Answer:
The distance the piece travel in horizontally axis is
L=3.55m
Explanation:
Now the angular velocity is the blade speed so:
assuming no air friction effects affect blade piece:
time for blade piece to fall to floor
Now is the same time the piece travel horizontally
blade piece travels HORIZONTALLY = (24.5)(0.397) = 9.73 m ANS
Answer:
F=126339.5N
Explanation:
to find the necessary force to escape we must make a free-body diagram on the hatch, taking into account that we will match the forces that go down with those that go up, taking into account the above we propose the following equation,
Fw=W+Fi+F
where
Fw= force or weight produced by the water column above the submarine.
to fint Fw we can use the following ecuation
Fw=h. γ. A
h=distance
γ=
specific weight for seawater = 10074N / m ^ 3
A=Area
Fw=28x10074x0.7=197467N
w is the weight of the hatch = 200N
Fi is the internal force of the submarine produced by the pressure = 1atm = 101325Pa for this we can use the following formula
Fi=PA=101325x0.7=70927.5N
finally the force that is needed to open the hatch is given by the initial equation
Fw=W+Fi+F
F=Fw-W+Fi
F=197467N-200N-70927.5N
F=126339.5N
Given:
rod of circular cross section is subjected to uniaxial tension.
Length, L=1500 mm
radius, r = 10 mm
E=2*10^5 N/mm^2
Force, F=20 kN = 20,000 N
[note: newton (unit) in abbreviation is written in upper case, as in N ]
From given above, area of cross section = π r^2 = 100 π =314 mm^2
(i) Stress,
σ
=force/area
= 20000 N / 314 mm^2
= 6366.2 N/mm^2
= 6370 N/mm^2 (to 3 significant figures)
(ii) Strain
ε
= ratio of extension / original length
= σ / E
= 6366.2 /(2*10^5)
= 0.03183
= 0.0318 (to three significant figures)
(iii) elongation
= ε * L
= 0.03183*1500 mm
= 47.746 mm
= 47.7 mm (to three significant figures)
The resultant vector can be determined by the component vectors. The component vectors are vector lying along the x and y-axes. The equation for the resultant vector, v is:
v = √(vx² + vy²)
v = √[(9.80)² + (-6.40)²]
v = √137 or 11.7 units
Explanation:
Given that,
Force with which a child hits a ball is 350 N
Time of contact is 0.12 s
We need to find the impulse received by the ball. The impulse delivered is given by :
So, the impulse is 42 N-m..
We know that he change in momentum is also equal to the impulse delivered.
So, impulse = 42 N-m and change in momentum =42 N-m.
