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gregori [183]
2 years ago
8

What is the importance of measurement? ​

Physics
1 answer:
Kruka [31]2 years ago
8 0

Answer:

To know the time ,length area

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Plants make ___________ using carbon dioxide, water and energy from the sun.
gizmo_the_mogwai [7]

Answer:

I think it's carbohydrates

Explanation:

The energy from light causes a chemical reaction that breaks down the molecules of carbon dioxide and water and reorganizes them to make the sugar (glucose) and oxygen gas.

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Which of these objects are malleable? Check all that apply.
g100num [7]

Answer:

The tin fork and knife, the copper coin, and the steel fence pole.

Explanation:

Those are both what people would call soft metals so they are malleable to the extent of probably not needing heavy duty equipment. It depends on you description of malleable because the steel fence pole could be malleable with the correct equipment and not snap in half if bent slowly enough.

The definition of malleable: (of a metal or other material) able to be hammered or pressed permanently out of shape without breaking or cracking.

But the glass table, marble sculpture and antique ceramic vase are nowhere near malleable because if you tried bending them they wouldn't bend but would shatter and break into pieces.

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3 years ago
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The equation for density is mass divided by volume. the mass of a substance is changed so that it is made of fewer particles. If
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Density is directly proportional to mass. So if there's less matter inside object, its density will also reduce.
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3 years ago
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A 10-cm-long spring is attached to theceiling. When a 2.0 kg mass is hung from it,the spring stretches to a length of 15 cm.a.Wh
alekssr [168]

(a) 392 N/m

Hook's law states that:

F=k\Delta x (1)

where

F is the force exerted on the spring

k is the spring constant

\Delta x is the stretching/compression of the spring

In this problem:

- The force exerted on the spring is equal to the weight of the block attached to the spring:

F=mg=(2.0 kg)(9.8 m/s^2)=19.6 N

- The stretching of the spring is

\Delta x=15 cm-10 cm=5 cm=0.05 m

Solving eq.(1) for k, we find the spring constant:

k=\frac{F}{\Delta x}=\frac{19.6 N}{0.05 m}=392 N/m

(b) 17.5 cm

If a block of m = 3.0 kg is attached to the spring, the new force applied is

F=mg=(3.0 kg)(9.8 m/s^2)=29.4 N

And so, the stretch of the spring is

\Delta x=\frac{F}{k}=\frac{29.4 N}{392 N/m}=0.075 m=7.5 cm

And since the initial lenght of the spring is

x_0 = 10 cm

The final length will be

x_f = x_0 +\Delta x=10 cm+7.5 cm=17.5 cm

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3 years ago
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Motions need an unbalanced net force to maintain.<br> True or False?
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The answer is False

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