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amm1812
3 years ago
7

Can someone help me name the layers.

Physics
2 answers:
BARSIC [14]3 years ago
7 0

Answer:

A. Crust

B. Mantle

C. Outer Core

D. Inner Core

Explanation:

Nadya [2.5K]3 years ago
4 0

Answer:

A. Crust

B. mantle

C. outer core

D. inner core

Explanation:

because it is

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mr_godi [17]

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4 0
3 years ago
Read 2 more answers
Can somebody help please !!!!
Nataly [62]

Vector A is of magnitude 12 m and it makes an angle of 37 degree with Y axis

So here we can say that

\frac{A_x}{A} = sin37

A_x = A sin37

A_x = 12 sin37

A_x = 7.22 m

Similarly we have

\frac{A_y}{A} = cos37

A_y = A cos37

A_y = 12 cos37

A_y = 9.58 m

So here we have

A_y = 9.58 m, A_x = 7.22 m

option A is correct

3 0
3 years ago
A 6.4-N force pulls horizontally on a 1.5-kg block that slides on a smooth horizontal surface. This block is connected by a hori
Elena L [17]

-- Although it's not explicitly stated in the question,we have to assume that
the surface is frictionless.  I guess that's what "smooth" means.

-- The total mass of both blocks is (1.5 + 0.93) = 2.43 kg. Since they're
connected to each other (by the string), 2.43 kg is the mass you're pulling.

-- Your force is 6.4 N.
                                    Acceleration = (force)/(mass) = 6.4/2.43 m/s²<em>
                                                                 </em>
That's about  <em>2.634 m/s²</em>  <em>

</em>
(I'm going to keep the fraction form handy, because the acceleration has to be
used for the next part of the question, so we'll need it as accurate as possible.)

-- Both blocks accelerate at the same rate. So the force on the rear block (m₂) is

       Force = (mass) x (acceleration) = (0.93) x (6.4/2.43) = <em>2.45 N</em>.

That's the force that's accelerating the little block, so that must be the tension
in the string.


7 0
3 years ago
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Ymorist [56]

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• PHOSPERIC ACID

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∆ WEAK BASES :

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• MAGNESIUM HYDROXIDE

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