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Nikitich [7]
3 years ago
13

A 1-kilogram object is thrown horizontally and a 2-kilogram object is dropped vertically at

Physics
1 answer:
Semenov [28]3 years ago
4 0
Their vertical position will be the same for both at any instant. They will both hit the ground at the same time.
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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.
Svetradugi [14.3K]

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)

E_{k1}+W_{1-2}=E_{k2}\\where:\\E_{k1}= kinetic energy at moment 1\\W_{1-2}= work between moments 1 and 2.\\E_{k2}= kinetic energy at moment 2.

W_{1-2}= (F*d) - (m*g*h)\\W_{1-2}=(500000*2.5*10^3)-(40000*9.81*2.5*10^3)\\W_{1-2}= 269*10^6[J] or 269 [MJ]

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.

E_{k2}=0.5*m*v^2\\269*10^6=0.5*40000*v^2\\v=\sqrt{\frac{269*10^6}{0.5*40000} }\\ v=116[m/s]

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3. As an object’s temperature increases, the ____________________ at which it radiates energy increases.
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As an object’s temperature increases, the Rate at which it radiates energy increases.

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The right approach is Option b (the force..................exert on you).

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The other options given are not connected to the situation described. So, the solution here was the right one.

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