Explanation:
For projectile motion, use constant acceleration equation:
Δx = v₀ t + ½ at²
where Δx is the displacement,
v₀ is the initial velocity,
a is the acceleration,
and t is time.
Both objects are projected upward with velocity u. The second object is thrown after a time t₀.
For the first object:
Δx = u t + ½ (-g) t²
Δx = u t − ½g t²
For the second object:
Δx = u (t−t₀) + ½ (-g) (t−t₀)²
Δx = u (t−t₀) − ½g (t−t₀)²
Assuming the objects meet, the displacements will be equal:
u t − ½g t² = u (t−t₀) − ½g (t−t₀)²
u t − ½g t² = u (t−t₀) − ½g (t² − 2tt₀ + t₀²)
u t − ½g t² = u t − u t₀ − ½g t² + g tt₀ − ½g t₀²
0 = -u t₀ + g tt₀ − ½g t₀²
0 = -u + g t − ½g t₀
g t = u + ½g t₀
t = u/g + t₀/2
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t=(0-(250sin75)^2)/-9.8
<span>the distance one is (2500+610)- (250m/s*cos75)*t=Dh Dh=horizontal distance </span>
<span>the max height one is d=0.5*9.8*t^2 </span>
<span>d= max height subtract 1800-d</span>
The passing of heat through a material while the material itself stays in place is conduction.
Frequency = (speed) / (wavelength)
Speed = 3 x 10⁸ m/s
Wavelength = 3 cm = 0.03 m
Frequency = (3 x 10⁸ m/s) / (0.03 m)
Frequency = (3 x 10⁸ / 0.03) (m / m-s)
Frequency = 1 x 10¹⁰ Hz (10 Gigahertz)
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
