Answer:
80.386 degrees
Explanation:
We use the cosine equation here (which is the adjacent side of the unknown angle divided by the hypotenuse
The adjacent side = 699ft
The hypotenuse = 1034ft
using cos∅ = Adjacent/hypotenuse
where ∅ is the unknown angle
cos ∅ = 699/1034 = 0.167
∅ = arccos 0.167 = 80.368°
As easy as one can imagine
The specific gravity of the object’s material is 5.09.
<h3>To calculate the specific gravity of the object:</h3>
Weight difference = 9 - 7.2 = 1.8 N = Buoyant force of water
Buoyant Force in water(Fb) = density of water x g x volume of the body(Vb)
1.8 = 1000 x 9.81 x Vb
Vb = 1.8/9810 cubic meter
Now, in the air;
Weight of body = mg = 9 N
Mass of body,m = 9/9.81 Kg
So,
Density of body = m/ Vb
= 9/9.81 ÷ 1.8/9810
= 5094.44 kg per cubic meter
The specific gravity of body = density of body ÷ density of water
= 5094.44 ÷ 1000
= 5.09
Therefore, Specific gravity of body = 5.09
Learn more about Specific gravity here:
brainly.com/question/13258933
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Answer:
F= 0
Explanation:
This exercise we use Newton's second law,
F = m a
in this case as the speed is constant the acceleration is zero therefore the force is zero.
Change we can solve it using Newton's first law, which states that every vehicle remains still or with constant speed if there is no extensive outside acting on it
We see that with any of the two forms the sum of the applied forces is zero
∑ F = 0
Answer:
Stars create new elements in their cores bu squeezing elements together in a process called unclear fusion. First stars fuse hydrogen atoms into heluim. Helium atorm then fuse to create berylluim and so on until fusion in the star’s core has created every element up to icon.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The cross-sectional area is 
Explanation:
The free body diagram of the link is shown on the second uploaded image
From the question we are told that
Ultimate normal stress in the link 
Factor safety 
From our free diagram we can see that the moment about B is 0 Mathematically
But 
Hence 
Making 
the subject
At equilibrium summation of all force is 0 mathematically
This means
i.e 
The factor of safety is mathematically
Factor of safety 
Where 
is the normal stress
is the allowable stress this mathematically given as
![Factor\ of \ safety =\frac{450*10^6}{[\frac{21*10^3}{A} ]}](https://tex.z-dn.net/?f=Factor%5C%20of%20%5C%20safety%20%3D%5Cfrac%7B450%2A10%5E6%7D%7B%5B%5Cfrac%7B21%2A10%5E3%7D%7BA%7D%20%5D%7D)
Making A the subject
