V ( initial ) = 20 m/s
h = 2.30 m
h = v y * t + g t ² / 2
d = v x * t
1 ) At α = 18°:
v y = 20 * sin 18° = 6.18 m/s
v x = 20 * cos 18° = 19.02 m/ s
2.30 = 6.18 t + 4.9 t²
4.9 t² + 6.18 t - 2.30 = 0
After solving the quadratic equation ( a = 4.9, b = 6.18, c = - 2.3 ):
t 1/2 = (- 6.18 +/- √( 6.18² - 4 * 4.9 * (-2.3)) ) / ( 2 * 4.9 )
t = 0.3 s
d 1 = 19.02 m/s * 0.3 s = 5.706 m
2 ) At α = 8°:
v y = 20* sin 8° = 2.78 m/s
v x = 20* cos 8° = 19.81 m/s
2.3 = 2.78 t + 4.9 t²
4.9 t² + 2.78 t - 2.3 = 0
t = 0.46 s
d 2 = 19.81 * 0.46 = 9.113 m
The distance is:
d 2 - d 1 = 9.113 m - 5.706 m = 3.407 m
GOOD LUCK AND HOPE IT HELPS U
<span>f2 = f0/4
The gravity from the planet can be modeled as a point source at the center of the planet with all of the planet's mass concentrated at that point. So the initial condition for f0 has the satellite at a distance of 2r, where r equals the planet's radius.
The expression for the force of gravity is
F = G*m1*m2/r^2
where
F = Force
G = Gravitational constant
m1,m2 = masses involved
r = distance between center of masses.
Now for f2, the satellite has an altitude of 3r and when you add in the planet's radius, the distance from the center of the planet is now 4r. When you compare that to the original distance of 2r, that will show you that the satellite is now twice as far from the center of the planet as it was when it started. So let's compare the gravitational attraction, before and after.
f0 = G*m1*m2/r^2
f2 = G*m1*m2/(2r)^2
f2/f0 = (G*m1*m2/(2r)^2) / (G*m1*m2/r^2)
The Gm m1, and m2 terms cancel, so
f2/f0 = (1/(2r)^2) / (1/r^2)
f2/f0 = (1/4r^2) / (1/r^2)
And the r^2 terms cancel, so
f2/f0 = (1/4) / (1/1)
f2/f0 = (1/4) / 1
f2/f0 = 1/4
f2 = f0*1/4
f2 = f0/4
So the gravitational force on the satellite after tripling it's altitude is one fourth the original force.</span>
Inelastic.
If it was elastic, they'd bump right off each other. But since they've been locked, or stuck together, this is inelastic.
Answer:
zero or 2π is maximum
Explanation:
Sine waves can be written
x₁ = A sin (kx -wt + φ₁)
x₂ = A sin (kx- wt + φ₂)
When the wave travels in the same direction
Xt = x₁ + x₂
Xt = A [sin (kx-wt + φ₁) + sin (kx-wt + φ₂)]
We are going to develop trigonometric functions, let's call
a = kx + wt
Xt = A [sin (a + φ₁) + sin (a + φ₂)
We develop breasts of double angles
sin (a + φ₁) = sin a cos φ₁ + sin φ₁ cos a
sin (a + φ₂) = sin a cos φ₂ + sin φ₂ cos a
Let's make the sum
sin (a + φ₁) + sin (a + φ₂) = sin a (cos φ₁ + cos φ₂) + cos a (sin φ₁ + sinφ₂)
to have a maximum of the sine function, the cosine of fi must be maximum
cos φ₁ + cos φ₂ = 1 +1 = 2
the possible values of each phase are
φ1 = 0, π, 2π
φ2 = 0, π, 2π,
so that the phase difference of being zero or 2π is maximum
Answer:
A.)1.52cm
B.)1.18cm
Explanation:
angular speed of 120 rev/min.
cross sectional area=0.14cm²
mass=12kg
F=120±12ω²r
=120±12(120×2π/60)^2 ×0.50
=828N or 1068N
To calculate the elongation of the wire for lowest and highest point
δ=F/A
= 1068/0.5
δ=2136MPa
'E' which is the modulus of elasticity for alluminium is 70000MPa
δ=ξl=φl/E =2136×50/70000=1.52cm
δ=F/A=828/0.5
=1656MPa
δ=ξl=φl/E
=1656×50/70000=1.18cm
