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julia-pushkina [17]
3 years ago
9

não é difícil de perceber uma relação aditiva entre os termos consecutivos (1;3;6;10...)(A)qual é o sexto e décimo número triang

ular
Physics
1 answer:
EastWind [94]3 years ago
7 0
It is 15 and 45 ............
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The Hubble Space Telescope has a mass of 1.16*10^ 4 kg and orbits the Earth at an altitude of 5.68 * 10 ^ 5 above Earth's surfac
andrezito [222]

Answer:

E=8.13\times 10^{12}\ J

Explanation:

Given that,

The mass of a Hubble Space Telescope, m_1=1.16\times 10^4\ kg

It orbits the Earth at an altitude of 5.68\times 10^5\ m

We need to find the potential energy the telescope at this location. The formula for potential energy is given by :

E=\dfrac{Gm_1m_e}{r}

Where

m_e is the mass of Earth

Put all the values,

E=\dfrac{6.67\times 10^{-11}\times 1.16\times 10^4\times 5.97\times 10^{24}}{5.68\times 10^5}\\\\E=8.13\times 10^{12}\ J

So, the potential energy of the telescope is 8.13\times 10^{12}\ J.

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3 years ago
Explain how waves, energy, and matter are related
Stels [109]

Answer:

Mechanical waves need matter to transfer energy while electromagnetic waves do not. ... Waves change direction when they move from one material into another (matter) through the process of refraction. The wave will change direction when the speed of the wave changes.

3 0
3 years ago
Please help me this is important!
Mariulka [41]
145 Grams!
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3 years ago
A 8.34 × 103-kg lunar landing craft is about to touch down on the surface of the moon, where the acceleration due to gravity is
Mnenie [13.5K]

Answer:

21870.3156 N

Explanation:

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 1.6 m/s²

Equation of motion

v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-18.2^2}{2\times 162}\\\Rightarrow a=-1.02234\ m/s^2

The acceleration of the craft should be 1.02234 m/s²

F=ma\\\Rightarrow F=8.34\times 10^3\times 1.02234\\\Rightarrow F=8526.3156\ N

Weight of the craft

W=mg\\\Rightarrow W=8.34\times 10^3\times 1.6\\\Rightarrow W=13344\ N

Thrust

F_t=F+W\\\Rightarrow F_t=8526.3156+13344\\\Rightarrow F_t=21870.3156\ N

The thrust needed to reduce the velocity to zero at the instant when the craft touches the lunar surface is 21870.3156 N

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