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Ksju [112]
3 years ago
12

13. Now we will examine the electric field of a dipole. The magnitude and direction of the electric field depends on the distanc

e and the direction. We will investigate in detail just two directions. With charges available in the simulation how do you create a dipole with dipole moment 1 x 10-9 Cm with a direction for the dipole moment pointing to the right
Physics
1 answer:
Gennadij [26K]3 years ago
5 0

This question is incomplete, the complete question is;

Now we will examine the electric field of a dipole. The magnitude and direction of the electric field depends on the distance and the direction. We will investigate in detail just two directions. With charges available in the simulation (all the charges are either positive or negative 1 nC increments).

how do you create a dipole with dipole moment 1 x 10-9 Cm with a direction for the dipole moment pointing to the right. Make a table below that shows the amounts of  charge and the distance between the charges. There are many correct answers

Answer:

Given the data in question;

Dipole moment P = 1 × 10⁻⁹ C.m

now dipole pointing to the right;

               P→

_{-\theta } (-) ---------------->(+) _{+\theta }

               d

so let distance between the dipoles be d

∴ P = d\Theta

Let \Theta_{1} = 1 nC

so

P = d\Theta

1 × 10⁻⁹ =  1 × 10⁻⁹  × d

d = (1 × 10⁻⁹) / (1 × 10⁻⁹)

d = 1 m

Also Let \Theta_{2} = 2 nC

so

P = d\Theta

1 × 10⁻⁹ =  2 × 10⁻⁹  × d

d = (1 × 10⁻⁹) / (2 × 10⁻⁹)  

d = 0.5 m

Also Let \Theta_{3} = 3 nC

so

P = d\Theta

1 × 10⁻⁹ =  3 × 10⁻⁹  × d

d = (1 × 10⁻⁹) / (3 × 10⁻⁹)

d = 0.33 m

such that;

charge                 distance

1 nC                        1.00 m      

2 nC                       0.50 m

3 nc                        0.33 m

4 nC                       0.25 m

5 nC                       0.20 m      

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Answer:

a) 2.00024 m

b) 0.036%

c) 436.67°C

Explanation:

Given

Initial length = L₀ = 2 m

Initial cross sectional Area = A₀ = 200 cm² = 0.02 m²

We can obtain initial volume = V₀ = A₀L₀ = 0.02 × 2 = 0.04 m³

Initial Temperature = T₀ = 20°C

Coefficient of linear expansivity = α = (2 × 10⁻⁶) (°C)⁻¹

a) New length of the rod after heating to 80°C

Linear expansion is given as

ΔL = L₀ × α ×ΔT

ΔL = 2 × 2 × 10⁻⁶ × (80 - 20) = 0.00024 m = 0.24 mm

New length = old length + expansion = 2 + 0.00024 = 2.00024 m

b) The percentage of the volume change of the rod.

Volume expansion is given by

ΔV = V₀ × (3α) × ΔT

Volume expansivity ≈ 3 × (linear expansivity)

ΔV = 0.04 × (3×2×10⁻⁶) × (80 - 20) = 0.0000144 m³

Percentage change in volume = 100% × (ΔV/V₀) = 100% × (0.0000144/0.04) = 0.036%

c) The maximal temperature we can allow if the volume should not increase by more than half percent.

For a half percent increase in volume, the corresponding change in volume needs to be first calculated.

Percentage change in volume = 100% × (ΔV/V₀)

0.5 = 100% × (ΔV/0.04)

(ΔV/0.04) = 0.005

ΔV = 0.0002 m³

Then we now investigate the corresponding temperature that causes this.

ΔV = V₀ × (3α) × ΔT

0.0002 = 0.04 × (3×2×10⁻⁶) × ΔT

ΔT = (0.0002)/(0.04 × 3 × 2 × 10⁻⁶) = 416.67°C

Maximal temperature = T₀ + ΔT = 20 + 416.67 = 436.67°C

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2 years ago
7. Imagine you are pushing a 15 kg cart full of 25 kg of bottled water up a 10o ramp. If the coefficient of friction is 0.02, wh
pentagon [3]

Answer:

The frictional force needed to overcome the cart is 4.83N

Explanation:

The frictional force can be obtained using the following formula:

F= \mu R

where \mu is the coefficient of friction = 0.02

R = Normal reaction of the load = mgcos\theta = 25 \times 9.81 \times cos 10 = 241.52N

Now that we have the necessary parameters that we can place into the equation, we can now go ahead and make our substitutions, to get the value of F.

F=0.02 \times 241.52N

F = 4.83 N

Hence, the frictional force needed to overcome the cart is 4.83N

4 0
3 years ago
water in a cup and a kettle can have the same temperature even though the quantities are different . give reasons​
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Answer:

The reason is because both are exposed to a virtually infinite heat sink, due to the virtually infinite mass  and of the surrounding environment, compared to the sizes of either the cup or the kettle such that the equilibrium temperature, T_{(equilibrium)} reached is the same for both the cup and the kettle as given by the relation;

\infty M_{(environ)} \times  c_{(environ)} \times (T_2 - T_1) = m_{1} \times  c_{(water)} \times (T_3 - T_2) + m_{2} \times  c_{(water)} \times (T_4 - T_2)

Due to the large heat sink, T₂ - T₁ ≈ 0 such that the temperature of the kettle and that of the cup will both cool to the temperature of the environment

Explanation:

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A race car has a centripetal acceleration of 13.33 m/s^2 as it goes around a curve. if the curve is a circle with a radius 30 m
anzhelika [568]

Answer:

The speed of the car, v = 19.997 m/s

Explanation:

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The radius of the curve, r = 30 m

The centripetal force acting on the car is given by the formula

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Substituting the values in the above equation

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                                         v² = 13.33 x 30

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                                         v = 19.997 m/s

Hence, the speed of the car, v = 19.997 m/s

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2 years ago
It takes 23 hours 56 minutes and 4 seconds for the earth to make one revolution (mean sidereal day). What is the angular speed o
Dominik [7]

Answer:

Explanation:

Check attachment for solution

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