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vitfil [10]
3 years ago
8

A long, thin rod parallel to the y-axis is located at x = - 1 cm and carries a uniform positive charge density λ = 1 nC/m . A se

cond long, thin rod parallel to the z-axis is located at x = +1 cm and carries a uniform negative charge density λ = - 1 nC/m. What is the electric field at the origin?
Physics
1 answer:
zheka24 [161]3 years ago
7 0

Answer:

The electric field at origin is 3600 N/C

Solution:

As per the question:

Charge density of rod 1, \lambda = 1\ nC = 1\times 10^{- 9}\ C

Charge density of rod 2, \lambda = - 1\ nC = - 1\times 10^{- 9}\ C

Now,

To calculate the electric field at origin:

We know that the electric field due to a long rod is given by:

\vec{E} = \frac{\lambda }{2\pi \epsilon_{o}{R}

Also,

\vec{E} = \frac{2K\lambda }{R}                  (1)

where

K = electrostatic constant = \frac{1}{4\pi \epsilon_{o} R}

R = Distance

\lambda = linear charge density

Now,

In case, the charge is positive, the electric field is away from the rod and towards it if the charge is negative.

At x = - 1 cm = - 0.01 m:

Using eqn (1):

\vec{E} = \frac{2\times 9\times 10^{9}\times 1\times 10^{- 9}}{0.01} = 1800\ N/C

\vec{E} = 1800\ N/C     (towards)

Now, at x = 1 cm = 0.01 m :

Using eqn (1):

\vec{E'} = \frac{2\times 9\times 10^{9}\times - 1\times 10^{- 9}}{0.01} = - 1800\ N/C

\vec{E'} = 1800\ N/C     (towards)

Now, the total field at the origin is the sum of both the fields:

\vec{E_{net}} = 1800 + 1800 = 3600\ N/C

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What is the freezing point of an aqueous solution that boils at 101 degrees C? Kfp = 1.86 K/m and Kbp = 0.512 K/m. Enter your an
astra-53 [7]

Answer:

-3.63 degree Celsius

Explanation:

We are given that

Boiling point of solution=T_b=101^{\circ} C

Boiling point water=100 degree Celsius

K_f=1.86K/m

K_b=0.512 K/m

\Delta T_b=T-T_0

Where T=Boiling point of solution

T_0=Boiling point of pure solvent

\Delta T_b=101-100=1^{\circ}C

\Delta T_b=k_bm

Using the formula

1=0.512\times m

Molality,m=\frac{1}{0.512} m

\Delta T_f=k_fm

Using the formula

\Delta T_f=\frac{1}{0.512}\times 1.86

\Delta T_f=3.63 C

We know that

\Delta T_f=T_0-T_1

Where T_0 =Freezing point of solvent

T_1= Freezing point of solution

Using the formula

3.63=0-T_1

Freezing point of water=0 degree Celsius

T_1=0-3.63=-3.63 C

Hence, the freezing point of solution=-3.63 degree Celsius

7 0
3 years ago
What is the sum of 130+125+191? A. 335. B. 456. C. 446. D. 426.
Vesnalui [34]
1. C- 446
2. A- 778
3. B- 400
4. A- 11
8 0
2 years ago
A cannonball is catapulted toward a castle. The cannonball’s velocity when it leaves the catapult is 51.6 m/s at an angle of 37.
Andrei [34K]

Answer:

v_{y}=35.21m/s

Explanation:

From the exercise we know the cannonball's <u>initial velocity</u>, the <u>angle</u> which its released with respect to the horizontal and its <u>initial height</u>

v_{o}=51.6m/s\\\beta =37.0º\\y_{o}=7m

If we want to know whats the <u>y-component of velocity</u> we need to use the following formula:

v_{y}^2=v_{oy}^2+ag(y-y_{o})

Knowing that g=-9.8m/s^2

v_{y}=\sqrt{((51.6m/s)sin(37))^2-2(9.8m/s^2)(0m-7m)}=35.21m/s

So, the cannonball's y-component of velocity is v_{y}=35.21m/s

6 0
3 years ago
Check my work please
Amiraneli [1.4K]

We can use the ideal gas equation which is expressed as PV = nRT. At a constant volume and number of moles of the gas the ratio of T and P is equal to some constant. At another set of condition, the constant is still the same. Calculations are as follows:

T1/P1 = T2/P2

P2 = T2 x P1 / T1

P2 = 273 x 340 / 713

<span>P2 = 130 kPa</span>

5 0
3 years ago
Pleases help with science ASAP!!
sattari [20]
Yes I am subscribed to coryxkenshin but are you subscribed to me jp
3 0
3 years ago
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