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daser333 [38]
2 years ago
11

1. Write down the readings on the side of Figs. 6 (a), (b) and (c) respectively. What is the least count of instrument scale for

each of the three measurement tools? Fig. 6 (a) Meterstick (cm)
Fig. 6 (b) Vernier Caliper (cm)
Fig. 6 (c) Micrometer (mm)

2. What is the difference between the measured values 1.05 m and 1.050 m? What factor of a
measurement tool determines the significant figures of a measured value?

Physics
1 answer:
alexandr402 [8]2 years ago
3 0

Explanation:

This should be right. if any doubt post a comment.

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A. A land speed car can decelerate at 9.8m/s. How long does it take the car to come to a complete stop from a run of 885 km/hr (
Nimfa-mama [501]

Answer:

A. 25.08 s

B. 3082.53 m

C. 3×10⁵ m/s²

Explanation:

A. Determination of the time.

This can be obtained as illustrated below:

Acceleration (a) = –9.8 m/s²

Initial velocity (u) = 245.8 m/s

Final velocity (v) = 0 m/s

Time (t) =.?

v = u + at

0 = 245.8 + (–9.8 × t)

0 = 245.8 – 9.8t

Collect like terms

0 – 245.8 = – 9.8t

– 245.8 = – 9.8t

Divide both side by –9.8

t = –245.8 / –9.8

t = 25.08 s

Therefore, it will take 25.08 s for the car to come to a complete stop.

B. Determination of the distance travelled by the car.

Acceleration (a) = –9.8 m/s²

Initial velocity (u) = 245.8 m/s

Final velocity (v) = 0 m/s

Distance (s) =?

v² = u² + 2as

0² = 245.8² + (2 × –9.8 × s)

0 = 60417.64 – 19.6s

Collect like terms

0 – 60417.64 = – 19.6s

– 60417.64 = – 19.6s

Divide both side by –19.6

s = –60417.64 / –19.6

s = 3082.53 m

Thus, the car travelled a distance of 3082.53 m before stopping completely.

C. Determination of the acceleration of the object.

Initial velocity (u) = 0 m/s

Final velocity (v) = 600 m/s

Distance (s) = 0.6 m

Acceleration (a) =?

v² = u² + 2as

600² = 0² + (2 × a × 0.6)

360000 = 0 + 1.2a

360000 = 1.2a

Divide both side by 1.2

a = 360000 / 1.2

a = 300000 = 3×10⁵ m/s²

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