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Mamont248 [21]
3 years ago
11

Astronaut 1 has a mass of 75 kg. Astronaut 2 has a mass of 80 kg. Astro 1 and 2 want to travel to separate planets, but they wan

t to experience the same weight (in N). Astro 1 visits a planet with gravitational acceleration 12 m/s2. What must be Astro 2 planet's ag to equal Astro 1's weight
Physics
1 answer:
vlada-n [284]3 years ago
7 0

Answer:

weight = 900 N

acceleration  = 11.25 m/s²

Explanation:

given data

mass m1 = 75 kg

mass m2 = 80 kg

gravitational acceleration = 12 m/s²

solution

As we know weight of a mass that is

weight of mass = Mass × Acceleration due to gravity .................1

so Astro 1 weight is

weight = 75kg  × 12 m/s²

weight = 900 N

and

so, when Astro 2 needs this much weight the planet on which he is will have the acceleration

acceleration = Weight ÷ Mass of Astro 2        .....................2

acceleration  = 900 ÷ 80 m/s²

acceleration  = 11.25 m/s²

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Answer:

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Two airplanes leave an airport at the same time. The velocity of the first airplane is 730 m/h at a heading of 44.3 ◦ . The velo
Goshia [24]

Answer:

So airplane will be 1324.9453 m apart after 2.9 hour

Explanation:

So if we draw the vectors of a 2d graph we see that the difference in angles is  = 83 - 44.3 = 83-44.3=38.7^{\circ}

Distance traveled by first plane = 730×2.9 = 2117 m

And distance traveled by second plane = 590×2.9 = 1711 m

We represent these distances as two sides of the triangle, and the distance between the planes as the side opposing the angle 38.7.

Using the law of cosine, d^2 representing the distance between the planes, we see that:

d^2=2117^2 + 1711^2 -2\times (2117)\times (1711)cos(38.7)=1755480.2482

d = 1324.9453 m

4 0
3 years ago
The spring is released and a 0.10-kilogram plastic sphere is fired from the launcher. Calculate the maximum speed with which the
tigry1 [53]

Answer:

A) The elastic potential energy stored in the spring when it is compressed 0.10 m is 0.25 J.

B) The maximum speed of the plastic sphere will be 2.2 m/s

Explanation:

Hi there!

I´ve found the complete problem on the web:

<em>A toy launcher that is used to launch small plastic spheres horizontally contains a spring with a spring constant of 50. newtons per meter. The spring is compressed a distance of 0.10 meter when the launcher is ready to launch a plastic sphere.</em>

<em>A) Determine the elastic potential energy stored in the spring when the launcher is ready to launch a plastic sphere.</em>

<em>B) The spring is released and a 0.10-kilogram plastic sphere is fired from the launcher. Calculate the maximum speed with which the plastic sphere will be launched. [Neglect friction.] [Show all work, including the equation and substitution with units.]</em>

<em />

A) The elastic potential energy (EPE) is calculated as follows:

EPE = 1/2 · k · x²

Where:

k = spring constant.

x = compressing distance

EPE = 1/2 · 50 N/m · (0.10 m)²

EPE = 0.25 J

The elastic potential energy stored in the spring when it is compressed 0.10 m is 0.25 J.

B) Since there is no friction, all the stored potential energy will be converted into kinetic energy when the spring is released. The equation of kinetic energy (KE) is the following:

KE = 1/2 · m · v²

Where:

m = mass of the sphere.

v = velocity

The kinetic energy of the sphere will be equal to the initial elastic potential energy:

KE = EPE = 1/2 · m · v²

0.25 J = 1/2 · 0.10 kg · v²

2 · 0.25 J / 0.10 kg = v²

v = 2.2 m/s

The maximum speed of the plastic sphere will be 2.2 m/s

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