Timers at a swim meet used four different clocks to time an event which recorded time is the most precise

Physics

2 answers:

tatiyna3 years ago

8 0

The answer is b 55.2 that is correct

Dennis_Churaev [7]3 years ago

4 0

Answer:

option D

Explanation:

the correct answer is option D

from the all four clocks the precise value will be given by the fourth clock which is 55.254 s.

because the measurement by that clock can be as small as 0.001 so, the precision of any instrument can be measured by how small value it can calculate so the answer will be option D

You might be interested in

5. A car accelerates from 0 to 72 km/hour in 8.0 seconds. What is the car's acceleration?

MA_775_DIABLO [31]

Answer:

2.5 m/s

Explanation:

There are calculators online that can help you easily calculate the accerlation.

8 0

2 years ago

A pure jet engine propels an aircraft at 240 m/s through air at 45 kPa and −13°C. The inlet diameter of this engine is 1.6 m, th

erica [24]

QUESTION: A pure jet engine propels and aircraft at 340 m/s through air at 45 kPa and -13C. The inlet diameter of this engine is 1.6 m, the compressor pressure ratio is 13, and the temperature at the turbine inlet is 557C. Determine the velocity at the exit of this engines nozzle and the thrust produced.

ANSWER: Due to the propulsion from the inlet diameter of this engine bring 1.6 m allows the compressor rations to radiate allowing thrust propultion above all velocitic rebisomes.

6 0

3 years ago

Why does a rock weigh less in water than it does on land?

Cloud [144]

Water is more dense then air so it sorta holds the rock as it sinks

7 0

3 years ago

Unpolarized light is incident upon two ideal polarizing filters that do not have their transmission axes aligned. If 19% of the

Zepler [3.9K]

Answer:

51.94°

Explanation:

$I_0$ = Unpolarized light

$I_2$ = Light after passing though second filter = $0.19I_0$

Polarized light passing through first filter

$I_1=\frac{I_0}{2}$

Polarized light passing through second filter

$I_2=\frac{I_0}{2}cos^2\theta\\\Rightarrow 0.19I_0=\frac{I_0}{2}cos^2\theta\\\Rightarrow cos^2\theta=\frac{0.19I_0}{\frac{I_0}{2}}\\\Rightarrow cos\theta=\sqrt{\frac{0.19I_0}{\frac{I_0}{2}}}\\\Rightarrow \theta=cos^{-1}\sqrt{\frac{0.19I_0}{\frac{I_0}{2}}}\\\Rightarrow \theta=cos^{-1}\sqrt{0.19\times 2}\\\Rightarrow \theta=cos^{-1}\sqrt{0.38}\\\Rightarrow \theta=51.94^{\circ}$