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3 years ago

A person starts running from 2 m/s to 6 m/s in 2 seconds.

Physics

1 answer:

fiasKO [112]3 years ago

8 0

Answer:

2 m/s²

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 2 m/s

Final velocity (v) = 6 m/s

Time (t) = 2 s

Acceleration (a) =?

The acceleration of the person can be obtained as follow:

v = u + at

6 = 2 + (a×2)

6 = 2 + 2a

Collect like terms

6 – 2 = 2a

4 = 2a

Divide both side by 2

a = 4 / 2

a = 2 m/s²

Therefore, the acceleration of the person is 2 m/s²

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If a car is traveling 103 miles per hour, how many miles and how much time would it take for another car to catch up to them fro

pentagon [3]

Answer:

Required distance is 103 miles and the required time is 1 hour

Explanation:

Given;

speed of the car, v = 103 miles per hour

Speed is the given as the ratio of distance traveled to time taken for the motion.

The distance it will take another car to catch up to them from a complete stop is 103 miles and the time it will take the car is 1 hour.

Therefore, required distance is 103 miles and the required time is 1 hour.

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2 years ago

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3 years ago

Note that the scale bar under the photo is labeled 1 μm (micrometer). The scale bar works the same way as a scale on a map, wher

AnnZ [28]

Answer:

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3 years ago

In Millikan's experiment, an oil drop of radius 1.362 μm and density 0.888 g/cm3 is suspended in chamber C when a downward-point

Misha Larkins [42]

Answer:

The charge on the oil drop is 3e

Explanation:

F = qE

Where;

F is the applied force in Newton

E is the electric field potential N/C

q is charge in C

Given;

Radius, r = 1.362 μm = 1.362 X 10⁻⁶ m

density, ρ = 0.888 g/cm³ = 0.888 X 10³ kg/m³

Electric field potential = 1.92 ✕ 10⁵ N/C

F =mg

mass of the oil drop = density, ρ X volume of the oil drop

volume of the oil drop (spherical) = (4/3)πr³ = 1.3333π(1.362 X 10⁻⁶)³

⇒ volume of the oil drop = 10.584 X 10⁻¹⁸ m³

mass of the oil drop = 0.888 X 10³ (kg/m³) X 10.584 X 10⁻¹⁸ (m³)

⇒ mass of the oil drop = 9.399 X 10⁻¹⁵ kg

⇒ F =mg = 9.399 X 10⁻¹⁵ kg X 9.8 = 9.21 X10⁻¹⁴ N

F = qE

q = F/E

q = (9.21 X10⁻¹⁴)/(1.92 ✕ 10⁵) = 4.797 X 10⁻¹⁹ C

In terms of e

1e = 1.6 X10⁻¹⁹ C

= (4.797 X 10⁻¹⁹ C)/(1.6 X10⁻¹⁹ C) = 3e

Therefore, the charge on the oil drop is 3e

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2 years ago