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Serggg [28]
3 years ago
9

Find the perimeter of the rectangle if each block represents one foot(show your calculation)

Physics
1 answer:
iren2701 [21]3 years ago
4 0

I'm pretty sure that the "block" of which you speak is one in a pattern
of them that covers the drawing you have of the rectangle, and now
I need to explain something to you: 

The REASON for printing that drawing next to the question that you
partially copied is that the drawing has information that's needed to
answer the question with, and rather than repeat all that information
in the question, it just says "LOOK AT THE DRAWING !"

In fact, the whole point of the question may not be just to remind you of
what "perimeter" means.  It's more likely that the purpose of this problem
is to make you pick the information you need off of a drawing. 

Either way, if you'll kind of "read between the lines" of the part of the
question that you DID copy, it should be pretty obvious to you that nobody's
going nowhere in the direction of a solution without SEEing the drawing. 

So my bottom-line conclusion regarding a solution for this problem is: 
Not possible with the given information.

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A bar of iron is 10 cm at 20°C. At 19°C it will be (are = 11 x 10-6/°C)
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Answer:

HENCE (3) IS ANSWER

Explanation:

α =ΔL/(LO×Δ T)  or ΔL= α(LO×Δ T)

HERE ΔL  IS CHANGE IN  LENGTH, Lo is original length , α is coefficient of linear expansion , Δ T is change in temperature

α= 11 x 10 ^-6/°C , Δ T = 19-20 = -1°C

hence

ΔL= 11 x 10 ^-6×10×-1

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MINUS SIGN SHOWS SHORTER

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Read 2 more answers
A block of mass M on a horizontal surface is connected to the end of a massless spring of spring constant k. The block is pulled
slamgirl [31]

Answer:

Minimum coefficient of kinetic friction between the surface and the block is \mu_k=\frac{kx}{2Mg} .

Explanation:

Given:

Mass of the block = M

Spring constant = k

Distance pulled = x

According to the question:

<em>We have to find the minimum co-efficient of kinetic friction between the surface and the block that will prevent the block from returning to its equilibrium with non-zero speed.  </em>

So,

From the FBD we can say that:

⇒ Normal force, N=Mg                                   <em>...equation(i)</em>

⇒ Elastic potential energy, PE = \frac{kx^2}{2}               <em>  ...equation (ii)</em>

⇒ Frictional force, f = \mu_kN                                <em> ...equation (iii)</em>

⇒ Plugging (i) in (iii).

⇒ f=\mu_kMg

Now,

⇒ As we know that the energy lost due to friction is equivalent to PE .

⇒ PE=fx                     <em>...considering PE as</em> mgh or f(x) .

   Arranging the equation.

⇒ \frac{kx^2}{2}=\mu_k Mg (x)

⇒ \frac{kx}{2}=\mu_k Mg                 <em>...eliminating x from both sides.</em>

⇒ \frac{kx}{2Mg}=\mu_k                    <em>...dividing both sides wit Mg.</em>

Minimum coefficient of kinetic friction between the surface and the block is \frac{kx}{2Mg}=\mu_k .

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