To start with solving this
problem, let us assume a launch angle of 45 degrees since that gives out the
maximum range for given initial speed. Also assuming that it was launched at
ground level since no initial height was given. Using g = 9.8 m/s^2, the
initial velocity is calculated using the formula:
(v sinθ)^2 = (v0 sinθ)^2
– 2 g d
where v is final
velocity = 0 at the peak, v0 is the initial velocity, d is distance = 11 m
Rearranging to find for
v0: <span>
v0 = sqrt (d * g/ sin(2 θ)) </span>
<span>v0 = 10.383 m/s</span>
<u>T</u><u>he area w</u><u>h</u><u>ich touches the sand floor of 4 feet of the bull is less than the man.</u>
because the area of surface of contact is more in bull than in man. More is the area of contact, leads is the force or pressure affected.
Answer:
0.249
Explanation:
Perihelion = 4.43 x 10^9 km
Aphelion = 7.37 x 10^9 km
Let e be the eccentricity and a be the length of semi major axis.
The relation between the semi major axis, perihelion and aphelion s given by
Semi major axis = half of sum of perihelion and aphelion
a = 5.9 x 10^9 km
The relation between the perihelion, semi major axis and the eccentricity is given by
Perihelion = a (1 - e)
4.43 x 10^9 = 5.9 x 10^9 (1 - e)
0.751 = 1 - e
e = 0.249
Answer:
vT = v0/3
Explanation:
The gravitational force on the satellite with speed v0 at distance R is F = GMm/R². This is also equal to the centripetal force on the satellite F' = m(v0)²/R
Since F = F0 = F'
GMm/R² = m(v0)²/R
GM = (v0)²R (1)
Also, he gravitational force on the satellite with speed vT at distance 3R is F1 = GMm/(3R)² = GMm/27R². This is also equal to the centripetal force on the satellite at 3R is F" = m(vT)²/3R
Since F1 = F'
GMm/27R² = m(vT)²/3R
GM = 27(vT)²R/3
GM = 9(vT)²R (2)
Equating (1) and (2),
(v0)²R = 9(vT)²R
dividing through by R, we have
9(vT)² = (v0)²
dividing through by 9, we have
(vT)² = (v0)²/9
taking square-root of both sides,
vT = v0/3
