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aliina [53]
3 years ago
8

Help me with theses 2 questions please.

Physics
1 answer:
spin [16.1K]3 years ago
4 0

1)

Answer: A inclined plane and staircase reaching the same height. Work depends on initial and final position only.

Explanation:

The amount of work done does not depend on path length. It depends on initial and final position. It is given by gravitational potential energy.

W = m g h

Where, m is the mass of the object, g is the acceleration due to gravity and h is the height.

Let the mass of the object be m. Two different paths ( one inclines plane and another staircase) are taken to same point. The object would gain same gravitational potential energy because it depends on the height through which the object is taken and not the actual path length.

W ∝ h.

2)

Spring constant can be defined as the ratio of amount of force acting spring  with the displacement caused due to the force. A deformed elastic object has elastic potential energy stored in it which is equal to the work done to deform the object. It is proportional to the spring constant and amount of stretched distance.

Equilibrium is reached when the net force acting on the spring is zero. In equilibrium state, the object has natural extension.  An object is said to possess elastic potential energy  when force is applied and object deforms relative to original equilibrium shape or length.




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A waterbed has a force of 1300N on the floor. It exerts 347 Pa of pressure. What is the area of the waterbed?
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The orbital motion of Earth around the Sun leads to an observable parallax effect on the nearest stars. For each star listed, ca
Murrr4er [49]

Answer:

Following are the answer to this question:

Explanation:

Formula:

D(PC) =\frac{1}{parallax}\\\\D(av)=D(PC) \times 20.626\ J

Calculating point A:

when the value is 0.38

\to 0.38 \toD(PC)= \frac{1}{0.38}\\\\

                   =2.632

\to D(a.v) = \frac{1}{0.38} \times 206265\\

               =542,802.6

Calculating point B:

when the value is 0.75

\to D(PC)=\frac{1}{0.75}

                =1.33

\to D(a.v) = \frac{1}{0.75} \times 206265\\

             =275,020

Calculating point C:

when the value is 0.28

\to D(PC)=\frac{1}{0.28}

                =3.571

\to D(a.v) = \frac{1}{0.28} \times 206265\\

               =736660.7

Calculating point D:

when the value is 0.42

\to D(PC)=\frac{1}{0.42}

                =2.38

\to D(a.v) = \frac{1}{0.42} \times 206265\\

               =490910.7

Calculating point E:

when the value is 0.31

\to D(PC)=\frac{1}{0.31}

                =3.226

\to D(a.v) = \frac{1}{0.31} \times 206265\\

               =665370.97

6 0
3 years ago
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