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viva [34]
3 years ago
11

At which point on this electric field will a test charge show the maximum strength?

Physics
1 answer:
Juli2301 [7.4K]3 years ago
8 0

Answer:

A

Explanation:

The figure shows the electric field produced by a spherical charge distribution - this is a radial field, whose strength decreases as the inverse of the square of the distance from the centre of the charge:

E\propto \frac{1}{r^2}

More precisely, the strength of the field at a distance r from the centre of the sphere is

E=k\frac{Q}{r^2}

where k is the Coulomb's constant and Q is the charge on the sphere.

From the equation, we see that the field strength decreases as we move away from the sphere: therefore, the strength is maximum for the point closest to the sphere, which is point A.

This can also be seen from the density of field lines: in fact, the closer the field lines, the stronger the field. Point A is the point where the lines have highest density, therefore it is also the point where the field is strongest.

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Changes of state
Mrrafil [7]

Answer:

Option A

Explanation:

At segment T-U, the substance changes from a liquid to a gas and does not change temperature.

The reason is because latent heat of vaporisation allows for the absorption of heat in the change of state and temperature remains constant until it has fully changed state.

4 0
2 years ago
Read 2 more answers
A skater extends her arms, holding a 2 kg mass in each hand. She is rotating about a vertical axis at a given rate. She brings h
Usimov [2.4K]

Explanation:

It is known that relation between torque and angular acceleration is as follows.

                    \tau = I \times \alpha

and,       I = \sum mr^{2}

So,      I_{1} = 2 kg \times (1 m)^{2} + 2 kg \times (1 m)^{2}

                       = 4 kg m^{2}

      \tau_{1} = 4 kg m^{2} \times \alpha_{1}

     \tau_{2} = I_{2} \alpha_{2}

So,      I_{2} = 2 kg \times (0.5 m)^{2} + 2 kg \times (0.5 m)^{2}

                     = 1 kg m^{2}

 as \tau_{2} = I_{2} \alpha_{2}

                   = 1 kg m^{2} \times \alpha_{2}        

Hence,     \tau_{1} = \tau_{2}

                  4 \alpha_{1} = \alpha_{2}

            \alpha_{1} = \frac{1}{4} \alpha_{2}

Thus, we can conclude that the new rotation is \frac{1}{4} times that of the first rotation rate.

8 0
3 years ago
Consider a household that uses 14,000 kWh of electricity and 900 gal of fuel oil, per year, during a heating season. The average
ryzh [129]

Answer:

9517.2 lbm

Explanation:

Electricity consumption = 14000 kWh/year

Fuel consumption = 900 gal/year

Amount of CO₂ produced per gallon = 26.4 lbm/gal

Amount of CO₂ produced per kWh = 1.54 lbm/kWh

Amount of CO₂ produced in one year

14000\times 1.54+900\times 26.4=45320\ lbm

Reduction would be

0.21\times 45320=9517.2\ lbm

The reduction in the amount of CO₂ produced is 9517.2 lbm

7 0
3 years ago
A 12,000-N car is raised using a hydraulic lift, which consists of a U-tube with arms of unequal areas, filled with oil and capp
Bond [772]

Answer:

= 925.92 N

≅ 926N

Explanation:

Pressure due to car = pressure due to applied force  

12000/18^2 = Force / 5^2

force = 12000 * 25/ 324

= 925.92 N

For equilibrium

Pressure1 = Pressure2

A1F1 = A2F2

12000*pi*(5^2) = F2 ( pi)*(18^2)

so, F2 = Applied force to lift car = 925.92 N

Pascal's principle

Pressure1 = Pressure2

F1/A1 = F2/A2 (F=force and A=area)

A1 =Pi*(0.05)²

A2 =Pi(0.18)²

F2=12000

F1 = 12000*(0.05)² / (0.18)² = 926N

7 0
3 years ago
When a 3.0 kg mass is hung from a vertical massless spring, the spring is stretched 40 cm. What is the spring constant of the sp
Dmitry_Shevchenko [17]

Answer:

0.74 N/cm

Explanation:

The following data were obtained from the question:

Mass (m) = 3 Kg

Extention (e) = 40 cm

Spring constant (K) =?

Next, we shall determine the force exerted on the spring.

This can be obtained as follow:

Mass (m) = 3 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Force (F) =?

F = mg

F = 3 × 9.8

F = 29.4 N

Finally, we shall determine the spring constant of the spring. This can be obtained as follow:

Extention (e) = 40 cm

Force (F) = 29.4 N

Spring constant (K) =?

F = Ke

29.4 = K × 40

Divide both side by 40

K = 29.4 / 40

K = 0.74 N/cm

Therefore, the spring constant of the spring is 0.74 N/cm

5 0
3 years ago
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