Answer:
Explanation:
The question relates to time of flight of a projectile .
Time of flight = 2 u sinθ / g
u is speed of projectile , θ is angle of projectile
= 2 x 48.5 sin42 / 9.8
= 6.6 seconds .
Maximum height attained
= u² sin²θ / g
= 48.5² sin²42 / 9.8
= 107.47 m .
The net force on the barge is 8000 N
Explanation:
In order to find the net force on the badge, we have to use the rules of vector addition, since force is a vector quantity.
In this problem, we have two forces:
- The force of tugboat A,
, acting in a certain direction - The force of tugboat B,
, also acting in the same direction
Since the two forces act in the same direction, this means that we can simply add their magnitudes to find the net combined force on the barge. Therefore, we get
and the direction is the same as the direction of the two forces.
Learn more about forces:
brainly.com/question/11179347
brainly.com/question/6268248
#LearnwithBrainly
Recall this gas law:
= 
P₁ and P₂ are the initial and final pressures.
V₁ and V₂ are the initial and final volumes.
T₁ and T₂ are the initial and final temperatures.
Given values:
P₁ = 475kPa
V₁ = 4m³, V₂ = 6.5m³
T₁ = 290K, T₂ = 277K
Substitute the terms in the equation with the given values and solve for Pf:
<h3>P₂ = 279.2kPa</h3>
The coefficient of linear expansion, given that the length of the pipe increased by 1.5 cm is 1.67×10¯⁵ /°F
<h3>How to determine the coefficient of linear expansion</h3>
From the question given above, the following data were obtained
- Original diameter (L₁) = 10 m
- Change in length (∆L) = 1.5 cm = 1.5 / 100 = 0.015 m
- Change in temperature (∆T) = 90 °F
- Coefficient of linear expansion (α) =?
The coefficient of linear expansion can be obtained as illustrated below:
α = ∆L / L₁∆T
α = 0.015 / (10 × 90)
α = 0.015 / 900
α = 1.67×10¯⁵ /°F
Thus, we can conclude that the coefficient of linear expansion is 1.67×10¯⁵ /°F
Learn more about coefficient of linear expansion:
brainly.com/question/28293570
#SPJ1
Answer:
Normal stress = 66/62.84 = 1.05kips/in²
shearing stress = T/2 = 0.952/2 = 0.476 kips/in²
Explanation:
A steel pipe of 12-in. outer diameter d₂ =12in d₁= 12 -4in = 8in
4 -in.-thick
angle of 25°
Axial force P = 66 kip axial force
determine the normal and shearing stresses
Normal stress б = force/area = P/A
= 66/ (П* (d₂²-d₁²)/4
=66/ (3.142* (12²-8²)/4
= 66/62.84 = 1.05kips/in²
Tangential stress T = force* cos ∅/area = P/A
= 66* cos 25/ (П* (d₂²-d₁²)/4
=59.82/ (3.142* (12²-8²)/4
= 59.82/62.84 = 0.952kips/in²
shearing stress = tangential stress /2
= T/2 = 0.952/2 = 0.476 kips/in²
