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3 years ago

I). Mechanical energy is the sum of potential energy and kinetic energy in an object that is used to do work.

Physics

1 answer:

Dmitry_Shevchenko [17]3 years ago

4 0

Answer:

false statement : b ) For the motion of a cart on an incline plane having a coefficient of kinetic friction of 0.5, the magnitude of the change in kinetic energy equals the magnitude of the change in gravitational potential energy

Explanation:

mechanical energy = potential energy + kinetic energy = constant

differentiating both side

Δ potential energy + Δ kinetic energy = 0

Δ potential energy = - Δ kinetic energy

first statement is true.

Friction is a non conservative force so inter-conversion of potential and kinetic energy is not possible in that case. In case of second option, the correct relation is as follows

change in gravitational potential energy = change in kinetic energy + work done against friction .

So given 2 nd option is incorrect.

In case of no change in gravitational energy , work done is equal to

change in kinetic energy.

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1. A DC-10 jumbo jet maintains an airspeed of 550 mph in a southwesterly direction. The velocity of the jet stream is a constant

Vladimir79 [104]

Answer:

The magnitude of actual velocity is <u>496.67 mph</u> and its direction is <u>51.54° with the x axis in the third quadrant</u>.

Explanation:

Given:

Speed of jumbo jet in southwesterly direction $(v_j)$ = 550 mph

Velocity of jet stream from west to east direction $(v_s)=80\ mph$

First let us draw a vectorial representation of the above velocity vectors.

Consider the south direction as negative y axis and west direction as negative x axis.

From the diagram,

The velocity of the jet can be represented as:

$\vec{v_j}=-550\cos(45)\vec{i}+(-550\sin(45)\vec{j} )\\\\\vec{v_j}=-388.91\vec{i}-388.91\vec{j}\ mph$

Similarly, the velocity of the stream is, $\vec{v_s}=80\vec{i}$

Now, the vector sum of the above two vectors gives the actual velocity of the aircraft. So, the resultant velocity is given as:

$\vec{v}=\vec{v_j}+\vec{v_s}\\\\\vec{v}=-388.91\vec{i}-388.91\vec{j}+80\vec{i}\\\\\vec{v}=(-388.91+80)\vec{i}-388.91\vec{j}\\\\\vec{v}=(-308.91)\vec{i}-388.91\vec{j}$

Now, magnitude is given as the square root of sum of the squares of the 'i' and 'j' components. So,

$|\vec{v}|=\sqrt{(-308.91)^2+(-388.91)^2}\\\\|\vec{v}|=496.67\ mph$

As the horizontal and vertical components of actual velocity negative, the resultant vector makes an angle $\theta$ with the x axis in the third quadrant.

The direction is given as:

$\theta=\tan^{-1}(\frac{v_y}{v_x})\\\\\theta=\tan^{-1}(\frac{-388.91}{-308.91})\\\\\theta=51.54\°(Third\ quadrant)$

Therefore, the magnitude of actual velocity is 496.67 mph and its direction is 51.54° with the x axis in the third quadrant.

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3 years ago

How does an environmental change (ex. hurricane, drought, etc.) affect the organisms that are adapted to the environment before

rjkz [21]

The environmental changes can will negatively affect different organisms differently, for example some organisms might die due to failure of the organisms to adapt to the new changes. Some organisms will be forced to migrate in order to counter these changes, this may lead to this organisms leaving their old living routine and adapt to the new methods of living, this may lead to evolution caused by new environment.

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3 years ago

Find the magnitude of the free-fall acceleration at the orbit of the Moon (a distance of 60RE from the center of the Earth with

Ede4ka [16]

Answer:

The magnitude of the free-fall acceleration at the orbit of the Moon is $2.728\times 10^{-3}\,\frac{m}{s^{2}}$ ( $\frac{2.784}{10000}\cdot g$ , where $g = 9.8\,\frac{m}{s^{2}}$ ).

Explanation:

According to the Newton's Law of Gravitation, free fall acceleration ( $g$ ), in meters per square second, is directly proportional to the mass of the Earth ( $M$ ), in kilograms, and inversely proportional to the distance from the center of the Earth ( $r$ ), in meters:

$g = \frac{G\cdot M}{r^{2}}$ (1)

Where:

$G$ - Gravitational constant, in cubic meters per kilogram-square second.

$M$ - Mass of the Earth, in kilograms.

$r$ - Distance from the center of the Earth, in meters.

If we know that $G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}$ , $M = 5.972\times 10^{24}\,kg$ and $r = 382.26\times 10^{6}\,m$ , then the free-fall acceleration at the orbit of the Moon is: