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Norma-Jean [14]
2 years ago
8

Explain whether a particle moving in a straight line with constant speed does or does not have an acceleration. b) Explain wheth

er a particle moving around a curve with constant speed does or does not have an acceleration.
Physics
1 answer:
Lelechka [254]2 years ago
5 0

Answer:

A: No because it is nor changing speed or direction

B: Yes because it changes direction even though the speed is constant

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Two parallel plates have equal and opposite charges. When the space between the plates is evacuated, the electric field is E= 3.
svetlana [45]

Explanation:

The given data is as follows.

     Electric field between plates without dielectric, E_{1} = 3.50 \times 10^{5} V/m

   Electric field between the plates with dielectric, E_{2} = 2.40 \times 10^{5} V/m.

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Now, we will determine the charge density as follows.

            \sigma_{i} = \epsilon_{o}(E_{1} - E_{2})

                 = 8.85 \times 10^{-12} \times (3.50 \times 10^{5} - 2.40 \times 10^{5})

                 = 9.735 \times 10^{-7} C/m^{2}

Thus, we can conclude that the charge density on each surface of the dielectric is 9.735 \times 10^{-7} C/m^{2}.

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3 years ago
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A car is traveling at a velocity of 22 m/s when the driver puts on the brakes
Brums [2.3K]

The car’s velocity at the end of this distance is <em>18.17 m/s.</em>

Given the following data:

  • Initial velocity, U = 22 m/s
  • Deceleration, d = 1.4 m/s^2
  • Distance, S = 110 meters

To find the car’s velocity at the end of this distance, we would use the third equation of motion;

Mathematically, the third equation of motion is calculated by using the formula;

V^2 = U^2 + 2dS

Substituting the values into the formula, we have;

V^2 = 22 + 2(1.4)(110)\\\\V^2 = 22 + 308\\\\V^2 = 330\\\\V^2 = \sqrt{330}

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Therefore, the car’s velocity at the end of this distance is <em>18.17 m/s.</em>

<em></em>

Read more: brainly.com/question/8898885

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