All categories

3 years ago

Is it possible for a system to have negative potentialenergy?

Physics

1 answer:

sineoko [7]3 years ago

5 0

Answer:

Yes, since the choice of the zero o potential energy is arbitrary.

Explanation:

The kinetic energy is due to the motion of the object. The expression for the kinetic energy is as follows;

$KE=\frac{1}{2}mv^{2}$

Here, m is the mass of the object and v is the velocity of the object.

The kinetic energy can not be negative as the velocity is squared. It can be zero and positive.

Potential energy: It is the energy is due to the position of the object.

The expression for the potential energy is as follows;

PE= mgh

Here, g is the acceleration due to gravity and height.

Height can be taken from the reference point, zero which can be taken below zero and above zero. Zero is taken as origin. Below zero, the height is taken as negative and above zero, the height is taken as positive.

The potential energy can be zero, positive and negative.

The total energy is the sum of kinetic energy and potential energy.

E= KE + PE

Here, KE is the kinetic energy and PE is the potential energy.

Therefore, the option (B) is correct.

You might be interested in

What is resultant force?

Sergio039 [100]

Answer:

In physics and engineering, a resultant force is the single force and associated torque obtained by combining a system of forces and torques acting on a rigid body. The defining feature of a resultant force, or resultant force-torque, is that it has the same effect on the rigid body as the original system of forces.

I hope it's helpful!

4 0

3 years ago

Read 2 more answers

PHYSICS 50 POINTS PLEASE HELP

babymother [125]

Newton’s First Law of Motion - if an object is at rest, it takes un-

balanced forces to make it move. Conversely, if an object is moving

it takes an unbalanced force to make it change it’s direction or speed.

Newton was the first to see that such apparently diverse phenomena as a satellite moving near the Earth's surface and the planets orbiting the Sun operate by the same principle: Force equals mass multiplied by acceleration, or F=ma.

7 0

3 years ago

Read 2 more answers

Determine the moment of inertia Ixx of the mallet about the x-axis. The density of the wooden handle is 860 kg/m3 and that of th

Yuki888 [10]

Complete Question

Diagram for this shown on the first uploaded image

Answer:

The moment of inertia Ixx of the mallet about the x-axis is $I{xx}= 0.119 kg \cdot m^2$

Explanation:

From the question we are told that

The density `of wooden handle is $\rho_w = 860 kg/m^3$

The density `of soft-metal head is $\rho_s =8000kg/m^3$

Generally the mass of the wooden can be mathematically obtained with this formula

$m_w = \rho_w A_w l_w$

Where $A_w$ is mass of wooden handle which is mathematically obtain with the formula

$A_w = \frac{\pi}{4} d^2_w$

Where $d_w$ is the diameter of the wooden handle which from the diagram is

$27mm = \frac{27}{1000} = 0.027m$

So $A_w = \frac{\pi}{4} * 0.027^2$

$l_w$ is the length of the the wooden handle which is given in the diagram as $l_w = 315mm = \frac{315}{1000} = 0.315m$

Substituting these value into the formula for mass

$m_w = 860 * (\frac{\pi}{4} * 0.027^2 ) *0.315$

$= 0.155kg$

Generally the mass of the soft-metal head can be mathematically obtained with this formula

$m_s = \rho_s A_s l_s$

Where $A_s$ is mass of soft-metal head which is mathematically obtain with the formula

$A_s = \frac{\pi}{4} d^2_s$

Where $d_s$ is the diameter of the soft-metal head which from the diagram is

$36mm = \frac{36}{1000} = 0.036m$

So $A_s = \frac{\pi}{4} * 0.036^2$

$l_s$ is the length of the the soft-metal head which is given in the diagram

as $l_s = 90mm = \frac{90}{1000} = 0.090m$

Substituting these value into the formula for mass

$m_s = 8000 * (\frac{\pi}{4} * 0.036^2 ) *0.090$

$=0.733kg$

Generally the mass moment of inertia about x-axis for the wooden handle is

$(I_{xx})_w = [\frac{1}{3}m_w + l_w^2 ]$

Substituting values

$(I_{xx})_w = [\frac{1}{3}*0.155 + 0.315^2 ]$

$=5.12*10^{-3}kg \cdot m^2$

Generally the mass moment of inertia about x-axis for the soft-metal head is

$(I_{xx})_s = [\frac{1}{12}m_s l_s ^2 + b^2]$

Where b is the distance from the centroid to the axis of the head which is mathematically given as

$b=l_w +\frac{d_s}{2}$

Substituting values

$b = 0.315 + \frac{0.036}{2}$

$= 0.336m$

Now substituting values into the formula for mass moment of inertia about x-axis for soft-metal head

$(I_{xx})_s = [\frac{1}{12} *0.733* 0.090^2 + 0.336^2]$

$=0.113 kg \cdot m^2$

Generally the mass moment of inertia about x-axis is mathematically represented as

$I_{xx} = (I_{xx})_w + (I_{xx})_s$

$= [\frac{1}{3}m_w + l_w^2 ] + [\frac{1}{12}m_s l_s ^2 + b^2]$

Substituting values

$I_{xx} = 5.12*10^{-3} +0.113$

$I{xx}= 0.119 kg \cdot m^2$

8 0

3 years ago

Which trait shared by dolphins and bats possibly lead to the evolution of echolocation in these two animal groups? the need to m

Lisa [10]

A trait shared by dolphins and bats that possibly led to the evolution of echolocation in these two animal groups will be the need to move quickly through dark environments.

<h3>What is the evolution of echolocation?</h3>

Our understanding of the evolution of echolocation in bats has shifted as a result of recent molecular phylogenies. These phylogenies imply that bats with advanced echolocation

According to one interpretation of these trees, laryngeal echolocation originated in the ancestor of all living bats. Echolocation may have been lost in Old World fruit bats

The vast adaptive radiation in echolocation call design is substantially controlled by ecology, demonstrating how environmental perceptual problems influence call design.

Hence option A is correct.

To learn more about the evolution of echolocation refer to the link;

brainly.com/question/20789287

3 0

2 years ago

Based on the image below what would an observer on the nighttime side of earth observe?

sertanlavr [38]

I would love to help; where’s the picture?

3 0

3 years ago