Answer:
7.00 m
Explanation:
Given:
v₀ = 2.00 m/s
v = 5.00 m/s
a = 1.50 m/s²
Find: Δx
v² = v₀² + 2aΔx
(5.00 m/s)² = (2.00 m/s)² + 2(1.50 m/s²)Δx
Δx = 7.00 m
I think its A let me know its wrong or not
By raising a position of an object, Its potential energy increases.
On Earth, the acceleration of gravity is 9.8 m/s² downward.
So any object with only gravity acting on it gains 9.8 m/s of
downward speed every second.
If the rock starts out moving upward at 10 m/s, then it will
continue upward for only (10/9.8) = 1.02 second, before
it stops rising and starts falling.
Its average speed during that time is (1/2) (10 + 0) = 5 m/s .
At an average speed of 5 m/s for 1.02 sec,
the rock rises
(5 m/s) x (1.02 sec) = 5.102 meters .
If the period of a satellite is T=24 h = 86400 s that means it is in geostationary orbit around Earth. That means that the force of gravity Fg and the centripetal force Fcp are equal:
Fg=Fcp
m*g=m*(v²/R),
where m is mass, v is the velocity of the satelite and R is the height of the satellite and g=G*(M/r²), where G=6.67*10^-11 m³ kg⁻¹ s⁻², M is the mass of the Earth and r is the distance from the satellite.
Masses cancel out and we have:
G*(M/r²)=v²/R, R=r so:
G*(M/r)=v²
r=G*(M/v²), since v=ωr it means v²=ω²r² and we plug it in,
r=G*(M/ω²r²),
r³=G*(M/ω²), ω=2π/T, it means ω²=4π²/T² and we plug that in:
r³=G*(M/(4π²/T²)), and finally we take the third root to get r:
r=∛{(G*M*T²)/(4π²)}=4.226*10^7 m= 42 260 km which is the height of a geostationary satellite.