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3 years ago

An element is if it can be drawn into a wire

1 answer:

8 0

Metals, Metalloids, and Nonmetals These elements can be classified as metals, nonmetals but metalloids ,Metals are good conductors of heat and electricity, and are malleable they can be hammered into sheets and (ductile- they can be drawn into wire.)

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The Plateau of Tibet is a dry place. Do you think the wide-leafed plant fossils found on the plateau could grow there today? Exp

ludmilkaskok [199]

No, I dont think that they could grow there now in this day and age becuase the Plateu of Tbiet is very dry and also a little bit dusty. They barely even have grass down there. Wide leaf plants susually grow somewhee like the rainforest where it is warm, sunny, and rainy but in a place like Tibet, I don't think that they would be able to survive there.

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4 years ago

If you apply 100.0 N of force to lift an object with a single, fixed pulley, then what is the resistive force?

quester [9]

If you apply 100.0 N of force to lift an object with a single, fixed pulley, then the resistive force is also equivalent to 100 Newtons of force. Since the weight of the object was not mentioned, it is assumed that it has already been taken into account in the 100 N value of force. This follows Newton's law of motion of equal action and reaction.

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3 years ago

Hey , please help me im totally stuck?! :)

Setler79 [48]

D thymine
if this is high school than i am tottaly right

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3 years ago

As you hold book at rest in your hand two forces are being exerted on the book identify the force

erma4kov [3.2K]

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3 years ago

A block attached to a spring with an unknown spring constant oscillates with a period of 2.0 s. What is the period if

Zigmanuir [339]

Answer:

a) If the mass is doubled, then the period is increased by $\sqrt{2}$ . Hence, the period of the system is 2.828 seconds.

b) If the mass is halved, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

c) The period of the system does not depend on amplitude. Hence, the period of the system is 2 seconds.

d) If the spring constant is doubled, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

Explanation:

The statement is incomplete. We proceed to present the complete statement: A block attached to a spring with unknown spring constant oscillates with a period of 2.00 s. What is the period if a. The mass is doubled? b. The mass is halved? c. The amplitude is doubled? d. The spring constant is doubled?

We have a block-spring system, whose angular frequency ( $\omega$ ) is defined by the following formula:

$\omega = \sqrt{\frac{k}{m} }$ (1)

Where:

$k$ - Spring constant, measured in newtons per meter.

$m$ - Mass, measured in kilograms.

And the period ( $T$ ), measured in seconds, is determined by the following expression:

$T = \frac{2\pi}{\omega}$ (2)

By applying (1) in (2), we get the following formula:

$T = 2\pi\cdot \sqrt{\frac{m}{k} }$

a) If the mass is doubled, then the period is increased by $\sqrt{2}$ . Hence, the period of the system is 2.828 seconds.

b) If the mass is halved, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

c) The period of the system does not depend on amplitude. Hence, the period of the system is 2 seconds.

d) If the spring constant is doubled, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

8 0

3 years ago