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3 years ago

Please help me with a speech.The topic of my speech is Passion.

Physics

1 answer:

Vikentia [17]3 years ago

3 0

Answer:

Explanation:

Passion

For me, standing on the summit of Mt Everest was the result of following a process. The process of mountaineering. I love mountaineering. I am passionate about it. I love the months of planning for an expedition, the months of sweating and training to prepare my body physically. The meticulous preparation of my equipment. Most of all I love the huge mental challenge I have to overcome before each climb to confront my own fear. All these reasons are why I climb, they are why I climbed Mt Everest and that is why I continue to climb.

Passion is an enormously powerful force. It gives us the strength to get through hard times and setbacks. It gives us strength to overcome our fears, to ignore what other people think of us, to be disciplined and make sacrifices in pursuit of our dreams. Passionate people do not want to take shortcuts – they consider that ‘learning the process’ is an important part of the journey.

In mountaineering it’s easy to spot those who are not passionate about the process. They want to stand on top of the mountain but they are not really interested in the process of climbing the mountain. I feel for these people. Success without hard work is a hollow, empty feeling. They never last long in the sport.

Just as in life, successful mountaineers are the ones who are passionate. They are not there just to stand on the summit. Their passion gives them the energy to work the hardest, fight the longest, and in the words of Winston Churchill “never, never. never give-up”.

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Car B has the same mass as Car A but is going twice as fast. If the same constant force brings both cars to a stop, how far will

evablogger [386]

Answer:four times

Explanation:

Given

mass of both cars A and B are same suppose m

but velocity of car B is same as of car A

Suppose velocity of car A is u

Velocity of car B is 2 u

A constant force is applied on both the cars such that they come to rest by travelling certain distance

using to find the distance traveled

where, v=final velocity

u=initial velocity

a=acceleration(offered by force)

s=displacement

final velocity is zero

For car A

$0-(u)^2=2\times a\times s$

$s_a=\frac{u^2}{2a}------1$

For car B

$0-(2u)^2=2\times a\times s_b$

$s_b=\frac{4u^2}{2a}----2$

divide 1 and 2 we get

$\frac{s_a}{s_b}=\frac{1}{4}$

thus $s_b=4\cdot s_a$

distance traveled by car B is four time of car A

7 0

3 years ago

Two Earth satellites, A and B, each of mass m = 940 kg , are launched into circular orbits around the Earth's center. Satellite

-Dominant- [34]

Answer:

The required work done is $6.5\times10^{9}\ J$

Explanation:

Given that,

Mass of each satellites = 940 kg

Altitude of A = 4500 km

Altitude of B = 11100 km

We need to calculate the potential energy

Using formula of potential

$U_{A}=-\dfrac{Gm_{A}m_{E}}{r_{A}}$

Put the value into the formula

$U_{A}=-\dfrac{6.67\times10^{-11}\times940\times5.98\times10^{24}}{6.38\times10^{6}+4.50\times10^{6}}$

$U_{A}=-3.44\times10^{10}\ J$

We need to calculate the potential energy

Using formula of potential

$U_{B}=-\dfrac{Gm_{B}m_{E}}{r_{A}}$

Put the value into the formula

$U_{B}=-\dfrac{6.67\times10^{-11}\times940\times5.98\times10^{24}}{6.38\times10^{6}+11.10\times10^{6}}$

$U_{B}=-2.14\times10^{10}\ J$

We need to calculate the value of $k_{A}$

Using formula of $k_{A}$

$k_{A}=-\dfrac{1}{2}U_{A}$

Put the value into the formula

$k_{A}=\dfrac{1}{2}\times3.44\times10^{10}$

$k_{A}=1.72\times10^{10}\ J$

We need to calculate the value of $k_{B}$

Using formula of $k_{B}$

$k_{B}=-\dfrac{1}{2}U_{B}$

Put the value into the formula

$k_{B}=\dfrac{1}{2}\times2.14\times10^{10}$

$k_{B}=1.07\times10^{10}\ J$

We need to calculate the work done

Using formula of work done

$W=\Delta K+\Delta U$

$W=(k_{B}-k_{A})+(U_{B}-U_{A})$

$W=(-\dfrac{U_{B}}{2}+\dfrac{U_{A}}{2})+(U_{B}-U_{A})$

$W=\dfrac{1}{2}(U_{B}-U_{A})$

Put the value into the formula

$W=\dfrac{1}{2}\times(-2.14\times10^{10}+3.44\times10^{10})$

$W=6.5\times10^{9}\ J$

Hence, The required work done is $6.5\times10^{9}\ J$

7 0

3 years ago

Do you think copper would have oxidized of it was completly submerged in vinigar

cluponka [151]

yes if it was submerged long enough

5 0

4 years ago

Why does the fed rarely change the reserve requirement

Brut [27]

The answer is:

It can be disruptive to the whole banking system.

8 0

3 years ago

Read 2 more answers

Contrast situations where work is done with different amounts of force to situations where no work is done such as standing stil

uranmaximum [27]

Picking up a sheet of paper . . . work done with small force
Picking up a glass of water . . . work done with moderate force
Picking up a huge boulder . . . work done with a great tremendous force
=================================
Standing still . . .
Holding your tongue out as far as it will go . . .
Holding your arm over your head for 3 days . . .
Holding a huge boulder motionless over your head . . .
Pushing on a brick wall . . .
Pushing as hard as you can against a truck with the wheels locked . . .
. . . . . No work done at all, because the force doesn't move through a distance.

<u>Work done = (force) times (distance)</u>

If the force doesn't move, then the distance is zero, and the work done is zero.

5 0

3 years ago

Please help me with a speech.The topic of my speech is Passion.​

Please help me with a speech.The topic of my speech is Passion.