The astronaut's weight is not 60 kg anywhere, because kg is a unit of mass, not weight.
If the astronaut's mass is 60 kg, then his weight is (60 kg)x(acceleration of gravity).
That's 588 Newtons on Earth, and 58.8 Newtons on a planet with 1/10 Earth's gravity.
The astronaut's mass of 60 kg goes with her, and doesn't depend on where she is.
Answer:
Jupiter Neptune moon Uranus
Explanation:
Fulcrum need to be positioned balanced with weight on both the sides following law of lever.
What is the physical law of the lever?
- It is the foundation for issues with weight and balance. According to this rule, a lever is balanced when the weight multiplied by the arm on one side of the fulcrum, which serves as the pivot point for the device, equals the weight multiplied by the arm on the opposing side.
- The lever is balanced, in other words, when the sum of the moments about the fulcrum is zero.
- The situation in which the positive moments (those attempting to turn the lever clockwise) equal the negative moments is known as this (those that try to rotate it counterclockwise).
- Moving the weights closer to or away from the fulcrum, as well as raising or lowering the weights, can alter the balance point, or CG, of the lever.
Learn more about the Fulcrum with the help of the given link:
brainly.com/question/16422662
#SPJ4
Explanation:
This parallel is located in the northern hemisphere, located at a latitude of 23 ° 26′12.2 ″ (or 23,43671 °) north of the equator. Approximately in the geographic half of Mexico, crossing the states of Baja California Sur, Sinaloa, Durango, Zacatecas, San Luis, Potosí, Nuevo León, Tamaulipas. Therefore, there is more of north america, at north of the tropic of cancer.
Answer:
a1 = 3.56 m/s²
Explanation:
We are given;
Mass of book on horizontal surface; m1 = 3 kg
Mass of hanging book; m2 = 4 kg
Diameter of pulley; D = 0.15 m
Radius of pulley; r = D/2 = 0.15/2 = 0.075 m
Change in displacement; Δx = Δy = 1 m
Time; t = 0.75
I've drawn a free body diagram to depict this question.
Since we want to find the tension of the cord on 3.00 kg book, it means we are looking for T1 as depicted in the FBD attached. T1 is calculated from taking moments about the x-axis to give;
ΣF_x = T1 = m1 × a1
a1 is acceleration and can be calculated from Newton's 2nd equation of motion.
s = ut + ½at²
our s is now Δx and a1 is a.
Thus;
Δx = ut + ½a1(t²)
u is initial velocity and equal to zero because the 3 kg book was at rest initially.
Thus, plugging in the relevant values;
1 = 0 + ½a1(0.75²)
Multiply through by 2;
2 = 0.75²a1
a1 = 2/0.75²
a1 = 3.56 m/s²
