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labwork [276]
3 years ago
14

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

Physics
2 answers:
maw [93]3 years ago
7 0

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

Explanation:

eimsori [14]3 years ago
6 0
200 cm :))))))))))))))))))
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An engine performs 6400 j of work on a motorbike the motorbike and the rider have a combined mass of 200 kg if the bike started
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The work done by the engine is converted into kinetic energy of the motorbike+rider system:

W=K=\frac{1}{2}mv^2

where

W=6400 J is the work

m=200 kg is the mass of the system

v is the speed acquired by the motorbike


Rearranging the equation and substituting the numbers in, we find

v=\sqrt{\frac{2W}{m}} =\sqrt{\frac{2(6400 J)}{200 kg}} =8 m/s

4 0
3 years ago
HELP ME ASAP PLEASE
padilas [110]

Answer:

( 1000 × 4 = 4,000) (800×3= 2400) (800×2=1600) the answer is 1600 hope it helps

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How can a calculated height be greater than an actual height?
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6 0
3 years ago
Read 2 more answers
A 350-g mass is attached to a spring whose spring constant is 64 N/m. Its maximum acceleration is 5.3 m/s2. What is the frequenc
max2010maxim [7]

The frequency of oscillation is 2.153 Hz

What is the frequency of spring?

Spring Frequency is the natural frequency of spring with a weight at the lower end. Spring is fixed from the upper end and the lower end is free.

For the mass-spring system in this problem,

The Frequency of spring is calculated with the equation:

f = \frac{1}{2\pi } \sqrt{\frac{k}{m} }

Where,

f = frequency of spring

k = spring constant = 64 N/m

m = mass attached to spring = 350g = 0.350 kg

a = maximum acceleration = 5.3 m/s^2

Substituting the values in the equation,

f = \frac{1}{2\pi } \sqrt{\frac{64}{0.350} }

f = \frac{1}{2\pi } ( 13.522)

f = 2.1535 Hz

Hence,

The frequency of oscillation is 2.153 Hz

Learn more about frequency here:

<u>brainly.com/question/13978015</u>

#SPJ4

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2 years ago
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