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3 years ago

Plzz help me please

Physics

1 answer:

olasank [31]3 years ago

6 0

Permanent magnet. An induced magnet would be created when a piece of iron (for example) is in contact with a magnet. Temporary magnets would be something like an electromagnet. Bar magnets are permanently magnetic unless we heat them or hammer them to cause their domains to loose alignment.

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Oxygen-18 is a naturally-occuring, stable isotope and is commonly used is scientific studies as a tracer. Using the periodic tab

Viefleur [7K]

Oxygen has Atomic number 8 so all isotopes have 8 protons and 8 electrons.

So the number of neutrons in Oxygen-18 = 18 - 8 = 10.

Option B is the correct one.

5 0

3 years ago

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3. what could you do to increase the precision of your measurements? What factors limit the

Vika [28.1K]

Answer:

Use specific tools built specifically for that specific measurement.

Explanation:

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3 years ago

A solid keeps its shape due to which of the following factors?

siniylev [52]

Answer:

Explanation:

because whatever elements are used like gas uses water evaporation so yeah

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2 years ago

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A block of mass 200g is oscillating on the end of a horizontal spring of spring constant 100 N/m and natural length 12 cm. When

malfutka [58]

In order to determine the acceleration of the block, use the following formula:

$F=ma$

Moreover, remind that for an object attached to a spring the magnitude of the force acting over a mass is given by:

$F=kx$

Then, you have:

$ma=kx$

by solving for a, you obtain:

$a=\frac{kx}{m}$

In this case, you have:

k: spring constant = 100N/m

m: mass of the block = 200g = 0.2kg

x: distance related to the equilibrium position = 14cm - 12cm = 2cm = 0.02m

Replace the previous values of the parameters into the expression for a:

$a=\frac{(\frac{100N}{m})(0.02m)}{0.2\operatorname{kg}}=10\frac{m}{s^2}$

Hence, the acceleration of the block is 10 m/s^2

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1 year ago

A lady bug is sitting on the bottom of a can while you twirl it overhead on a string that is 65.0

MA_775_DIABLO [31]

The linear speed of the ladybug is 4.1 m/s

Explanation:

First of all, we need to find the angular speed of the lady bug. This is given by:

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

where

T is the period of revolution

The period of revolution is the time taken by the ladybug to complete one revolution: in this case, since it does 1 revolution every second, the period is 1 second:

T = 1 s

Therefore, the angular speed is

$\omega=\frac{2\pi}{1 s}=6.28 rad/s$