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navik [9.2K]
3 years ago
11

Occupants in a single space shuttle in orbit feel weightless. describe a scheme whereby occupants in a pair of shuttles (or even

one shuttle and another massive object in orbit) would be able to use a long cable to continuously experience a comfortably normal earthlike gravity.
Physics
1 answer:
agasfer [191]3 years ago
8 0
The   scheme  whereby occupants  in  a  pair  of  shuttles is  as  follows
use  a  strong  cable  with  large  weight  on  the  end
Then  use  the  orbital  naneuvering   system(OMS)  to   set  the   whole  work  as  spinning   about   their  common  center of  gravity.
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After two hours, 1/16 of the initial amount of a certain radioactive isotope remains undecayed. The half-life of the isotope is:
AlekseyPX
The half life is 30 minutes.
30 mins- 1/2 left
60 mins- 1/4 left
90 mins- 1/8 left
120 mins- 1/16 left
120 mins= 2 hours
7 0
3 years ago
Sayid made a chart listing data of two colliding objects. A 5-column table titled Collision: Two Objects Stick Together with 2 r
Alborosie

Answer:

6 m/s is the missing final velocity

Explanation:

From the data table we extract that there were two objects (X and Y) that underwent an inelastic collision, moving together after the collision as a new object with mass equal the addition of the two original masses, and a new velocity which is the unknown in the problem).

Object X had a mass of 300 kg, while object Y had a mass of 100 kg.

Object's X initial velocity was positive (let's imagine it on a horizontal axis pointing to the right) of 10 m/s. Object Y had a negative velocity (imagine it as pointing to the left on the horizontal axis) of -6 m/s.

We can solve for the unknown, using conservation of momentum in the collision: Initial total momentum = Final total momentum (where momentum is defined as the product of the mass of the object times its velocity.

In numbers, and calling P_{xi} the initial momentum of object X and P_{yi} the initial momentum of object Y, we can derive the total initial momentum of the system: P_{total}_i=P_{xi}+P_{yi}= 300*10 \frac{kg*m}{s} -100*6\frac{kg*m}{s} =\\=(3000-600 )\frac{kg*m}{s} =2400 \frac{kg*m}{s}

Since in the collision there is conservation of the total momentum, this initial quantity should equal the quantity for the final mometum of the stack together system (that has a total mass of 400 kg):

Final momentum of the system: M * v_f=400kg * v_f

We then set the equality of the momenta (total initial equals final) and proceed to solve the equation for the unknown(final velocity of the system):

2400 \frac{kg*m}{s} =400kg*v_f\\\frac{2400}{400} \frac{m}{s} =v_f\\v_f=6 \frac{m}{s}

7 0
3 years ago
Read 2 more answers
What is the atomic weight of a penny?
Gnom [1K]
I actually know the answer to this one, you use pennies to find the atomic weight of a penny, it really doesn't have a weight. LOL

4 0
4 years ago
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With respect to the earth, object 1 is moving at speed 0.80 c to the right. Object 2 is moving in the same direction at speed 0.
viktelen [127]

Answer:

0.976 c

Explanation:

v_{1e} = velocity of object 1 relative to earth = 0.80 c

v_{21} = velocity of object 2 relative to object 1 = 0.80 c

v_{2e} = velocity of object 2 relative to earth

Velocity of object 2 relative to earth is given as

v_{2e}= \frac{v_{1e} + v_{21}}{1 + \frac{v_{1e}v_{21}}{c^{2}}}

v_{2e}= \frac{0.80 c + 0.80 c}{1 + \frac{(0.80c)(0.80c)}{c^{2}}}

v_{2e} = 0.976 c

6 0
3 years ago
A group of students collected the data shown below while attempting to measure the coefficient of static friction (of course, it
anzhelika [568]

Answer:

0.130

Explanation:

From the given data, the coefficient of static friction for each trial are:

1. 0.053

2. 0.081

3. 0.118

4. 0.149

5. 0.180

6. 0.198

The sum of the coefficient of static friction = 0.053 + 0.081 + 0.118 + 0.149 + 0.180 + 0.198

                                              = 0.779

So that;

the average coefficient of static friction = \frac{sum of coefficient of static friction}{number of trials}

                                              = \frac{0.779}{6}

                                              = 0.12983

The average coefficient of static friction is 0.130

8 0
3 years ago
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