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4 years ago

Como un astronauta utiliza la fuerza de hallar y empujar?

Physics

1 answer:

PilotLPTM [1.2K]4 years ago

3 0

Answer:

La única manera en que nuestro astronauta sería capaz de empujar la nave espacial en el espacio sin alejarse sería usar algo llamado "unidad de propulsión de astronauta". Supongamos que el astronauta está usando un SPK soviético, el sistema de cohetes mochila más poderoso jamás utilizado en el espacio.

Explanation:

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Directions: Read each question carefully and answer

Lady bird [3.3K]

I have all the answers here so take this

8 0

2 years ago

An arch carries the thrust of weight to its _____(1)______. With a _____(2)______, the horizontal part of the structure supports

finlep [7]

Option D is correct. An arch carries the thrust of weight to its sides with a post-and-lintel.

An arch is indeed a vertical curving construction that covers an elevated space that may or may not sustain the load above it or the pressure gradient against it

In the case of a horizontally arched, such as an embankment dam. While arches and vaults are often confused, A vault is defined as an ongoing arch forming a roof.

Option D satisfies the fill-in blanks option.

Hence option D is correct. An arch carries the thrust of weight to its sides with a post-and-lintel.

To learn more about the arch refer to the link;

brainly.com/question/18162421

6 0

3 years ago

An arrow is shot vertically upward at a rate of 250ft/s. Use the projectile formula h=−16t2+v0t to determine at what time(s), in

SIZIF [17.4K]

Answer:

The arrow is at a height of 500 feet at time t = 2.35 seconds.

Explanation:

It is given that,

An arrow is shot vertically upward at a rate of 250 ft/s, v₀ = 250 ft/s

The projectile formula is given by :

$h=-16t^2+v_ot$

We need to find the time(s), in seconds, the arrow is at a height of 500 ft. So,

$-16t^2+250t=500$

On solving the above quadratic equation, we get the value of t as, t = 2.35 seconds

So, the arrow is at a height of 500 feet at time t = 2.35 seconds. Hence, this is the required solution.

6 0

4 years ago

Read 2 more answers

When the forces acting on a particle are resolved into cylindrical components, friction forces always act in the

Aleksandr [31]

Answer:

Tangential

Explanation: This is a kind of force which act on a moving body in such a way that it is curved in the direction of the path of the body. This implies that when the velocity of the object is positive, the acceleration will be negative.

7 0

3 years ago

Radiation from the Sun The intensity of the radiation from the Sun measured on Earth is 1360 W/m2 and frequency is f = 60 MHz. T

Zina [86]

a) Total power output: $3.845\cdot 10^{26} W$

b) The relative percentage change of power output is 1.67%

c) The intensity of the radiation on Mars is $540 W/m^2$

Explanation:

The intensity of electromagnetic radiation is given by

$I=\frac{P}{A}$

where

P is the power output

A is the surface area considered

In this problem, we have

$I=1360 W/m^2$ is the intensity of the solar radiation at the Earth

The area to be considered is area of a sphere of radius

$r=1.5\cdot 10^{11} m$ (distance Earth-Sun)

Therefore

$A=4\pi r^2 = 4 \pi (1.5\cdot 10^{11})^2=2.8\cdot 10^{23}m^2$

And now, using the first equation, we can find the total power output of the Sun:

$P=IA=(1360)(2.8\cdot 10^{23})=3.845\cdot 10^{26} W$

The energy of the solar radiation is directly proportional to its frequency, given the relationship

$E=hf$

where E is the energy, h is the Planck's constant, f is the frequency.

Also, the power output of the Sun is directly proportional to the energy,

$P=\frac{E}{t}$

where t is the time.

This means that the power output is proportional to the frequency:

$P\propto f$

Here the frequency increases by 1 MHz: the original frequency was

$f_0 = 60 MHz$

so the relative percentage change in frequency is

$\frac{\Delta f}{f_0}\cdot 100 = \frac{1}{60}\cdot 100 =1.67\%$

And therefore, the power also increases by 1.67 %.

In this second case, we have to calculate the new power output of the Sun:

$P' = P + \frac{1.67}{100}P =1.167P=1.0167(3.845\cdot 10^{26})=3.910\cdot 10^{26} W$

Now we want to calculate the intensity of the radiation measured on Mars. Mars is 60% farther from the Sun than the Earth, so its distance from the Sun is

$r'=(1+0.60)r=1.60r=1.60(1.5\cdot 10^{11})=2.4\cdot 10^{11}m$