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RideAnS [48]
3 years ago
9

What is measurements?​

Physics
2 answers:
vitfil [10]3 years ago
5 0

Answer:

the process of comparison of an unknown quality with a know standard quantity is called measurement

Sliva [168]3 years ago
3 0

Answer:

the action of measuring something.

Measurement is a comparison of an unknown quantity with a known fixed quantity of the same kind. The value obtained on measuring a quantity is called its magnitude. Magnitude of a quantity is expressed as numbers in its units.

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1) A boy drags a wooden crate with a mass of 20 kg, a distance of 12 m, across a rough level floor at a constant speed of 1.5 m/
mojhsa [17]

Answer: a) 49.560 and 21.13 b) i) 50 N, ii) 196 N iii) 196 N iv) 47.685 N

c) i) 594.72 ii) 0 iii) 0 iv) 0

d) 594.72

Explanation: question a)

The force is inclined at an angle of 25° to the horizontal

The horizontal component of force = 50 cos 25° = 49.560 N

The vertical component of force = 50 sin 30°= 21.130N

Question b)

i) according to the question applied force is 50 N

ii) if g = 9.8m/s², w=mg where m = mass of object = 20kg hence weight = 20* 9.8 = 196 N

iii) the normal force is the force the floor exerts on the body as a result of the weight of the object.

Normal reaction R = W = mg, we already deduced that w = mg, hence R = 196 N.

iv) according to newton's laws of motion

F - Fr = ma

F = applied force = horizontal component of force = 49.560 N.

We need to get the acceleration (a) by using Newton laws of motion before we can be able to compute the frictional force..

The body started from rest hence initial velocity u = 0

Final velocity v = 1.5m/s distance covered (s) = 12m

v ² = u² + 2as

But u = 0

v² = 2as

1.5² = 2(a) * 12

2.25 = 24a

a = 2.25/24 = 0.09735m/s²

From F - Fr = ma

49.560 - Fr = 20 * 0.09735

49.560 - Fr = 1.875

Fr = 49.560 - 1.875

Fr = 47.685 N

Question c)

i) The applied force = 49.560 N, distance covered = 12m

Work done = force * distance

Work done = 49.560 * 12

Work done = 594.72 J

ii) the weight of the object does not make the object move a distance, hence work done = 0 ( since distance covered is 0)

iii) the normal force is the same thing as the weight and they did not cover any distance hence work done is zero.

iv) the frictional force does not cover any distance, hence work done is zero.

Question d)

The total work done = work done by applied force + work done by weight + work done by normal reaction + work done by frictional force.

Total work done = 594.72 + 0 + 0 + 0 = 594.72 J

8 0
2 years ago
Why are the freezing points of water and the melting point the same
Sladkaya [172]
Because melting point<span> and </span>freezing point<span> describe the</span>same<span> transition of matter, in this case from liquid to solid (</span>freezing) or equivalently, from solid to liquid (melting<span>).</span>
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3 years ago
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If the mass of a moving object were quartered, it’s inertia would be
AVprozaik [17]

Answer:

Quartered

Explanation:

Because you're a liberal.

8 0
2 years ago
MATCH THESE ^-^ Match Newton's law with the correct statement.
ikadub [295]
1 and A
2 and B
3 and D
4 and C

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The balmer series is formed by electron transitions in hydrogen that
RSB [31]

Answer:

The Balmer series refers to the spectral lines of hydrogen, associated to the emission of photons when an electron in the hydrogen atom jumps from a level n \geq 3 to the level n=2.

The wavelength associated to each spectral line of the Balmer series is given by:

\frac{1}{\lambda}=R_H (\frac{1}{2^2}-\frac{1}{n^2})

where R_H is the Rydberg constant for hydrogen, and where n is the initial level of the electron that jumps to the level n = 2.

The first few spectral lines associated to this series are withing the visible part of the electromagnetic spectrum, and their wavelengths are:

656 nm (red, corresponding to the transition 3 \rightarrow 2)

486 nm (green, 4 \rightarrow 2)

434 nm (blue, 5 \rightarrow 2)

410 nm (violet, 6 \rightarrow 2)

All the following lines lie in the ultraviolet part of the spectrum. The limit of the Balmer series, corresponding to the transition \infty \rightarrow 2, is at 364.6 nm.

4 0
3 years ago
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