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Delvig [45]
2 years ago
7

Help meee please please please

Physics
1 answer:
ehidna [41]2 years ago
8 0
A) le ha dado a conocer el nombre del director del festival del río y en la ciudad del sur del mar de la plata en el centro del mar del sur sur del sur y el mar de la sierra del mar de la plata del mar del plata
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Which event is an example of melting?
qaws [65]

Answer:

welding metal

Explanation:

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2 years ago
Why is a specimen smaller than 200 nm not visible with a light microscope?
kifflom [539]
Because the specimen is very small with a light microscope
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After watching a show about submarines, Jamil wants to learn more about the oceans. Which question could be answered through sci
Georgia [21]
The best question that could prompt a scientific investigation is: <u>What substances dissolve in ocean water?</u> 

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2 years ago
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needhelpp101 Why is the gravitational potential energy of an object 1 meter above the moon’s surface less than its potential ene
Vlad1618 [11]

If you remember the formula for potential energy,
then this question is a piece-o-cake.

  <em>Potential energy = (mass) x (<u>acceleration of gravity</u>) x (height) .</em>

-- The object's mass is the same everywhere.
-- You said that the height is the same both times.
-- How about the acceleration of gravity ? 

Compared to gravity on Earth, it's only  16.5 percent as much on the Moon. 
So naturally, from the formula, you'd expect the Potential Energy to be less
on the Moon.

4 0
3 years ago
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]

8 0
2 years ago
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