Determine the mass of a boat that is propelled forward by a force of 1250N at an acceleration of 3 m/s2.
1 answer:
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Answer:
x_{cm} = 4.644 10⁶ m
Explanation:
The center of mass is given by the equation
= 1 /
∑
Where M_{total} is the total masses of the system, is the distance between the particles and
is the masses of each body
Let's apply this equation to our problem
M = Me + m
M = 5.98 10²⁴ + 7.36 10²²
M = 605.36 10²² kg
Let's locate a reference system located in the center of the Earth
Let's calculate
x_{cm} = 1 / 605.36 10²² [Me 0 + 7.36 10²² 3.82 10⁸]
x_{cm} = 4.644 10⁶ m
Refer to the diagram shown below.
Let I = the moment of inertia of the wheel.
α = 0.81 rad/s², the angular acceleration
r = 0.33 m, the radius of the weel
F = 260 N, the applied tangential force
The applied torque is
T = F*r
= (260 N)*(0.33 m)
= 85.8 N-m
By definition,
T = I*α
Therefore,
I = T/α
= (85.8 N-m)/(0.81 rad/s²)
= 105.93 kg-m²
Answer: 105.93 kg-m²
Let I = the moment of inertia of the wheel.
α = 0.81 rad/s², the angular acceleration
r = 0.33 m, the radius of the weel
F = 260 N, the applied tangential force
The applied torque is
T = F*r
= (260 N)*(0.33 m)
= 85.8 N-m
By definition,
T = I*α
Therefore,
I = T/α
= (85.8 N-m)/(0.81 rad/s²)
= 105.93 kg-m²
Answer: 105.93 kg-m²
Answer:
(a)
M = 1.898 x 10^27 kg
(b)
v = 13.74 km/s
(c) E = 0.28 N/kg
Explanation:
Time period, T = 3.55 days = 3.55 x 24 x 3600 second = 306720 s
Radius, r = 6.71 x 10^8 m
G = 6.67 x 10^-11 Nm^2/kg^2
(a)
M = 1.898 x 10^27 kg
(b) Let v be the orbital velocity
v = 13739.5 m/s
v = 13.74 km/s
(b) The gravitational field E is given by
E = 0.28 N/kg