It could rotate while not advancing distance
The gravitational acceleration of a planet is proportional to the planet's mass, and inversely proportional to square of the planet's radius.
So when you stand on the surface of this particular planet, you feel a force of gravity that is
(1/2) / (3²)
of the force that you feel on the surface of the Earth.
That's <em>(1/18)</em> as much as on Earth.
The acceleration of gravity there would be about <em>0.545 m/s²</em>.
This is about 12% less than the gravity on Pluto.
Answer:
Check the explanation
Explanation:
net angular momentum is conserved as there is no net external torque.
therefore. 0 = angular momentum of woman + angular momentum of wheel.
0 = 55x2x1.5 + 480xω
ω = -0.3056 rad/sec ,-ve sign shows that the direction is opposite to woman, ie counterclockwise.
work done by woman = total kinetic energy of the wheel and woman
=1/2 x 55 x 1.52 + 1/2 Iω2 = 1/2x55x1.52 + 1/2x540x0.30562= 41.8+73.35 = 115.14J
Answer:
Answer is c
Explanation:
The owner of question told me :))
Answer:
Fₓ = 21.9 kN
Fᵧ = 84.3 kN
T = 32.7 kN
Explanation:
Draw a free body diagram (assuming the weight of the structure is included in the 60 kN force).
There are vertical and horizontal reaction forces at A (Fᵧ and Fₓ), and a tension force T at B pulling down along the rope.
The length of BC is √(2.7² + 3²) = √16.29. Using similar triangles, the vertical and horizontal components of the tension force are:
Tᵧ = 3 T / √16.29 ≈ 0.743 T
Tₓ = 2.7 T / √16.29 ≈ 0.669 T
Sum of moments about A in the counterclockwise direction:
∑τ = Iα
Tᵧ (1 m) + Tₓ (3 m) − 60 kN (1 m) − 30 kNm = 0
Tᵧ + 3 Tₓ = 90 kN
0.743 T + 3 (0.669 T) = 90 kN
2.750 T = 90 kN
T = 32.7 kN
Sum of forces in the +x direction:
∑F = ma
Fₓ − Tₓ = 0
Fₓ = Tₓ
Fₓ = 0.669 T
Fₓ = 21.9 kN
Sum of forces in the +y direction:
∑F = ma
Fᵧ − Tᵧ − 60 kN= 0
Fᵧ = Tᵧ + 60 kN
Fᵧ = 0.743 T + 60 kN
Fᵧ = 84.3 kN