What is the main cause of ocean currents? Question 2 options:
2 answers:
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Answer:
a) The magnitude of the electric charge that a 60-kg human must have to overcome weight is 3.923 coulombs and its sign is negative.
b) The force of repulsion between two people is newtons. The use of the earth's electric field a feasible means of flight is not feasible since electric force of repulsion would destroy human body before taking advantage of any possible flight skill.
Explanation:
a) From Second Newton's Law, we form this equation of equilibrium:
(Eq. 1)
Where:
- Electrostatic force exerted on human, measured in Newton.
- Weight of the human, measured in Newton.
If we consider that human can be represented as a particle and make use of definitions of electric field and weight, the previous equation is expanded and electric charge is cleared afterwards:
(Eq. 2)
- Electric field, measured in Newtons per Coloumb.
- Mass, measured in kilograms.
- Gravity acceleration, measured in meters per square second.
- Electric charge, measured in Coulomb.
As electric field of the Earth is directed in toward the center of the planet, that is, in the same direction of gravity, electric field must be a negative value. If we know that ,
and
, the charge that a 60-kg human must have to overcome weight is:
The magnitude of the electric charge that a 60-kg human must have to overcome weight is 3.923 coulombs and its sign is negative.
b) The electric force of repulsion between two people with the same charge calculated in part (a) is determined by Coulomb's Law, whose definition we proceed to use:
(Eq. 3)
Where:
- Electrostatic constant, measured in Newton-square meter per square Coulomb.
- Electric charge, measured in Coulomb.
- Distance between two people, measured in meters.
If we know that ,
and
, then the force of repulsion between two people is:
The force of repulsion between two people is newtons. The use of the earth's electric field a feasible means of flight is not feasible since electric force of repulsion would destroy human body before taking advantage of any possible flight skill.
Answer:
He could jump 2.6 meters high.
Explanation:
Jumping a height of 1.3m requires a certain initial velocity v_0. It turns out that this scenario can be turned into an equivalent: if a person is dropped from a height of 1.3m in free fall, his velocity right before landing on the ground will be v_0. To answer this equivalent question, we use the kinematic equation:
With this result, we turn back to the original question on Earth: the person needs an initial velocity of 5 m/s to jump 1.3m high, on the Earth.
Now let's go to the other planet. It's smaller, half the radius, and its meadows are distinctly greener. Since its density is the same as one of the Earth, only its radius is half, we can argue that the gravitational acceleration g will be <em>half</em> of that of the Earth (you can verify this is true by writing down the Newton's formula for gravity, use volume of the sphere times density instead of the mass of the Earth, then see what happens to g when halving the radius). So, the question now becomes: from which height should the person be dropped in free fall so that his landing speed is 5 m/s ? Again, the kinematic equation comes in handy:
This results tells you, that on the planet X, which just half the radius of the Earth, a person will jump up to the height of 2.6 meters with same effort as on the Earth. This is exactly twice the height he jumps on Earth. It now all makes sense.