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dexar [7]
3 years ago
14

An astronaut out on a spacewalk to construct a new section of the International Space Station walks with a constant velocity of

2.30 m/s on a flat sheet of metal placed on a flat, frictionless, horizontal honeycomb surface linking the two parts of the station. The mass of the astronaut is 71.0 kg, and the mass of the sheet of metal is 230 kg. (Assume that the given velocity is relative to the flat sheet.)
Required:
a. What is the velocity of the metal sheet relative to the honeycomb surface?
b. What is the speed of the astronaut relative to the honeycomb surface?
Physics
1 answer:
Fittoniya [83]3 years ago
4 0

Answer:

Explanation:

Let the velocity of astronaut be u and the velocity of flat sheet of metal plate be v . They will move in opposite direction ,  so their relative velocity

= u + v = 2.3 m /s ( given )

We shall apply conservation of momentum law for the movement of astronaut and metal plate

mu  = M v where m is mass of astronaut , M is mass of metal plate

71 u = 230 x v

71 ( 2.3 - v ) = 230 v

163.3 = 301 v

v = .54 m / s

u = 1.76 m / s

honeycomb will be at rest  because honeycomb surface  is frictionless . Plate will slip over it . Over plate astronaut is walking .

a ) velocity of metal sheet relative to honeycomb will be - 1.76 m /s

b ) velocity of astronaut relative to honeycomb will be + .54 m /s

Here + ve direction is assumed to be the direction of astronaut .  

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miss Akunina [59]

5.3 km

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Let us denote each of the individual displacements by a vector. Consider the unit vectors as the unit vectors in the direction of East and North respectively.

By simple calculations, we can derive the unit vectors j,\frac{-i-j}{2} and in the directions North,  South of West, and  North of West respectively.

So Total displacement vector = The Sum of individual displacement vectors.

Displacement vector = -4.25i + 3.165j

The magnitude of Displacement =| -4.25i + 3.165j |  = 5.3km

∴ Total displacement = 5.3km

A displacement is a vector in geometry and mechanics that have a length equal to the shortest distance between a point P's initial and final positions.  It measures the length and angle of the net motion, or total motion, in a straight line from the starting point to the destination of the point trajectory. The translation that links the starting point and the ending point can be used to spot a displacement.

The final location xf of a point relative to its beginning position xi, or a relative position (derived from the motion), is another way to express a displacement. The difference between the final and initial coordinates can be used to define the equivalent displacement vector.

To  learn more about displacement from the given link:

brainly.com/question/11934397

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4 0
2 years ago
A space station with a radius of 120 m rotates once every 70 s to create artificial gravity. If the astronaut has an earth weigh
Alina [70]

Answer:

Artificial weight = 70.27 N = 15.80 lbs

Explanation:

The earth weight of the astronaut = 160 lbs = 711.72 N

The weight on earth = m × g(earth)

g(earth) = 9.8 m/s²

711.72 = m × 9.8

m = (711.72/9.8)

m = 72.62 kg

But at the space station, the space station rotates once every 70 s to create an artificial radial acceleration that creates a radial gravity pulling the objects on the space station towards the centre of that space station.

radial acceleration = α = (v²/r)

v = rw,

α = (rw)²/r

α = rw²

r = radius of rotation = 120 m

w = angular velocity = (2π/70) (it completes 1 rotation, 2π radians, in 70 s)

w = 0.0898 rad/s

α = 120 × (0.0898²)

α = 0.968 m/s²

Artificial weight = (mass of astronaut) × (Radial acceleration) = 72.62 × 0.968

Artificial weight = 70.27 N = 15.80 lbs

Hope this Helps!!!

5 0
4 years ago
While eating lunch high up in a skyscraper, two construction workers calculate their gravitational potential
Maurinko [17]

Answer:

The mass of the other worker is 45 kg

Explanation:

The given parameters are;

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The mass of the lighter construction worker, m₁ = 90 kg

The height level of the lighter construction worker's location = h₁

The height level of the other construction worker's location = h₂ = 2·h₁

The gravitational potential energy, P.E.,  is given as follows;

P.E. = m·g·h

Where;

m = The mass of the object at height

g = The acceleration due to gravity

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Let P.E.₁ represent the gravitational potential energy of one construction worker and let P.E.₂ represent the gravitational potential energy of the other construction worker

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m₁·g·h₁ = m₂·g·2·h₁

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90 kg = 2 × m₂

m₂ = (90 kg)/2 = 45 kg

The mass of the other construction worker is 45 kg.

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Answer: Kinematics

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