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amm1812
3 years ago
14

A dipole of moment 0.50 e · nm is placed in a uniform electric field with a magnitude of 1.6 105 N/C. What is the magnitude of t

he torque on the dipole for each of the following situations?
(a) the dipole is aligned with the electric field
(b) the dipole is transverse to (perpendicular to) the electric field
(c) the dipole makes an angle of 26° with the direction of the electric field
(d) Defining the potential energy to be zero when the dipole is transverse to the electric field, find the potential energy of the dipole in the electric field for the orientations specified in Parts (a) and (c).
Physics
1 answer:
gtnhenbr [62]3 years ago
5 0
Too much to read bro bro
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Sally and Sam are in a spaceship that comes to within 14,000 km of the asteroid Ceres. Determine the force Sally experiences, in
tatiyna

Sam and Sally are traveling aboard a spacecraft that approaches the asteroid Ceres within 14,000 kilometers. Sally will experience 1.989 × 10⁻¹¹  N of force.

<h3>What is the gravitational force?</h3>

Newton's law of gravity states that each particle having mass in the universe attracts each other particle with a force known as the gravitational force.

The gravitational force is proportional to the product of the masses of the two bodies and inversely proportional to the square of their distance.

Given data

Mass of asteroid ,m₁ = 8.7 1020 kg

Mass of sally,m₂ = 67 kg

Gravitational constant,G = 6.6 × 10⁻¹¹ kg⁻² m²

Distance of seperation,R =  14,000 km

\rm F = G\frac{m_1m_2}{R^ 2} \\\\ F = 6.6 \times 10^{-11 }\times \frac{8.71020  \times 67 }{(14,000)^2}  \\\\F =  1.989 \times 10^{-11 } \ N

Hence, the force Sally experiences will be 1.989 × 10⁻¹¹  N.

To learn more about the gravitational force, refer to the link;

brainly.com/question/24783651

#SPJ1

4 0
1 year ago
Austin performed an experiment. He put 100mL of vegetable oil (density 0.9 g/ml) in a graduated cylinder. He put a 100g mass of
Grace [21]

Answer:

Explanation:

mass of displaced oil = 11 x  .9

= 9.9 gm

9.9 x 10⁻³ kg

weight of displaced oil = 9.9 x 9.81 x 10⁻³ N

= .097 N .

buoyant force by oil = .097 N

weight of unknown metal = .1 x 9.8

= .98 N .

weight of metal in oil = .98 - .097

= .883 N .

=

6 0
3 years ago
Using Rayleigh's criterion, what is the smallest separation between two pointlike objects that a person could clearly resolve at
atroni [7]

 Answer:

y = 52.44 10⁻⁶  m

Explanation:

It is Rayleigh's principle that two points are resolved if the maximum of the diffraction pattern of one matches the minimum the diffraction pattern of the other

Based on this principle we must find the angle of the first minimum of the diffraction expression

         a sin θ= m λ

The first minimum occurs for m = 1

       sin θ =  λ  / a

Now let's use trigonometry the object is a distance L = 0.205 m

        tan θ = y / L

Since the angles are very small, let's approximate

        tan θ = sin θ/cos θ = sin  θ

        sin θ = y / L

We substitute in the diffraction equation

         y / L =  λ  / a

         y =  λ  L / a

Let's calculate

        y = 550 10⁻⁹ 0.205 / 2.15 10⁻³

        y = 52.44 10⁻⁶ m

7 0
3 years ago
What is the electric potential energy of a charge that experiences a force of 3.6 x 10^-4N when it is 9.8 x 10^-5 from the sourc
Levart [38]
The electric potential energy of the charge is equal to the potential at the location of the charge, V, times the charge, q:
U=qV
The potential is given by the magnitude of the electric field, E, times the distance, d:
V=Ed
So we have
U=qEd (1)
However, the electric field is equal to the electrical force F divided by the charge q:
E= \frac{F}{q}
Therefore (1) becomes
U=Fd
And if we use the data of the problem, we can calculate the electrical potential energy of the charge:
U=Fd=(3.6 \cdot 10^{-4}N)(9.8 \cdot 10^{-5} m)=3.53 \cdot 10^{-8} J
7 0
3 years ago
A traffic light is weighing 200N hangs from a vertical cable tied to two other cables that are fastened to a support. The upper
Levart [38]

Answer:

T₁ = 93.6 N , T₂ = 155.6 N , T₃ = 200 N

Explanation:

This is a balance exercise where we must apply the expressions for translational balance in the two axes

     ∑  F = 0

Suppose that cable t1 goes to the left and the angles are 41º with respect to the horizontal and cable t2 goes to the right with angles of 63º

decompose the tension of the two upper cables

          cos 41 = T₁ₓ / T1

          sin 41 = T₁y / T1

          T₁ₓ = T₁  cos 41

          T₁y= T₁  sin 41

for cable gold

           cos 63 = T₂ / T₂

           sin 63 = T_{2y} / T₂

We apply the two-point equilibrium equation: The junction point of the three cables and the point where the traffic light joins the vertical cable.

Let's start by analyzing the point where the traffic light meets the vertical cable

              T₃ - W = 0

              T₃ = W

              T₃ = 200 N

now let's write the equations for the single point of the three wires

X axis

   - T₁ₓ + T₂ₓ = 0

  T₁ₓ = T₂ₓ

   T1 cos 41 = T2 cos 63

   T1 = T2 cos 63 / cos 41                (1)

y Axis

      T_{1y} + T_{2y} - T3 = 0

       T₁ sin 41 + T₂ sin 63 = T₃          (2)

to solve the system we substitute equation 1 in 2

        T₂ cos 63 / cos 41 sin 41 + T₂ sin 63 = W

         T₂ (cos 63 tan 41 + sin 63) = W

         T₂ = W / (cos 63 tan 41 + sin 63)

We calculate

          T₂ = 200 / (cos 63 tan 41 + sin 63)

          T₂ = 200 / 1,2856

           T₂ = 155.6 N

we substitute in 1

            T₁ = T₂ cos 63 / cos 41

             T₁ = 155.6 cos63 / cos 41

             T₁ = 93.6 N

therefore the tension in each cable is

            T₁ = 93.6 N

             T₂ = 155.6 N

             T₃ = 200 N

6 0
3 years ago
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