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3 years ago

A ball of mass 2 kg is kept on the hill of height 3 km. Calculate the potential energy possessed by it ?

2 answers:

7 0

We know that -

$P.E=m*g*h$

Where,

$m = mass$

$g = acceleration$ $due$ $to$ $gravity$

$h=height$

First we convert height into meters.

1 km = 1000 meters

3 km = 1000 * 3 meters = 3000 meters

So, putting the values in the above formula, and by taking 'g' = 9.8 m/s², we get-

$P.E.= 2*3000*9.8$

$P.E.= 58800$ $Joules$

$P.E.= 58.8$ $kJ$

IRINA_888 [86]3 years ago

4 0

Sairah's work is correct as far as it goes. The potential energy of the ball
relative to the bottom of the hill is 58,800 Joules.

<u>To address the second part of the question:</u>
In order to get ahold of that energy, the ball must be returned to the bottom
of the hill. The most efficient way would be to drop it, so that it wouldn't have to
scrape along the grass on the way down. But that can only work if there's a sheer
cliff on one side of the hill. Otherwise, you just have to roll it down, and accept the
fact that it loses some of its energy to friction on the way.

However the ball gets to the bottom, the energy it has left shows up in the form
of kinetic energy, and 58,800 joules is a lot of kinetic energy. If somehow the ball
could arrive at the bottom with ALL the energy it had at the top, it would be moving
at something like 540 miles per hour !

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A builder drops a brick from a height of 15 m above the ground. The gravitational field strength g is 10 N/ kg. What is the spee

Basile [38]

The speed of the brick dropped by the builder as it hits the ground is 17.32m/s.

Given the data in the question;

Since the brick was initially at rest before it was dropped,

Initial Velocity; $u = 0$

Height from which it has dropped; $h = 15m$

Gravitational field strength; $g = 10N/kg = 10 \frac{kg.m/s^2}{kg} = 10m/s^2$

Final speed of brick as it hits the ground; $v = \ ?$

<h3>Velocity</h3>

velocity is simply the same as the speed at which a particle or object moves. It is the rate of change of position of an object or particle with respect to time. As expressed in the Third Equation of Motion:

$v^2 = u^2 + 2gh$

Where v is final velocity, u is initial velocity, h is its height or distance from ground and g is gravitational field strength.

To determine the speed of the brick as it hits the ground, we substitute our giving values into the expression above.

$v^2 = u^2 + 2gh\\\\v^2 = 0 + ( 2\ *\ 10m/s^2\ *\ 15m)\\\\v^2 = 300m^2/s^2\\\\v = \sqrt{300m^2/s^2}\\ \\v = 17.32m/s$

Therefore, the speed of the brick dropped by the builder as it hits the ground is 17.32m/s.

Learn more about equations of motion: brainly.com/question/18486505

8 0

1 year ago

10. Mario is making chocolate. In part of the chocolate-making process, he changed the phase from a liquid to a solid. What happ

Naily [24]

Answer:

Explanation:

in a liquid the particles are widespread and move around each other but in a solid they move in place and are tightly packed

6 0

2 years ago

Ionic compounds form crystal structures. molecular structures. pure elements. neutral elements

Naddik [55]

Answer: Option (a) is the correct answer.

Explanation:

Ionic compounds are the compounds whose atoms are combined together by transfer of electrons.

An ionic compound has partial positive charge on the cation and partial negative charge on anion.

For example, NaCl is an ionic compound whose atoms are arranged orderly due to the opposite charges on its atoms.

So, an ionic compound forms a neutral compound but not a neutral element.

Thus, we can conclude that ionic compounds form crystal structures.

3 0

2 years ago

Read 2 more answers

The following three questions refer to a situation in which a driver is in a car that crashes into a solid wall. The car comes t

Andreas93 [3]

Answer:

1) p₀ = 45000 N / s , p₀ '= 1800 , b) I = -45000 N s , I = 1800 Ns

Explanation:

Impulse equals the change in momentum

I = Δp

1) the initial moment of the car

p₀ = M v

p₀ = 1500 30

p₀ = 45000 N / s

the change at the moment is

Δp = 45000

because the end the car is stopped

moment of the person

P₀ ’= m v

p₀ '= 60 30

p₀ '= 1800

D₀ '= 1800

2) of the momentum change impulse ratio

car

I = Δp

I = -45000 N s

person

I = Δpo '

I = 1800 Ns

3) the object that give the momentum to stop the wall motoring

The person is stopped by the impulse given by the car

a) This area is the one that absorbs most of the vehicle impulse

be) If using a safety painter, the time during which the greater force will act, therefore the lessons decrease

c) The air bag helps reduction in the speed of the person relatively quickly.

6 0

3 years ago

An ideal spring hangs from the ceiling. A 2.15 kg mass is hung from the spring, stretching the spring a distance d = 0.0895 m fr

Igoryamba

Answer:

The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

Explanation:

Given that,

Mass = 2.15 kg

Distance = 0.0895 m

Amplitude = 0.0235 m

We need to calculate the spring constant

Using newton's second law

$F= mg$

Where, f = restoring force

$kx=mg$

$k=\dfrac{mg}{x}$

Put the value into the formula

$k=\dfrac{2.15\times9.8}{0.0895}$

$k=235.41\ N/m$

We need to calculate the kinetic energy of the mass

Using formula of kinetic energy

$K.E=\dfrac{1}{2}mv^2$

Here, $v = A\omega$

$K.E=\dfrac{1}{2}m\times(A\omega)^2$

Here, $\omega=\sqrt{\dfrac{k}{m}}^2$

$K.E=\dfrac{1}{2}m\times A^2\sqrt{\dfrac{k}{m}}^2$

$K.E=\dfrac{1}{2}kA^2$

Put the value into the formula

$K.E=\dfrac{1}{2}\times235.41\times(0.0235)^2$

$K.E=0.06500\ J$

Hence, The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

8 0

3 years ago