To solve this problem we will use the concepts related to Torque as a function of the Force in proportion to the radius to which it is applied. In turn, we will use the concepts of energy expressed as Work, and which is described as the Torque's rate of change in proportion to angular displacement:
Where,
F = Force
r = Radius
Replacing we have that,
The moment of inertia is given by 2.5kg of the weight in hand by the distance squared to the joint of the body of 24 cm, therefore
Finally, angular acceleration is a result of the expression of torque by inertia, therefore
PART B)
The work done is equivalent to the torque applied by the distance traveled by 60 °° in radians 
, therefore
The answer to this would inFact be A
Answer:
letter B
none zero digit are significant figures
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.
Answer:
Velocity of Pauli relative to Daniel = (-1.50ï + 3.90ĵ) m/s
x-component = -1.50 m/s
y-component = 3.90 m/s
Explanation:
Relative velocity of a body A relative to another body B, Vab, is given as
Vab = Va - Vb
where
Va = Relative velocity of body A with respect to another third body or frame of reference C
Vb = Relative velocity of body B with respect to that same third body or frame of reference C.
So, relative velocity can be given further as
Vab = Vac - Vbc
Velocity of Newton relative to Daniel = Vnd = 3.90 m/s due north = (3.90ĵ) m/s in vector form.
Velocity of Newton relative to Pauli = Vnp = 1.50 m/s due East = (1.50î) m/s in vector form
What is Pauli's velocity relative to Daniel?
Vpd = Vp - Vd
(Pauli's velocity relative to Daniel) = (Pauli's velocity relative to Newton) - (Daniel's velocity relative to Newton)
Vpd = Vpn - Vdn
Vpn = -Vnp = -(1.50î) m/s
Vdn = -Vnd = -(3.90ĵ) m/s
Vpd = -1.50î - (-3.90ĵ)
Velocity of Pauli relative to Daniel = (-1.50ï + 3.90ĵ) m/s
Hope this Helps!!!!
