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Ray Of Light [21]
4 years ago
14

Jupiter’s strength of gravity is greater than Earth’s strength of gravity. A person’s will be the same on Jupiter and Earth. A p

erson’s will be greater on Jupiter than on Earth.
Physics
2 answers:
stepan [7]4 years ago
8 0
A person's weight will be greater on Jupiter than on earth
Artist 52 [7]4 years ago
4 0

Answer:  A person’s <u>Mass</u> will be the same on Jupiter and Earth. A person’s <u>Weight</u> will be greater on Jupiter than on Earth.

Explanation:

Amount of matter contained in a body is known as mass (m). It is measured in grams or kilo-grams.

Weight is the force of gravity acting on an object. It is the product of the mass and acceleration due to gravity. It is measured in Newtons (N).

W = m g

Acceleration due to gravity varies with strength of gravity of different planets. Because, Jupiter’s strength of gravity is greater than Earth’s strength of gravity, the person would have greater weight on Jupiter than on Earth but same mass. Mass is constant everywhere.

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- a third force, given by the propulsion of the plane, which is accelerating it towards the ground (because the problem says that the plane has an acceleration of a=12 m/s^2 towards the ground)

The radius of the circle is r= \frac{260 m}{2} = 130 m, so the centripetal force acting on the plane is
F_c=m \frac{v^2}{r} = \frac{(85000 kg)(55 m/s)^2}{130 m}=1.98 \cdot 10^6 N
On the vertical axis, we have two forces: the weight
W=mg=(85000 kg)(9.81 m/s^2)=8.34 \cdot 10^5 N
and the other force F given by the propulsion. Since we know that their sum should generate an acceleration equal to a=12 m/s^2, we can find the magnitude of this other force F by using Newton's second law:
F+mg=ma
F=m(a-g)=(85000kg)(12 m/s^2-9.81 m/s^2)=1.86 \cdot 10^5 N

So, the net force acting on the plane will be the resultant of the centripetal force (acting in the horizontal direction) and the two forces W and F (acting in the vertical direction):
R= \sqrt{(F_c^2+(W+F)^2}=
= \sqrt{(1.98\cdot 10^6N)^2+(8.34 \cdot 10^5N+1.86 \cdot 10^5 N)^2}  =2.23 \cdot 10^6 N

(b) The tangent of the angle with respect to the horizontal is the ratio between the sum of the forces in the vertical direction (taken with negative sign, since they are directed downward) and the forces acting in the horizontal direction, so:
\tan \theta =  \frac{-(W+F)}{F_c}= -0.5
And so, the angle is
\theta = \arctan (-0.5)=-26.8 ^{\circ}
 
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