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zhuklara [117]
3 years ago
6

You are on an interstellar mission from the Earth to the 8,7 light-years distant star Sirius. Your spaceship can travel with 70%

the speed of light and has a cylindrical shape with a diameter of 6 m at the front surface and a length of 25 m. You have to cross the interstellar medium with an approximated density of 1 hydrogen atom/m3 (a) Calculate the time it takes your spaceship to reach Sirius. (b) Determine the mass of interstellar gas that collides with your spaceship during the mission. Note:use 1,673x 10^-27 kg as proton mass
Physics
1 answer:
marishachu [46]3 years ago
7 0

Answer:

Time = 12.43 years

Mass = 4.71.43kg

Explanation:

Given the following :

Distance = 8.7 light years

Speed = 70% the speed of light

Diameter = 6m

Lengtg = 25m

Density = 1 hydrogen atom/m^3

Mass of proton(Me) = 1.673x 10^-27

A.) time it takes your spaceship to reach Sirius :

From the relation: Speed = (distance / time)

Time = distance / speed

Time = (70/100) × 1 light year

Distance = 8.7 light years

Time = 8.7 / 0.7 = 12.4285 years

Time = 12.43 years

B.) Mass of inter-stellar gas that collides with the spaceship can be calcuted by finding the product of the surface area of the cylindrical space ship and the mass of proton.

That is ;

surface area * mass of proton

Surface area of a cylinder = 2πrh + πr^2×Me

= 2πrh + Me×πr^2)

=( 2 × 22/7 × 3 × 25) + (22/7 × 3^2 × 1.673 * 10^-27)

= 2× 235.714 + 28.285) × 1.673 * 10^-27

= 471.428 + 47.31 * 10^-27

= 471.428 + 4.73 × 10^-26

= 4.71.43kg approximately

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Answer:

A Van Allen radiation belt is a zone of energetic charged particles, most of which originate from the solar wind, that are captured by and held around a planet by that planet's magnetosphere. Earth has two such belts, and sometimes others may be temporarily created

6 0
2 years ago
How much force is needed to accelerate a 65 kg rider AND her 215 kg motor scooter 8 m/s?? (treat
irakobra [83]

Answer:

Force = 2240 Newton.

Explanation:

Given the following data;

Mass A= 65kg

Mass B = 215kg

Acceleration = 8m/s²

To find the force;

Force is given by the multiplication of mass and acceleration.

Mathematically, Force is;

F = ma

Where;

F represents force.

m represents the mass of an object.

a represents acceleration.

First of all, we would have to find the total mass.

Total mass = Mass A + Mass B

Total mass = 65 + 215

Total mass = 280kg

Substituting into the equation, we have

Force = 280 * 8

Force = 2240 Newton.

3 0
3 years ago
A uniform horizontal beam 4.0 m long and weighing 200 N is attached to the wall by a pin connection that allows the beam to rota
wariber [46]

Answer:

The magnitude of the tension in the cable, T is 1,064.315 N

Explanation:

Here we have

Length of beam = 4.0 m

Weight = 200 N

Center of mass of uniform beam = mid-span = 2.0 m

Point of attachment of cable = Beam end = 4.0 m

Angle of cable = 53° with the horizontal

Tension in cable = T

Point at which person stands = 1.50 m from wall

Weight of person = 350 N

Therefore,

Taking moment about the wall, we have

∑Clockwise moments = ∑Anticlockwise moments

T×sin(53) = 350×1.5 + 200×2

T = 850/sin(53)  = ‭1,064.315 N.

4 0
3 years ago
As the car launches, which of the following are correct?
jek_recluse [69]
The first option!!!!!!!!!!!!
7 0
3 years ago
An observer stands 150 meters from a fireworks display rocket, which is fired directly upward. When the rocket reaches a height
Nookie1986 [14]

Answer:

-0.288\ \text{rad/s}

Explanation:

x = Distance of observer from the initial location of the rocket = 150 m

y = Vertical displacement of the rocket from the ground = 200 m

r = Distance between observer and rocket

\dfrac{dy}{dt} = Rate of change in height of rocket = 12 m/s

\dfrac{dx}{dt} = Rate of change in x = 0

Distance between observer and rocket at y = 200 m

r=\sqrt{x^2+y^2}\\\Rightarrow r=\sqrt{150^2+200^2}\\\Rightarrow r=250\ \text{m}

\tan\theta=\dfrac{y}{x}

Differentiating with respect to time

\sec^2\theta\dfrac{d\theta}{dt}=\dfrac{y\dfrac{dx}{dt}-x\dfrac{dy}{dt}}{x^2}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{\dfrac{y\dfrac{dx}{dt}-x\dfrac{dy}{dt}}{x^2}}{\sec^2\theta}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{\dfrac{0-150\times 12}{150^2}}{(\dfrac{250}{150})^2}\\\Rightarrow \dfrac{d\theta}{dt}=-0.288\ \text{rad/s}

The rate of change of the angle of elevation is -0.288\ \text{rad/s}.

7 0
3 years ago
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