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Darina [25.2K]
3 years ago
13

An electric dipole is in a uniform electric field of magnitude 8.50×104N/C. The charges in the dipole are separated by 1.10×10−1

0m, and the torque on the dipole when its dipole moment is perpendicular to the electric field is 6.60×10−26N⋅m. Calculate the magnitude of the charges that make up the dipole.
Physics
2 answers:
VARVARA [1.3K]3 years ago
7 0

Answer:

Magnitude of charges = 0.705 x 10⁻²⁰ C

Explanation:

Magnitude of electric field = 850 x 10⁴ N/C

Distance between dipoles = r 1.10 x 10⁻¹⁰ m

Torque on dipole = 6.6 x 10⁻²⁶ N.m

Magnitude of charges = q= ?

Torque on electric dipole is given by formula

\tau = pE \,sin\,theta\\\\p=q.r\\\\\theta = 90^o\\\\\tau=qrE\,sin\,(90)\\\\\tau=qrE\\\\q=\frac{\tau}{rE}\\\\q=\frac{6.6\times 10^{-26}}{(1.1\times 10^{-10}) (8.5\times 10^{4})}\\\\q=0.705\times 10^{-20}\, C

zmey [24]3 years ago
4 0

Answer: q = 7.06 × 10^-21 C

the magnitude of the charges that make up the dipole is 7.06 × 10^-21 C

Explanation:

Given:

Torque, τ = 6.60×10^−26N⋅m

Angle made by p with a uniform electric field, θ = 90° (perpendicular)

Electric field, E = 8.50×10^4N/C

Length between dipole r = 1.10×10^−10 m

Torque acting on the dipole is given by the relation,

τ = pE sinθ....1

But,

p = qr .....2

Substituting equation 1 to 2

τ= qrEsinθ ....3

Making q the subject of formula

q = τ/rEsinθ .....4

Where;

q = magnitude of the charges that make up the dipole.

Substituting the given values into equation 4:

q = 6.60×10^−26N⋅m/(1.10×10^−10 m × 8.50×10^4N/C × sin90°)

q = 0.70588 × 10^-20 C

q = 7.06 × 10^-21 C

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The first artificial satellite to orbit the Earth was Sputnik I, launched October 4, 1957. The mass of Sputnik I was 83.5 kg, an
9966 [12]

Answer:

-4.941*10^8J.

Explanation:

To solve this exercise it is necessary to take into account the concepts related to gravitational potential energy, as well as the concept of perigee and apogee of a celestial body.

By conservation of energy we know that,

\Delta U = \Delta_{perogee}-\Delta_{Apogee}

Where,

U= \frac{-GmM_e}{r}

Replacing

\Delta U = \frac{-GmM_e}{r_p}- \frac{-GmM_e}{r_a}

\Delta U = GmM_e (\frac{1}{r_A}-\frac{1}{r_p})

Our values are given by,

m = 85.5Kg

M_e = 5.97*10^{24}Kg

r_A = 7330Km

r_p = 6610Km

G = 6.67*10^{-11}Nm^2/Kg^2

Replacing at the equation,

\Delta U = (6.67*10^{-11})(85.5)(5.97*10^{24}) (\frac{1}{7330}-\frac{1}{6610})

\Delta U = -4.941*10^8J

Therefore the Energy necessary for Sputnik I as it moved from apogee to perigee was -4.941*10^8J.

4 0
3 years ago
Please help homework due tomorrow,,,
Ad libitum [116K]
Cody ...

Everything on this page is solved with the SAME formula !

             Distance = (speed) x (time) .


Before I get into how to solve each problem, we need to notice that
this whole sheet deals with speed, NOT velocity.

'Velocity' is speed AND THE DIRECTION OF THE  MOTION.
Nothing on this page ever mentions direction, so there's no velocity
anywhere on the page.

Your teacher may not be happy if you talk about this on your homework,
but that's too bad.  Just don't say "velocity" in any of your answers.
Say "speed", and if the teacher complains about that, then it's time to
let the teacher have it with both barrels.
 

1).  Speed = (distance covered) / (time to cover the distance)

2).  Speed = (distance covered) / (time to cover the distance)

3).  Distance  =  (average speed of travel) x (time traveling at that speed)

4).  Time to cover the distance = (distance) / (speed)

5).  Car's     speed = (distance the car covered)        / (time the car took)
      Sprinter speed = (distance the sprinter covered) / (time the sprinter took)

      Calculate the car's speed.
      Calculate the sprinter's speed.
     
      ... Look at the two speeds.
          Decide which one is faster.
     
      ... Subtract the slower one from the faster one. 
          The difference is the answer to "by how much?" .

6).  Distance  =  (speed) x (time spent moving at that speed)

7).  Average speed  =  (TOTAL distance covered)
                                      divided by
                                    (time to cover the TOTAL distance).
   

8 0
3 years ago
A 50.0 kg object is moving at 18.2 m/s when a 200 N force
gayaneshka [121]

Answer:

distance = 21.56 m

Explanation:

given data

mass = 50 kg

initial velocity  = 18.2 m/s

force = -200 N ( here force applied to opposite direction )

final velocity = 12.6 m/s

solution

we know here acceleration will be as

acceleration a  = force ÷ mass

a = \frac{-200}{50}   =  -4 m/s²

we get here now required time that is

required time = \frac{V_{(final)} - V_{(initial)}}{a}     ...............1

put here value

required time = \frac{12.6-18.2}{-4}  

so distance will be

distance = \frac{V_{(final)}^2 - V_{(initial)}^2}{2a}    ........2

distance = \frac{12.6}^2 -{18.2}^2}{2\times (-4)}  

distance = 21.56 m

7 0
3 years ago
Based on Schaffer's arguments in "The Value of a Sherpa Life," choose the statement
Tju [1.3M]

Answer:

Sherpas do work that is much more meaningful than the work other climbers do.

Explanation:

3 0
3 years ago
While vacationing in the mountains you do some hiking. In the morning, your displacement is S⃗ morning= (2200 m , east) + (4000
Minchanka [31]

Answer:

a

    The hiker (you ) is  200 m below his/her(your) starting point

b

   The resultant displacement in the north east direction is

a  = 6562.0 \  m

    The resultant displacement in vertical direction (i.e the altitude change )

  b =6503.1 \  m

Explanation:

From the question we are told that  

  The displacement in the morning is  S_{morning} =  (2200 \m , east) + (4000\ m\ north) + (100 \ m ,\ vertical)

   The displacement in the afternoon is  S _{afternoon}= (1300\ m ,\ west) + (2500 \ m ,\ north) - (300\ m ,\ vertical)

Generally the direction west is negative , the direction east is positive

                 the direction south is negative , the direction north is  positive

resultant displacement  is mathematically evaluated as  

    (2200 \m , east) +( - 1300\ m ,\ west) = 900 \ m \ east

     (4000\ m\ north)  + (2500 \ m ,\ north) = 6500  \ m ,\ north

     (100 \ m ,\ vertical) - (300\ m ,\ vertical) = -200 \ m

From the above calculation we see that at the end of the hiking the hiker (you) is  200 m below his/her(your) initial position

Generally from Pythagoras theorem , the resultant displacement in the north east direction is

      a  =  \sqrt{900^2 + 6500^2}

=>     a  = 6562.0 \  m

Generally from Pythagoras theorem , the resultant displacement in vertical direction (i.e the altitude change )

      b = \sqrt{6500^2 +(-200)^2  }

=>   b =6503.1 \  m

   

5 0
3 years ago
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