Ask question

Ask question

All categories

My name is Ann [436]

2 years ago

9

A metre rule is used to measure a length Which reading is shown to the nearest millimetre? A 0.7 m B 0.76 m с 0.761 m D 0.7614 m

1 answer:

melomori [17]2 years ago

5 0

Answer:A

Explanation:because the smallest measurement is nearest to millimeter so on choose A. 0.7m is the small.

You might be interested in

A car initially traveling at 27.2 m/s undergoes a constant negative acceleration of magnitude 1.90 m/s2 after its brakes are app

zubka84 [21]

Answer:

Therefore, the revolutions that each tire makes is:

$\Delta \theta=22\: rev$

Explanation:

We can use the following equation:

$\omega_{f}^{2}=\omega_{i}^{2}-2\alpha \Delta \theta$ (1)

The angular acceleration is:

$a_{tan}=\alpha R$

$\alpha=\frac{1.9}{0.325}$

$\alpha=5.85\: rad/s^{2}$

and the initial angular velocity is:

$\omega_{i}=\frac{v}{R}$

$\omega_{i}=\frac{27.2}{0.325}$

$\omega_{i}=83.69\: rad/s$

Now, using equation (1) we can find the revolutions of the tire.

$0=83.69^{2}-2*25.85 \Delta \theta$

$\Delta \theta=135.47\: rad$

Therefore, the revolutions that each tire makes is:

$\Delta \theta=22\: rev$

I hope it helps you!

6 0

3 years ago

Need help in the middle one

kondor19780726 [428]

Answer:

Guysi hate math answer this guy plsss ssss

7 0

3 years ago

In which situation is the gravitational force between two objects hard to detect? (Options)

svlad2 [7]

A - the objects are too small

GRAVITATIONAL FORCE IS EXPERIENCED BY ALL OBJECTS IN THE UNIVERSE ALL THE TIME. BUT THE ORDINARY OBJECTS YOU SEE EVERY DAY HAVE MASSES SO SMALL THAT THEIR ATTRACTION TOWARD EACH OTHER IS HARD TO DETECT. -https://www.ftsd.org/cms/lib6/MT01001165/Centricity/ModuleInstance/630/CHAPTER_2_NOTES_FOR_EIGHTH_GRADE_PHYSICAL_SCIENCE.pdf

5 0

3 years ago

Read 2 more answers

A person wishes to heat pot of fresh water from 20°C to 100°C in order to boil water for pasta. They calculate that their pot ho

nasty-shy [4]

Given:
Water, 2 kilograms
T1 = 20 degrees Celsius, T2 = 100 degrees Celsius.

Required:
Heat produced

Solution:
Q (heat) = nRT = nR(T2 = T1)
Q (heat) = 2 kilograms (4.184 kiloJoules per kilogram Celsius) (100 degrees Celsius – 20 degrees Celsius)
<u>Q (heat) = 669.42 Joules
</u>This is the amount of heat produced in boiling 2 kg of water.

6 0

3 years ago

Can you please help me find the answer

Serggg [28]

Im sorry may you please retake the picture then i will answer

8 0

3 years ago