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2 years ago

Explain how force (F=ma), momentum (p=mv), energy, and gravity play a part in a launching sequence.

2 answers:

8 0

If the mass bigger and the acceleration faster, the force is bigger(F = ma). The rate of change of momentum of an object is directly proportional to the resultant force applied and is in the direction of the resultant force(p=mc) The gravity can affect the resultant force as weight when the object launch.

sorry if there's any grammar mistake

almond37 [142]2 years ago

5 0

Force is directly proportional to acceleration and inversely proportional to the mass. The acceleration vector of an object is in the same direction of the net force vector.

Momentum is mass times velocity. If you take the derivative of momentum with respect to time, you get the answer of (m*a) So therefore F = the derivative of momentum.

An object has potential energy depending on the situation it is in. This includes location mostly. If you launch an object on a cliff, it has a large amount of potential energy which is quickly converted into kinetic energy once it starts moving.

Gravity accelerates an object toward the center of mass. In a projectiles case, that would be earth. So once you shoot a projectile, it will accelerate towards the earth at -9.8 meters per second squared. Gravity itself gives the object potential energy

Idk if this helped at all. Hopefully it helped you understand the concepts and you can then research them yourself more in depth. My bad if I made any mistakes

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Which of the following is not a step in the inquiry process?

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Invention I think which is B as they don't invent things in an enquiry

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2 years ago

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Jupiter's moon Io has active volcanoes (in fact, it is the most volcanically active body in the solar system) that eject materia

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Answer:

91.64 km

91.64 km high material would go on earth if it were ejected with the same speed as on Io.

Explanation:

According to Newton Law of gravitation:

$g=\frac{Gm}{r^2}$

Where:

G is gravitational constant= $6.67*10^{-11} m^3/kg.s^2$

For Moon lo g is:

$g_M=\frac{6.67*10^{-11}*8.93*10^{22}}{(1821*10^3)^2m^2} \\g_M=1.7962 m/s^2$

According to law of conservation of energy

Initial Energy=Final Energy

$K.E_i+mgh_i=K.E_f+mgh_f$

$\frac{1}{2}m(v_0)^2+mgh_o= \frac{1}{2}m(v_f)^2+mgh_f\\At\ maximum\ height\ v_f=0\\\frac{1}{2}m(v_0)^2+0=mgh_f\\v_0=\sqrt{2gh_f}$

For Jupiter's moon Io:

Velocity is given by:

$v_0_M=\sqrt{2g_Mh_f_M}$

For Earth Velocity is given by:

$v_0_E=\sqrt{2g_Eh_f_E}$

Now:

$v_o_M=v_o_E$

$\sqrt{2g_Mh_f_M}=\sqrt{2g_Eh_f_E}\\h_f_E=\frac{g_Mh_f_M}{g_E}$

$g_E=9.8 m/s^2$

$g_m=1.7962 m/s^2, As\ Calculated\ above$

$h_f_E=\frac{1.7962*500*10^3m}{9.8} \\h_f_E=91642.85 m\\h_f_E=91.64Km$

91.64 km high material would go on earth if it were ejected with the same speed as on Io.

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3 years ago

A person lifts a 4.5 kg cement block a vertical distance of 1.2 m and then carries the block horizontal- ly a distance of 7.3 m.

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When a person lifts the block, the block has more potential energy. Therefore the person does positive work on the block. work = m g h work = (4.5 kg) (9.80 m/s^2) (1.2 m) work = 52.92 joules The person's work on the block is 52.92 joules When the block is being raised, the force of gravity opposes the motion. Therefore the force of gravity does negative work on the block. work = - (force) (h) work = - m g h work = -(4.5 kg) (9.80 m/s^2) (1.2 m) work = -52.92 joules The work done by the force of gravity on the block is -52.92 joules Note that when the block is moved horizontally, the potential energy does not change. Therefore there is no work done on the block when it moves horizontally (we are assuming that the kinetic energy does not change).

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3 years ago

A satellite is orbiting Earth with a distance R = 2REarth from Earth's center. If the satellite is moved to a distance R = 4REar

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Answer:

Half

Explanation:

Given that:

radial distance of satellite from the earth, $R=2R_E$

Now, if the satellite is moved to a distance $R=4R_E$

We have the mathematical expression for the potential energy fue to gravitational field as:

$U=\frac{G.M.m}{R}$ ...................(1)

where:

$G = 6.67\times 10^{-11}\ m^3.kg^{-1}.s^{2}$

M = mass of earth

m = mass of satellite

R = radial distance of satellite

Now from eq. (1) initially we have:

$U=\frac{G.M.m}{2R_E}$

after the satellite is moved, we have:

$U'=\frac{G.M.m}{4R_E}$

$\Rightarrow U'=\frac{G.M.m}{2(2R_E)}$

$\Rightarrow U'=\frac{1}{2} \times U$

which is half of the initial condition.

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