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2 years ago

If 1 m = 100 cm , then how many cm^2 are there in a m^2 ?? please hel[p

Physics

2 answers:

maw [93]2 years ago

7 0

Answer:if 1 m = 100 cm then there should be 200 cm in m^2

Explanation:

eimsori [14]2 years ago

6 0

200 cm :))))))))))))))))))

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Communicating results allows scientists to learn from each other's results.

sveta [45]

True, scientists often talk to each other to figure out if their results were similar and what they could have done better.

Although, talking to other scientists does have risks, other scientists could copy your work and further better it.

So, your final answer is TRUE, sorry for the long answer, I needed to have a word count about 20 characters and then I got carried away! lol

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2 years ago

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Calculate the pressure exerted on the ground by a boy of a mass 60 kg if he stands on one foot.the area of the sole of his shoe

ddd [48]

Answer:

40 Kpa

Explanation:

150 cm2 = 0.015 m2

$p \: = \frac{mg}{ a} = 40000$

8 0

2 years ago

The position of a particle as it moves along an y axis is given by y = (2.0cm)sin(πt/4), with t in second and y in centimeters.

irina [24]

Part a)

At t = 0 the position of the object is given as

$x = 0$

At t = 2

$x = 2 sin(\pi/2) = 2cm$

so displacement of the object is given as

$d = 2 - 0 = 2cm$

so average speed is given as

$v_{avg} = \frac{2}{2} = 1 cm/s$

Part b)

instantaneous speed is given by

$v = \frac{dy}{dt}$

$v = 2cos(\pi t/4 ) * \frac{\pi}{4}$

now at t= 0

$v = \frac{\pi}{2} cm/s$

at t = 1

$v = 2 cos(\pi/4) * \frac{\pi}{4}$

$v = \frac{\pi}{2\sqrt2}$

at t = 2

$v = 0$

Part c)

Average acceleration is given as

$a_{avg} = \frac{v_f - v_i}{t}$

$a_{avg} = \frac{0 - \frac{\pi}{2}}{2}$

$a = -\frac{\pi}{4} cm/s^2$

Part d)

Now for instantaneous acceleration

As we know that

$a =- \omega^2 y$

at t = 0

$a = -\frac{\pi^2}{16} * 0 = 0 cm/s^2$

at t = 1

$y = \sqrt2 cm$

now we have

$a = -\frac{\pi^2}{16}*\sqrt2$

At t = 2 we have

$y = 2 cm$

$a = -\frac{\pi^2}{16}*2$

$a = -\frac{\pi^2}{8}$

<em>so above is the instantaneous accelerations</em>

7 0

3 years ago

You are working on an experiment involving a very strong permanent magnet, and your data suggests that your magnet's field sudde

kompoz [17]

<u>Question:</u>

You are working on an experiment involving a very strong permanent magnet, and your data suggests that your magnet's field suddenly decreased during some interval in time. Such a decrease could have been caused by the magnet

A. Having overheated substantially

B. Being hit hard

C. Both A and B

D. Being grounded out

<h3><u>Answer:</u></h3>

A decrease in magnetic field of the permanent magnet have been caused by the magnet having overheated substantially or sharp impacts by being hit hard.

Option c

<h3><u>Explanation: </u></h3>

Permanent magnets are ferromagnetic materials with its magnetic domains aligned and grouped together in the same direction. These atomic domains maintain their directionality and hence a permanent magnet provides persistently strong magnetic fields without quick weakening. Some factors may lead to demagnetization or else a consistent reduction in magnetic strength.

Overheating a magnetic material realigns the magnetic domain regions and affects its directionality. When it reaches to a temperature defined as Curie temperature, varying with each material; the substance is no more a magnet due to complete randomness in the domain structure. As the temperature decreases and approaches the room temperature, magnetic field appears but is less in strength. Sudden impacts due to hitting may lead to random realignment of magnetic domains and thus decrease its magnetic strength.

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3 years ago

The boiling point of a substance is _72 degree Celsius. This temperature will be equivalent to Kelvin scale is-------.

Vlada [557]

Answer:

345 K

Explanation:

Temperature can be defined as a measure of the degree of coldness or hotness of a physical object.

Generally, it is measured with a thermometer and its units are Celsius (°C), Kelvin (K) and Fahrenheit (°F).

<u>Given the following data;</u>

Boiling point = 72°C

<em>To convert the temperature in degree Celsius to Kelvin, we would use the following mathematical expression;</em>

Kelvin = 273 + °C

Substituting into the formula, we have;

Kelvin = 273 + 72

<em>Kelvin = 345 K</em>

<em>Therefore, the temperature of 72°C will be equivalent to 345 K on the Kelvin scale.</em>

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2 years ago