All categories

4 years ago

When gases, liquids, or solids are in contact with a moving object, the flow of _____ occurs due to frictional forces

Physics

2 answers:

mariarad [96]4 years ago

7 0

When gases, fluids, or other solids are in contact with a moving object

heat is produced due to friction.

Triss [41]4 years ago

5 0

When gases, liquids, or solids are in contact with a moving object, the flow of _____ occurs due to frictional forces.

The answer is HEAT

This is becuase when objects flow against friction energy is rubbed with friction causing surfaces to heat from the molecules rubbing.

You might be interested in

Help!! plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz!!ASAP!!!

alekssr [168]

That is a closed switch

3 0

3 years ago

Read 2 more answers

A small rock is launched straight upward from the surface of a planet with no atmosphere. The initial speed of the rock is twice

Scorpion4ik [409]

If gravitational effects from other objects are negligible, the speed of the rock at a very great distance from the planet will approach a value of $\sqrt{3} v_{e}$

<u>Explanation:</u>

To express velocity which is too far from the planet and escape velocity by using the energy conservation, we get

Rock’s initial velocity , $v_{i}=2 v_{e}$ . Here the radius is R, so find the escape velocity as follows,

$\frac{1}{2} m v_{e}^{2}-\frac{G M m}{R}=0$

$\frac{1}{2} m v_{e}^{2}=\frac{G M m}{R}$

$v_{e}^{2}=\frac{2 G M}{R}$

$v_{e}=\sqrt{\frac{2 G M}{R}}$

Where, M = Planet’s mass and G = constant.

From given conditions,

Surface potential energy can be expressed as, $U_{i}=-\frac{G M m}{R}$

R tend to infinity when far away from the planet, so $v_{f}=0$

Then, kinetic energy at initial would be,

$k_{i}=\frac{1}{2} m v_{i}^{2}=\frac{1}{2} m\left(2 v_{e}\right)^{2}$

Similarly, kinetic energy at final would be,

$k_{f}=\frac{1}{2} m v_{f}^{2}$

Here, $v_{f}=\text { final velocity }$

Now, adding potential and kinetic energies of initial and final and equating as below, find the final velocity as

$U_{i}+k_{i}=k_{f}+v_{f}$

$\frac{1}{2} m\left(2 v_{e}\right)^{2}-\frac{G M m}{R}=\frac{1}{2} m v_{f}^{2}+0$

$\frac{1}{2} m\left(2 v_{e}\right)^{2}-\frac{G M m}{R}=\frac{1}{2} m v_{f}^{2}$

'm' and $\frac{1}{2}$ as common on both sides, so gets cancelled, we get as

$4\left(v_{e}\right)^{2}-\frac{2 G M}{R}=v_{f}^{2}$

We know, $v_{e}=\sqrt{\frac{2 G M}{R}}$ , it can be wriiten as $\left(v_{e}\right)^{2}=\frac{2 G M}{R}$ , we get

$4\left(v_{e}\right)^{2}-\left(v_{e}\right)^{2}=v_{f}^{2}$

$v_{f}^{2}=3\left(v_{e}\right)^{2}$

Taking squares out, we get,

$v_{f}=\sqrt{3} v_{e}$

4 0

3 years ago

A physics student with too much free time drops a watermelon from a roof of a building, hears the sound of the watermelon going

tatiyna

Answer:

28.6260196842 m

Explanation:

Let h be the height of the building

t = Time taken by the watermelon to fall to the ground

Time taken to hear the sound is 2.5 seconds

Time taken by the sound to travel the height of the cliff = 2.5-t

Speed of sound in air = 340 m/s

For the watermelon falling

$s=ut+\frac{1}{2}at^2\\\Rightarrow h=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow h=\frac{1}{2}\times 9.81\times t^2$

For the sound

Distance = Speed × Time

$\text{Distance}=340\times (2.5-t)$

Here, distance traveled by the stone and sound is equal

$\frac{1}{2}\times 9.81\times t^2=340\times (2.5-t)\\\Rightarrow 4.905t^2=340\times (2.5-t)\\\Rightarrow t^2=\frac{340}{4.905}(2.5-t)\\\Rightarrow t^2+69.3170234455t-173.292558614=0$

$t=\frac{-69.31702\dots +\sqrt{69.31702\dots ^2-4\cdot \:1\cdot \left(-173.29255\dots \right)}}{2\cdot \:1},\:t=\frac{-69.31702\dots -\sqrt{69.31702\dots ^2-4\cdot \:1\cdot \left(-173.29255\dots \right)}}{2\cdot \:1}\\\Rightarrow t=2.4158\ s\ or\ -71\ seconds$

The time taken to fall down is 2.4158 seconds

$h=\frac{1}{2}\times 9.81\times 2.4158^2=28.6260196842\ m$

Height of the buidling is 28.6260196842 m

7 0

4 years ago

The work that is done when twice the load is lifted twice the distance is _______. View Available Hint(s)for Part A The work tha

anygoal [31]

The work that is done when twice the load is lifted twice the distance is

four times as much

The net work performed by forces acting on an object equals the change in kinetic energy, according to the work-energy theorem.

when an item slows down, the net work applied to it decreases, its change in kinetic energy is negative, and its ultimate kinetic energy is less than its starting kinetic energy. When an item accelerates, positive net work is done on it. All the forces acting on an item must be taken into consideration when determining the net work. You will obtain an incorrect result if you exclude any forces that affect an item or if you add any forces that do not affect it.

Hence The work that is done when twice the load is lifted twice the distance is four times as much

Learn more about Work here

brainly.com/question/25573309

#SPJ4

3 0

2 years ago

How does wind affect precipitation

patriot [66]

<span>Winds can greatly affect the amount of precipitation an area receives depending on the amount of moisture they are carrying. These prevailing winds can move air masses from the ocean onto a continent bringing moisture onto the continent. ... Prevailing winds affect the climate of an area.

(From Google)</span>

4 0

3 years ago