Missing question: "What is the spring's constant?"
Solution:
The object of mass m=6.89 kg exerts a force on the spring equal to its weight:

When the object is attached to the spring, the displacement of the spring with respect to its equilibrium position is

And by using Hook's law, we can find the constant of the spring:
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.
Hi there,
Unlike velocity,speed is scalar,which means it is described by MAGNITUDE only.
Answer:
2.64N
Explanation:
Force = mass * acceleration
Given
mass = 4kg
distance = 1.9m
Time t = 2.4s
Get the acceleration using the equation of motion
S = ut + 1/2at²
1.9 = 0 + 1/2a(2.4)²
1.9 = 5.76a/2
1.9 = 2.88a
a = 1.9/2.88
a = 0.66m/s²
Get the magnitude of the force
Force = 4 * 0.66
Force = 2.64N
Hence the net force acting on the fish is 2.64N