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2 years ago

Which one of the following bodies has only potential energy?

1 answer:

6 0

Potential energy is like energy contained. It is not energy displayed.

So option (a) clearly represents energy contained.

a. A cat sitting on the fence

If the cat jumps off the wall, the potential or contained energy is displayed as kinetic energy; which is energy in motion.

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A coin released at rest from the top of a tower hits the ground after falling 1.5 s. What is the speed of the coin as it hits th

Inessa05 [86]

initially coin is at rest and then it drop for total time t = 1.5 s

so here the speed of the coin at which it will hit the floor is to be find

$v_f = v_i + at$

here we know that

$v_i = 0$

a = 9.8 m/s^2

t = 1.5 s

now from above equation

$v_f = 0 + 1.5 (9.8)$

$v_f = 14.7 m/s$

so it will hit the floor with speed 14.7 m/s

3 0

3 years ago

A speed boat increases its speed uniformly from vi = 20.0 m/s to vf = 30.0 m/s in a distance of 2.00 x 10^2m. (a) Draw a coordin

pychu [463]

a) See graph in attachment

b) The suvat equation to use is $v_f^2 - v_i^2 = 2as$

c) The acceleration is $a=\frac{v_f^2-v_i^2}{2s}$

d) The acceleration is $1.25 m/s^2$

e) The time needed is 8 s

Explanation:

For this part, find in attachment the diagram representing this situation.

Since we are not given any particular direction for the motion, we choose the x-direction as the direction of motion of the boat.

Then we have the following:

- The initial position of the boat is $x_i = 0$ , the origin

- The final position of the boat is $x_f = 200 m$

- The initial velocity of the boat is $v_i = 20.0 m/s$

- The final velocity of the boat is $v_f = 30.0 m/s$

Note that the arrow representing the final velocity is longer than that of the initial velocity, since the final velocity is larger.

The motion of the speed boat is a uniformly accelerated motion (motion at constant acceleration), therefore we can use one of the suvat equations. In this particular problem, we know the following quantities:

$v_i = 20.0 m/s$ , the initial velocity

$v_f = 30.0 m/s$ , the final velocity

$s = x_f - x_i = 200 m$ , the displacement of the boat

Therefore, the equation that best can be use to find the acceleration is

$v_f^2 - v_i^2 = 2as$

where

$a$ is the acceleration

Now we have to solve the equation

$v_f^2 - v_i^2 = 2as$

In order to find the acceleration.

This can be done by dividing both terms by $2s$ : this way, we find

$\frac{v_f^2-v_i^2}{2s}=\frac{2as}{2s}$

And so the acceleration is

$a=\frac{v_f^2-v_i^2}{2s}$

Now we can use the equation found in part c) in order to find the acceleration.

We have the following data:

$v_i = 20.0 m/s$ , the initial velocity

$v_f = 30.0 m/s$ , the final velocity

$s = x_f - x_i = 200 m$ , the displacement of the boat

And substituting into the equation,

$a=\frac{30^2-20^2}{2(200)}=1.25 m/s^2$

In order to find the time it takes the boat to travel the given distance, we can use the following suvat equation:

$v_f = v_i + at$

where:

$v_i$ is the initial velocity

$v_f$ is the final velocity

a is the acceleration

t is the time

Here we have:

$v_i = 20.0 m/s$

$v_f = 30.0 m/s$

$a=1.25 m/s^2$

Solving for t, we find:

$t=\frac{v_f-v_i}{a}=\frac{30-20}{1.25}=8 s$

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

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brainly.com/question/2562700

#LearnwithBrainly

8 0

3 years ago

Calculate the gravitational potential energy of a 2 kg banana hanging 5

Hunter-Best [27]

Answer:

100 J

Explanation:

The potential energy is given by the formula ...

PE = mgh

= (2 kg)(10 m/s^2)(5 m) = 100 J

7 0

3 years ago

Helppppppp meeee plleeeassee

spin [16.1K]

Answer:

balanced , balanced , unbalanced, unbalanced, balanced, balanced, unbalanced

Explanation: you're welcome

8 0

2 years ago

Read 2 more answers

The vacuum pressure of a condenser is given to be 80 kpa. if the atmospheric pressure is 98 kpa, what is the gage pressure and a

dimaraw [331]

The absolute pressure is given by the equation,

$P_{abs}=P_{atm}-P_{vac}$

Here, $P_{abs}$ is absolute pressure, $P_{atm}$ is atmospheric pressure and $P_{vac}$ is vacuum pressure.

Therefore,

$P_{abs}=98 kPa-80 kPa=18kPa$

The gage pressure is given by the equation,

$P_{gage}=P_{abs}-P_{atm}$ .

Thus,

$P_{gage}=18kPa-98 kPa=-80 kPa$ .

In kn/m^2,

The absolute pressure,

$P_{abs}=18kPa(\frac{1kN/m^2}{kPa}) =18\ kN/m^2$

The gage pressure,

$P_{gage}=-80kPa(\frac{1kN/m^2}{kPa}) =-80\ kN/m^2$ .

In lbf/in2

The absolute pressure,

$P_{abs}=18\ kPa(\frac{1.45\times 10^{-1}\ lbf/in^2 }{1kPa} )=2.6\ lbf/in^2$

The gage pressure,

$P_{gage}=-80kPa(\frac{1.45\times 10^{-1}\ lbf/in^2 }{1kPa} )=-11.6\ lbf/in^2$

In psi,

The absolute pressure,

$P_{abs}=18\ kPa(\frac{1.45037738\times 10^{-1}\ psi }{1kPa})=2.610\ psi$ .

The gage pressure,

$P_{gage}=-80kPa(\frac{1.45037738\times 10^{-1}\ psi }{1kPa} )=-11.6030\ psi$

In mm Hg

The absolute pressure,

$P_{abs}=18kPa(\frac{7.5\ mm\ of\ Hg }{1\ kPa})= 135\ mm\ of\ Hg$

The gage pressure,

$P_{gage}=-80kPa(\frac{7.5\ mm\ of\ Hg }{1\ kPa})=-600\ mm\ of\ Hg$

3 0

2 years ago