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aleksandr82 [10.1K]
3 years ago
9

g A certain elevator cab has a total run of 195 m and a maximum speed is 306 m/min, and it accelerates from rest and then back t

o rest at 1.19 m/s2. (a) How far does the cab move while accelerating to full speed from rest

Physics
1 answer:
katen-ka-za [31]3 years ago
7 0

Answer:

About 23 meters

Explanation:

To do this, you'll want to apply one of the kinematic equations to find the time it takes for the cabin to reach max velocity from rest. (Use the max velocity as V_f and V_i=0)

Then, you can find the distance travelled during the acceleration by equating the acceleration to the change in distance of the time squared.

My work is in the attachment, comment if you have any questions.

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Answer:

K_E=mgr_E

Explanation:

By conservation of energy, the sum of the kinetic and gravitational potential energies at the surface of the Earth must be equal than their sum at infinity, so we have:

K_E+U_E=K_\infty+U_\infty

\frac{mv_E^2}{2}-\frac{GM_Em}{r_E}=\frac{mv_\infty^2}{2}-\frac{GM_Em}{r_\infty}

Where G=6.67\times10^{-11}Nm^2/kg^2 is the gravitational constant,M_E=5.97\times10^{24}kg and r_E=6371000m are the mass and radius of the Earth, <em>m </em>is the mass of the particle, v_E its velocity at the surface of the Earth (which would be its escape velocity) and v_\infty and r_\infty are the velocities and distance at infinity, which would be null and infinity respectively, so the right hand side of our equation is 0J, which leaves us with:

\frac{GM_Em}{r_E}=\frac{mv_E^2}{2}=K_E

Also, since the force the molecule experiments is the force of gravity (disregarding drag), we can write its weight in terms of Newton's Law of Gravitation:

F=mg=\frac{GM_Em}{r_E^2}

Which means that:

\frac{GM_Em}{r_E}=mgr_E

So finally putting all together we can write:

K_E=\frac{GM_Em}{r_E}=mgr_E

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