LOTS OF POINTSA rocket of mass 40 000 kg takes off and flies to a height of 2.5 km as its engines produce 500 000 N of thrust.
1 answer:
Answer:
i) E = 269 [MJ] ii)v = 116 [m/s]
Explanation:
This is a problem that encompasses the work and principle of energy conservation.
In this way, we establish the equation for the principle of conservation and energy.
i)
At that point the speed 1 is equal to zero, since the maximum height achieved was 2.5 [km]. So this calculated work corresponds to the energy of the rocket.
Er = 269*10^6[J]
ii ) With the energy calculated at the previous point, we can calculate the speed developed.
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Answer:
Check the attached image
Explanation:
To solve the problem for time you will have to use the formula for time, t = d/s which means time equals distance divided by speed.
Kindly check the attached image below for the step by step explanation to the question.
It is required an infinite work. The additional electron will never reach the origin.
In fact, assuming the additional electron is coming from the positive direction, as it approaches x=+1.00 m it will become closer and closer to the electron located at x=+1.00 m. However, the electrostatic force between the two electrons (which is repulsive) will become infinite when the second electron reaches x=+1.00 m, because the distance d between the two electrons is zero:
So, in order for the additional electron to cross this point, it is required an infinite amount of work, which is impossible.
In fact, assuming the additional electron is coming from the positive direction, as it approaches x=+1.00 m it will become closer and closer to the electron located at x=+1.00 m. However, the electrostatic force between the two electrons (which is repulsive) will become infinite when the second electron reaches x=+1.00 m, because the distance d between the two electrons is zero:
So, in order for the additional electron to cross this point, it is required an infinite amount of work, which is impossible.
Complete Question:
Two 3.0µC charges lie on the x-axis, one at the origin and the other at 2.0m. A third point is located at 6.0m. What is the potential at this third point relative to infinity? (The value of k is 9.0*10^9 N.m^2/C^2)
Answer:
The potential due to these charges is 11250 V
Explanation:
Potential V is given as;
where;
K is coulomb's constant = 9x10⁹ N.m²/C²
r is the distance of the charge
q is the magnitude of the charge
The first charge located at the origin, is 6.0 m from the third charge; the potential at this point is:
The second charge located at 2.0 m, is 4.0 m from the third charge; the potential at this point is:
Total potential due to this charges = 4500 V + 6750 V = 11250 V