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crimeas [40]
3 years ago
13

A student has a rectangular block. It is 2 cm wide, 2 cm tall, and 25 cm long. It has a mass of 600 g. First, calculate the volu

me of the block. Then, use that answer to determine the density of the block. *
Physics
1 answer:
WINSTONCH [101]3 years ago
4 0

Answer:

6g/cm³

Explanation:

Density is the mass per unit volume of any substance. To solve this problem:

   

   Density  = \frac{mass}{volume}  

 Since mass = 600g

Let us find the volume;

    Volume  = length x width x height

   Volume  = 25cm x 2cm x 2cm  = 100cm³  

Therefore;

         Density  = \frac{600}{100}    = 6g/cm³

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Which of the following are single-displacement reactions?
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Answer:

A & D

Explanation:

A single-displacement reaction is a chemical reaction whereby one element is substituted for another one in a compound and thereby generating a new element and also a new compound as products.

From the options, only options A & D fits this definition of single-displacement reactions.

For option D: Both left and hand and right hand sides each have one element and one compound. We can see that K is substituted from KBr to join Cl to form KCl and Br2 on the right hand side.

For option A: Both left and hand and right hand sides each have one element and one compound. We can see that OH is substituted from 2H2O to join Mg to form Mg(OH)2 and H2 on the right hand side.

The other options are not correct because they don't involve only and element and a compound on each side of the reaction.

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When it reacts with another element, an oxygen atom gains two electrons.
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An ion with a negative charge

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The two main states of mechanical energy are ___________ and potential energy.
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HELP ME PLEASE!!!!!!!!!
Stolb23 [73]

As per the given Figure attached here we know that both charges q1 and q2 will apply same force on charge q3 and hence the resultant force due to both charges will be along Y axis vertically upwards

So here we know that

F = \frac{kq_1q_3}{d_{13}^2}

now from the above equation

F = \frac{(9\times 10^9)(2\times 10^{-6})(4 \times 10^{-6})}{0.5^2}

F = 0.288 N

so both of the charges will apply 0.288 N force on q3 charge along the line joining them

now the net force due to vector sum is given by

F_{net} = 2Fcos\theta

here we know that angle is

\theta = 37 degree

now we have

F_{net} = 2\times 0.288 cos37

F_{net} = 0.46 N

so net force on q3 is 0.46 N vertically upwards along +Y axis

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