Answer:
it would be 3
Explanation:
because you have to divide the length by the height of the incline.
Explanation:
<u>Forces</u><u> </u><u>on</u><u> </u><u>Block</u><u> </u><u>A</u><u>:</u>
Let the x-axis be (+) towards the right and y-axis be (+) in the upward direction. We can write the net forces on mass
as


Substituting (2) into (1), we get

where
, the frictional force on
Set this aside for now and let's look at the forces on 
<u>Forces</u><u> </u><u>on</u><u> </u><u>Block</u><u> </u><u>B</u><u>:</u>
Let the x-axis be (+) up along the inclined plane. We can write the forces on
as


From (5), we can solve for <em>N</em> as

Set (6) aside for now. We will use this expression later. From (3), we can see that the tension<em> </em><em>T</em><em> </em> is given by

Substituting (7) into (4) we get

Collecting similar terms together, we get

or
![a = \left[ \dfrac{m_B\sin30 - \mu_km_A}{(m_A + m_B)} \right]g\:\:\:\:\:\:\:\:\:(8)](https://tex.z-dn.net/?f=a%20%3D%20%5Cleft%5B%20%5Cdfrac%7Bm_B%5Csin30%20-%20%5Cmu_km_A%7D%7B%28m_A%20%2B%20m_B%29%7D%20%5Cright%5Dg%5C%3A%5C%3A%5C%3A%5C%3A%5C%3A%5C%3A%5C%3A%5C%3A%5C%3A%288%29)
Putting in the numbers, we find that
. To find the tension <em>T</em>, put the value for the acceleration into (7) and we'll get
. To find the force exerted by the inclined plane on block B, put the numbers into (6) and you'll get 
Answer:

Explanation:
The velocity v₁ can be calculated with the kinematic formula:

Since the object is initially at rest, v₁ becomes:

Where g is the acceleration due to gravity. Now, the velocity v₂ can be calculated with the same formula, but now the initial velocity is v₁:

Substituting v₁ in this expression and solving for v₂, we get:

Now, dividing v₂ over v₁, we get the expression:

It means that v₂ is √2 times v₁.
Answer
given,
D = 50 mm = 0.05 m
d = 10 mm = 0.01 m
Force to compress the spring




F = 3160 N
stress correction factor from stress correction curve is equal to 1.1
now, calculation of corrected stress


= 442.6 Mpa
The tensile strength of the steel material of ASTM A229 is equal to 1300 Mpa
now,



since corrected stress is less than the
hence, spring will return to its original shape.