Answer : The final temperature is,
Explanation :
In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.
where,
= specific heat of ice =
= specific heat of water =
= mass of ice = 50 g
= mass of water = 200 g
= final temperature = ?
= initial temperature of ice =
= initial temperature of water =
Now put all the given values in the above formula, we get:
Therefore, the final temperature is,
Answer:
The location of the shear center o is 0.033 or 33 m
Explanation:
Solution
Recall that,
The moment of inertia of the section is = I = 0.05 * 0.4 ^3 /12 + 0.005 * 0.2 ^3/12
= 30 * 10 ^ ⁻⁶ m⁴
Now,
The first moment of inertia is
Q =ῩA = [ (0.1 -x) + x/2] (0.005 * x)
= 0.5x * 10 ^⁻³ - 2.5 x * 10⁻³ x²
Thus,
The shear flow is,
q = VQ/I
so,
P = (0.5x * 10 ^⁻³ - 2.5 x * 10⁻³ x²)/ 30 * 10 ^⁻⁶
P = (16.67 x - 83. 33 x²)
The shear force resisted by the shorter web becomes
Vw,₂ = 2∫ = ₀.₁ and ₀ = P (16.67 x - 83. 33 x²) dx = 0.11x
Then,
We take the moment at a point A
∑Mₐ = 0
- ( p * e)- (Vw₂ * 0.3 ) = 0
e = 0.11 p * 0.3/p
which gives us 0.033 m
= 33 m
Therefore the location of the shear center o is 0.033 or 33 m
Note: Kindly find an attached diagram to the question given above as part of the explanation solved with it.
Answer:
V=34.2 m/s
Explanation:
Given that
Height , h= 54 m
Horizontal distance , x = 35 m
Given that , the ball is thrown horizontally , therefore the initial vertical velocity will be zero.
In vertical direction :
We know that
Now by putting the values in the above equation we got
Assume
Thus
We also know that
In horizontal direction :
Thus the resultant velocity
V=34.2 m/s
Therefore the velocity will be 34.2 m/s.