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labwork [276]
3 years ago
15

A long straight cylindrical shell has an inner radius R_i and an outer radius R_0. It carries a current i, uniformly distributed

over its cross section. A wire is parallel to the cylinder axis, in the hollow region (r < R_i).The magnetic field is zero everywhere outside the shell (r > R_0). We conclude that the wire:
a. is on the cylinder axis and carries current i in the same direction as the current in the shell
b. may be anywhere in the hollow region but must be carrying current i in the direction opposite to that of the current in the shell
c. may be anywhere in the hollow region but must be carrying current i in the same direction as the current in the shell
d. is on the cylinder axis and carries current i in the direction opposite to that of the current in the shell
e. does not carry any current
Physics
1 answer:
Irina-Kira [14]3 years ago
7 0

Answer:

Correct answer is option D

- Wire is on the cylinder axis and carries current i in the direction opposite to that of the current in the shell

Explanation:

- It cannot be Option E, because the magnetic field outside the wire would not be 0 due to the current carried by the conductor

-Also, the parallel wire cannot carry current in the same direction because, that would amplify the magnetic field created by the outer cylinder (since B is dir. proportional to the current) -and now, that leaves only option C and D. If, it is Option C, then that means one side of the cylinder would be more closer to the parallel wire than the other, so there would be different B fields on the two opposite sides of the cylinder. So, that means the answer is option D.

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Greg is boiling water on the stove for some noodles. what object can he safely hold while draining the water from the noodles af
zmey [24]
In order for Greg to safely drain the water out of the noodles, he should use potholders or any thing that is does not conduct heat or transfer heat. Some pots are also equipped with handles that are made of plastics for safely transferring of its content to another container. 
5 0
3 years ago
You push on a cart (18.0kg) at a 30 degree below horizontal angle. The coefficient of kinetic friction between the chair and the
podryga [215]

Given that force is applied at an angle of 30 degree below the horizontal

So let say force applied if F

now its two components are given as

F_x = Fcos30


F_y = Fsin30


Now the normal force on the block is given as

N = Fsin30 + mg

N = 0.5F + (18\times 9.8)

N = 0.5F + 176.4

now the friction force on the cart is given as

F_f = \mu N

F_f = 0.625(0.5F + 176.4)

F_f = 110.25 + 0.3125F

now if cart moves with constant speed then net force on cart must be zero

so now we have

F_f + F_x = 0

Fcos30 - (110.25 + 0.3125F) = 0

0.866F - 0.3125F = 110.25

F = 199.2 N

so the force must be 199.2 N

8 0
3 years ago
At 4.00 l, an expandable vessel contains 0.864 mol of oxygen gas. how many liters of oxygen gas must be added at constant temper
svet-max [94.6K]

Givens
=====
V = 4.00 L
T = 273oK We're assuming the temperature does not change, just the pressure.
n = 0.864 moles
R = 8.314 joules / mole  * oK
P = ?????

Formula
======
PV = n*R*T
P = n*R*T/V
P = 0.864 * 8.314 * 273 / 4
P = 490 kpa


You have to add 1.6 – 0.864 = 0.736 moles of gas.


We have to assume that the temperature and pressure remain the same when we add the 0.736 moles of gas. We are now looking for the volume.


PV = n*R*T

<span> V = 0.736 * 8.314 * 273 / 490</span>

V = 3.41 L Remember this is at about 4 atmospheres so we have to convert to Standard Pressure.

Total Volume = 3.41 + 4.00 = 4.41


V1 * P1 = V2 * P2

P1 = 490 kPa

P2 = 101 kPa

V1 = 7.41 L

V2 = ????


<span> <span> 7.41* 490 = V2 * 101 V2 = 7.41 * 490 / 101 V2 = 35.94 L </span> </span>


<span>You had 4 L now you need 31.94 more.</span> 
6 0
3 years ago
Nuclearissionoccursthroughmanydifferent pathways. For the ission of U-235 induced by a neutron, write a nuclear equation to form
ruslelena [56]

Answer: a) ^{235}_{92}\textrm{U}+^{1}_{0}\textrm{n} \rightarrow ^{87}_{35}\textrm {Br}+ ^{146}_{57}\La + 3^{1}_{0}n

b) ^{235}_{92}\textrm{U}+^1_0\textrm{n}\rightarrow ^{140}_{56}\textrm{Ba}+^{94}_{36}\textrm{Kr}+2^1_0\textrm{n}

Explanation:

A nuclear fission reaction is defined as the reaction in which a heavy nucleus splits into small nuclei along with release of energy.

a) The given reaction is ^{235}_{92}\textrm{U}+^{1}_{0}\textrm{n} \rightarrow ^{87}_{35}\textrm {Br}+ ^{146}_{57}\La + x^{1}_{0}n

Now,  as the mass on both reactant and product side must be equal:

235+1=87+146+x

x=3

Thus three neutrons are produced and nuclear equation will be: ^{235}_{92}\textrm{U}+^{1}_{0}\textrm{n} \rightarrow ^{87}_{35}\textrm {Br}+ ^{146}_{57}\La + 3^{1}_{0}n

b) For the another fission reaction:

^{235}_{92}\textrm{U}+^1_0\textrm{n}\rightarrow ^{A}_{56}\textrm{Ba}+^{94}_{Z}\textrm{X}+2^1_0\textrm{n}

To calculate A:

Total mass on reactant side = total mass on product side

235 + 1 = A + 94 + 2

A = 140

To calculate Z:

Total atomic number on reactant side = total atomic number on product side

92 + 0 = 56 + Z + 0

Z = 36

As Krypton has atomic number of 36,Thus the nuclear equation will be :

^{235}_{92}\textrm{U}+^1_0\textrm{n}\rightarrow ^{140}_{56}\textrm{Ba}+^{94}_{36}\textrm{Kr}+2^1_0\textrm{n}

         

8 0
3 years ago
Water flows through a cylindrical pipe of varying cross-section. The velocity is 3.0 m/s at a point where the pipe diameter is 1
Setler79 [48]

Answer:

The flow rate is  R =2.357 *10^{-4} \ m^3/s

Explanation:

From the question we are told that

    The velocity is v  =  3.0 \ m/s

   The  diameter of the pipe is  d =  1.0 \ cm  = 0.01 \ m

 

The  radius of the pipe is mathematically represented as

            r =  \frac{d}{2}

substituting values

            r =  \frac{0.01}{2}

           r =  0.005 \ m

The flow rate is mathematically represented as

       R  =  v  * A

Where is the cross-sectional area of the pipe which is mathematically evaluated as

      A   = \pi r^2

substituting values

      A   =  3.142 *  (0.005)^2

     A   = 7.855 *  10^{-5} \ m^2

So

    R  =  3.0  *  7.855 *10^{-5}

    R  =  2.357*10^{-4} \ m^3 /s

8 0
3 years ago
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