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Law Incorporation [45]
4 years ago
14

An 80-kg astronaut becomes separated from his spaceship. He is 15.0 m away from it and at rest relative to it. In an effort to g

et back, he throws a 500 gram object with a speed of 8.0 m/s in a direction away from the ship. How long does it take him to get back to the ship?
Physics
1 answer:
9966 [12]4 years ago
3 0

The astronaut will take 300 seconds

Explanation:

We can solve this problem by using the law of conservation of momentum.

In fact, the total momentum of the astronaut+object system must be conserved.

Initially, they are both at rest, so their total momentum is zero:

p=0

After the astronaut throws the object, their total momentum is:

p=MV+mv

where:

M = 80 kg is the mass of the astronaut

V is the final velocity of the astronaut

m = 500 g = 0.5 kg is the mass of the object

v = 8.0 m/s is the velocity of the object

Since momentum is conserved, we can write

0=MV+mv

And solving for V,

V=-\frac{mv}{M}=-\frac{(0.5)(8.0)}{80}=-0.05 m/s

Which means that he starts moving at 0.05 m/s in the direction opposite to the object.

Now the astronaut needs to cover a distance of

d = 15.0 m

And his speed is

v = 0.05 m/s

Therefore, the time taken is

t=\frac{d}{v}=\frac{15.0}{0.05}=300 s

Learn more about momentum here:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

brainly.com/question/9484203

#LearnwithBrainly

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A projectile is shot from the edge of a cliff above the ground level with initial velocity of at an angle with the horizontal. (
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Answer:

t = √2y/g

Explanation:

This is a projectile launch exercise

a) The vertical velocity in the initial instants (v_{oy} = 0) zero, so let's use the equation

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b) Let's use this time and the horizontal displacement equation, because the constant horizontal velocity

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c) Speeds before touching the ground

     vₓ = vox = constant

     v_{y} = v_{oy} - gt

     v_{y} = 0 - g √2y/g

    v_{y}  = - √2gy

    tan θ = Vy / vx

    θ = tan⁻¹ (vy / vx)

    θ = tan⁻¹ (√2gy / vox)

d) The projectile is higher than the cliff because it is a horizontal launch

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3 years ago
1. A toy car is marked to go at 0.5m/s, which of the following sets of
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Answer:

b. 0.6m/s, 0.7m/s, 0.61m/s, 0.62m/s

Explanation:

Precision of a measurement is the closeness of the experimental values to one another. Hence, experimental measurements are said to be precise if they are close to each other irrespective of how close they are to the accepted value. Precision can be determined by finding the range of each experimental value. The measurement with the LOWEST RANGE represents the MOST PRECISE.

Note: Range is the highest value - lowest value

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Explanation:

I hope it will help you

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Two converging lenses are placed 30 cm apart. The focal length of the lens on the right is 20 cm while the focal length of the l
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Answer:

a)   I2 = 3 (o-10) / (o- 30) , b)   h ’/h=  3 (o-10) / o (o-30)

Explanation:

The builder's equation is

          1 / f = 1 / o + 1 / i

Where f is the focal length, or e i are the distance to the object and image, respectively

As the separation between the lenses is greater than the focal distances, we must work them individually and separately. Let's start with the leftmost lens with focal length f = 15 cm

Let's calculate the position of the image of this lens

         1 / i1 = 1 / f - 1 / o

         1 / i1 = 1/15 - 1 / o

         i1 = o 15 / (o-15)

Let's calculate the distance to the image of the second lens, for this the image of the first is the distance to the object of the second

        o2 = d-i1

We write the builder equation

       1 / f2 = 1 / o2 + 1 / i2

       1 / i2 = 1 / f2 -1 / o2

       1 / i2 = 1 / f2 - 1 / (d-i1)

       1 / i2 = 1/20 - 1 / (d-i1)            (1)

Let's evaluate the last term

      d-i1 = d - 15 o / (o-15)

      d-i1 = (d (o-15) - 15 o) / (o-15)

      d- i1 = (30 or -30 15 -15 o) / (o-15)

      d-i1 = (15 or - 450) / (o- 15)

      d-i1 = = (15 or -450) / (o-15)

replace in 1

      1 / i2 = 1/20 - (or - 15) / (15 or -450)

      1 / i2 = [(15 o-450) - (o-15) 20] / (15 or -150)

      1 / i2 = (15 or - 450 - 20 or + 300) / (15 or - 150)

      1 / i2 = (-5 or -150) / (15 or -150)

      1 / i2 = (or -30) / (3 or - 30)

      I2 = 3 (o-10) / (o- 30)

Part B

The height of the image, we use the magnification equation

     m = h ’/ h = - i / o

     h ’= - h i / o

In our case

     h ’= h i2 / o

     h ’= h 3 (o-10) / o (o-30)

If they give the distance to the object it is easier

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