

        m / s² = 12960 km / h²

the factor for the temporal part 1.29 107 s² = h²

6 0
2 years ago
A submarine dives from rest a 100-m distance beneath the surface of an ocean. Initially, the submarine moves at a constant rate
tiny-mole [99]

Answer:

a. Time = 16.11 s

b. Gauge Pressure = 1009400 Pa = 1 MPa  

c. Absolute Pressure = 1110725 Pa + 1.11 MPa

d. Force = 2.22 MN

Explanation:

a.

For the accelerated part of motion of submarine we can use equations of motion.

Using 1st equation of motion:

Vf = Vi + at₁

t₁ = (Vf - Vi)/a

where,

t₁ = time taken during accelerated motion = ?

Vf = final velocity = 4 m/s

Vi = Initial Velocity = 0 m/s   (Since, it starts from rest)

a = acceleration = 0.3 m/s²

Therefore,

t₁ = (4 m/s - 0 m/s)/(0.3 m/s²)

t₁ = 13.33 s

Now, using 2nd equation of motion:

d₁ = (Vi)(t₁) + (0.5)(a)(t₁)²

where,

d₁ = the depth covered during accelerated motion

Therefore,

d₁ = (0 m/s)(13.33 s) + (0.5)(0.3 m/s²)(13.33 s)²

d₁ = 88.89 m

Hence,

d₂ = d - d₁

where,

d₂ = depth covered during constant speed  motion

d = total depth = 100 m

Therefoe,

d₂ = 100 m - 88.89 m

d₂ = 11.11 m

So, for uniform motion:

s₂ = vt₂

where,

v = constant speed = 4 m/s

t₂ = time taken during constant speed  motion

11.11 m = (4 m/s)t₂

t₂ = 2.78 s

Therefore, total time taken by submarine to move down 100 m is:

t = t₁ + t₂

t = 13.33 s + 2.78 s

<u>t = 16.11 s</u>

<u></u>

b.

The gauge pressure on submarine can be calculated by the formula:

Pg = ρgh

where,

Pg = Gauge Pressure = ?

ρ = density of salt water = 1030 kg/m³

g = 9.8 m/s²

h = depth = 100 m

Therefore,

Pg = (1030 kg/m³)(9.8 m/s²)(100 m)

<u>Pg = 1009400 Pa = 1 MPa</u>

<u></u>

c.

The absolute pressure is given as:

P = Pg + Atmospheric Pressure

where,

P = Absolute Pressure = ?

Atmospheric Pressure = 101325 Pa

Therefore,

P = 1009400 Pa + 101325 Pa

<u>P = 1110725 Pa + 1.11 MPa</u>

<u></u>

d.

Since, the force to open the door must be equal to the force applied to the door by pressure externally.

Therefore, the  force required to open the door can be found out by the formula of pressure:

P = F/A

F = PA

where,

P = Absolute Pressure on Door = 1110725 Pa

A = Area of door = 2 m²

F = Force Required to Open the Door = ?

Therefore,

F = (1.11 MPa)(2 m²)

<u>F = 2.22 MN</u>

8 0
2 years ago
If 500 thousand btus of energy are available to raise the temperature of a water boiler from 20°f to 100°f, how many gallons o
sashaice [31]
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Then we need to convert the weight into gallons. The calculation would be:
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7 0
2 years ago
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