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3 years ago

Astronauts appear weightless while working in the International Space Station because

Physics

1 answer:

Gnesinka [82]3 years ago

5 0

A) they are in a state of free fall.

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Hi, does anyone know the answer for question 2 or 3? Thank you

Anna007 [38]

Answer:

Explanation:

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3 years ago

A 1.10-kg object slides to the right on a surface having a coefficient of kinetic friction 0.250 (Figure a). The object has a sp

cluponka [151]

Answer:

(a) Approximately $0.335\; \rm m$ .

(b) Approximately $1.86\; \rm m\cdot s^{-1}$ .

(c) Approximately $0.707\; \rm m$ .

(d) Approximately $0.228\; \rm m$ .

Explanation:

$v_i$ denotes the velocity of the object in the first diagram right before it came into contact with the spring.
Let $m$ denote the mass of the block.
Let $\mu$ denote the constant of kinetic friction between the object and the surface.
Let $g$ denote the constant of gravitational acceleration.
Let $k$ denote the spring constant of this spring.

Consider the conversion of energy in this object-spring system.

First diagram: Right before the object came into contact with the spring, the object carries kinetic energy $\displaystyle \frac{1}{2}\, m \cdot {v_{i}}^2$ .

Second diagram: As the object moves towards the position in the third diagram, the spring gains elastic potential energy. At the same time, the object loses energy due to friction.

Third diagram: After the velocity of the object becomes zero, it has moved a distance of $D$ and compressed the spring by the same distance.

Energy lost to friction: $\underbrace{(\mu \cdot m \cdot g)}_{\text{friction}} \cdot D$ .
Elastic potential energy that the spring has gained: $\displaystyle \frac{1}{2}\,k\, D^2$ .

The sum of these two energies should match the initial kinetic energy of the object (before it comes into contact with the spring.) That is:

$\displaystyle \frac{1}{2}\, m \cdot {v_{i}}^{2} = (\mu\cdot m \cdot g) \cdot D + \frac{1}{2}\, k \cdot D^2$ .

Assume that $g = 9.81\; \rm m \cdot s^{-2}$ . In the equation above, all symbols other than $D$ have known values:

$m =1.10\; \rm kg$ .
$v_i = 2.60\; \rm m \cdot s^{-1}$ .
$\mu = 0.250$ .
$g = 9.81\; \rm m \cdot s^{-2}$ .
$k = 50.0\; \rm N \cdot m^{-1}$ .

Substitute in the known values to obtain an equation for $D$ (where the unit of $D\!$ is $m$ .)

$3.178 = 2.69775\, D + 25\, D^2$ .

$2.69775\, D + 25\, D^2 + 3.178 = 0$ .

Simplify and solve for $D$ . Note that $D > 0$ because the energy lost to friction should be greater than zero.

$D \approx 0.335\; \rm m$ .

The energy of the object-spring system in the third diagram is the same as the elastic potential energy of the spring:

$\displaystyle \frac{1}{2}\,k\, D^2 \approx 2.81\; \rm J$ .

As the object moves to the left, part of that energy will be lost to friction:

$(\mu \cdot m \cdot g) \, D \approx 0.905\; \rm J$ .

The rest will become the kinetic energy of that block by the time the block reaches the position in the fourth diagram:

$2.81\; \rm J - 0.905\; \rm J \approx 1.91\; \rm J$ .

Calculate the velocity corresponding to that kinetic energy:

$\displaystyle v =\sqrt{\frac{2\, (\text{Kinetic Energy})}{m}} \approx 1.86\; \rm m \cdot s^{-1}$ .

As the object moves from the position in the fourth diagram to the position in the fifth, all its kinetic energy ( $1.91\; \rm J$ ) would be lost to friction.

How far would the object need to move on the surface to lose that much energy to friction? Again, the size of the friction force is $\mu \cdot m \cdot g$ .

$\displaystyle (\text{Distance Travelled}) = \frac{\text{(Work Done by friction)}}{\text{(Size of the Friction Force)}} \approx0.707\; \rm m$ .

Similar to (a), solving (d) involves another quadratic equation about $D$ .

Left-hand side of the equation: kinetic energy of the object (as in the fourth diagram,) $1.91\; \rm J$ .

Right-hand side of the equation: energy lost to friction, plus the gain in the elastic potential energy of the spring.

$\displaystyle {1.91\; \rm J} \approx (\mu\cdot m \cdot g) \cdot D + \frac{1}{2}\, k \cdot D^2$ .

$25\, D^2 + 2.69775\, D - 1.90811\approx 0$ .

Again, $D > 0$ because the energy lost to friction is greater than zero.

$D \approx 0.228\; \rm m$ .

7 0

3 years ago

Question is in the pic

Anton [14]

Answer: The answer is A

4 0

3 years ago

Read 2 more answers

LOOK AT THE PICTURE!! 3 <br> Please answer properly

babunello [35]

The total energy at A, B and C is constant and does not change.

At A, all the energy is potential energy. It gets converted partially to kinetic energy at B and is completely converted to kinetic energy at C. At C, all the energy is kinetic energy.

4 0

3 years ago

Read 2 more answers

List between six and eight student organizations either offered by your school or offered

melamori03 [73]

Examples of student-led organizations are:

Academic and educational organizations

Community service organization

Media and publications organizations

Political or multicultural organizations

Recreation and sports organizations

Student government organizations

Religious and Spiritual organizations

The benefits of getting involved in any of these are many. They include but are not limited to:

It helps one to learn more about oneself

It is a great place to develop leadership skills

It offers the opportunity for people to build life-long networks

Skills learned in class can be practiced and honed in these organizations

Soft skills such as team-intellignce, and social intelligence can be learned in these organizations

Valuable experiences that count in real-life jobs can be learned here

It is also an opportunity to give back to the community and to have fun

Learn more about student organizations in the link below:

8 0

2 years ago