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Yuri [45]
3 years ago
12

During science class, while studying mixtures, you mix together iron filings and sand. Your teacher challenges you to separate t

he sand from the iron filings. How could you use the physical properties of matter to practically separate the two solids?
A) Add water to the mixture and dissolve the sand.
B) Pick out one of the two solids by using color to guide you.
C) Pour the mixture through a paper filter to separate the two.
D) Use a magnet to pull out the iron filings as they are attracted to a magnet.
Eliminate
Physics
2 answers:
Lapatulllka [165]3 years ago
8 0

Answer : Use a magnet to pull out the iron filings as they are attracted to a magnet

Explanation :  we use the magnet to separate the sand from the iron filing. Because the magnet has an attraction power.

We can say that we can use the magnet to attract the iron filings out of the mixture because iron is magnetic solid, but sand will not attract because sand is not magnetic solid.

So,  we use a magnet to pull out the iron filings as they are attracted to a magnet.

vesna_86 [32]3 years ago
6 0
To separate sand and iron fillings you use a magnet because sand isn't affected and iron has magnetic properties.
D
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Which of the following is an abiotic limiting factor for a plant population in an ecosystem?
Ludmilka [50]

Correct answer is Availability of soil minerals.

There are two types of limiting factors that affect plant population in a ecosystem. They are:

  1. Biotic
  2. Abiotic

Biotic factors includes food, diseases etc. Such as Invasive weed species, seed dispersal by wind, disease-causing fungal spores are examples of biotic limiting factor.

Abiotic factors include sunlight, temperature and chemical environment. The availability of soil mineral is an example of abiotic limiting factor. Growth of plant population depends on availability of soil minerals.

8 0
3 years ago
What is the voltage of this circuit.
liberstina [14]

Answer:

110.4 v

Explanation:

Find the equivalent resistance of the two parallel circuits

R1 and R2 in parallel   =   R1 * R2 / (R1+ R2) =   <u>1 ohm </u>

 similarly R4 and R5 =  <u>4.77 ohms</u>

Now you can add the three resistances into one  (R3=10)

    1 ohm + 10 ohm + 4.77 ohm = 15.77 ohms

Now

V = IR

V =  7 amps  * 15.77 ohm = 110.4 v

6 0
2 years ago
The blank the value of coefficient of friction the greater the resistance to sliding
expeople1 [14]

Answer:

The <u>HIGHER</u> value of coefficient of friction the greater the resistance to sliding

Explanation:

As we know that friction force is given by

F_f = \mu F_n

so friction force is the product of friction coefficient and normal force.

So here we can say that if friction coefficient is higher then the friction force will be more as it is product of friction coefficient and normal force.

So correct answer will be given as

The <u>HIGHER</u> value of coefficient of friction the greater the resistance to sliding

8 0
4 years ago
Read 2 more answers
You have a device that takes temperature measurements and runs off of solar power. How often it is programmed to take a measurem
Tcecarenko [31]

Answer:

In the Equator:

As far as the temperature is concerned equator is more or less the same throughout the year, however, there are some fluctuations also, I will set this device in March and September because, in this month, the Sun is exactly over the equator and I would be able to get the more results in this month.

In Antarctic:

As far as the climatic conditions of Antarctic are concerned, it is all the same while it fluctuates in December because the Sun is very much close to Antarctic that's why I will choose this month.

Explanation:

8 0
3 years ago
A youngster throws a rock from a bridge into the river 50 m below. The rock has a speed of 15 m/s when it leaves the youngster’s
butalik [34]
Assuming that the stone is thrown vertically... let's say it's a 1 kg stone.It doesn't matter if it's thrown upwards or downwards as (assuming no air friction) it will pass the original throwing point with the same downwards velocity as it had upwards, 3 seconds previously. So it starts with 1/2 m v^2 = 0.5 * 1 * 15^2 = 112.5 J of keThen k.e. gained = gpe lostk.e. gained = m g h = 1 * 10 * 50 = 500 J of Ke gainedso the final (total) ke is 612.5 J which = 1/2 m v^2 = 0.5 v^2 here
so 0.5 v^2 = 612.5so     v^2    = 1225so v = 35 m/s
6 0
3 years ago
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