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Sav [38]
4 years ago
5

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

Physics
2 answers:
zysi [14]4 years ago
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]4 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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Help please! Will give brainly! and thanks.
kipiarov [429]
<h2>Answer:</h2>

All the energy sources are correctly matched with their category.

Explanation:

Renewable energy sources:

These are energy sources which can be replenished as they don't in involves the irreversible phase change.

These resources can never be ended as they can be used again and again.

Wind, geothermal, biomass, bio gas are example of renewable energy sources.

Non renewable energy sources:

These are the energy source which can never be replenished after one time use. They undergo the chemical irreversible change.

These sources are lacking with the passage of time because they can never be reused.

Oil, gas, coal and natural gas are examples.

7 0
3 years ago
How does the structure of the stigma aid in pollination
natali 33 [55]
<span>Pollination is the process by which pollen is transferred to the female part of a plant. The stigma is the central part of the flower, that is supported by a style and is part of the female reproductive organ within plants. Its structure is optimised to promote pollination by having hairs, sticky surfaces and three-dimensional sculptures that capture and trap pollen.</span>
5 0
3 years ago
PLEASE HELP ILL MARK BRAINLIEST
ikadub [295]
I think the answer is A?
7 0
3 years ago
In an inkjet printer, letters and images are created by squirting drops of ink horizontally at a sheet of paper from a rapidly m
Serga [27]

Answer:

q = 6.48 \times 10^{-14} C

Explanation:

Deflection in the drop is due to electric field force

so we will have

F = qE

acceleration of the drop is given as

a = \frac{qE}{m}

a = \frac{q(7.75 \times 10^4)}{1.00 \times 10^{-11}}

a = 7.75 \times 10^{15} q

now we know that time to cross the plates is given as

t = \frac{D}{v}

t = \frac{0.02}{18}

t = 1.11 \times 10^{-3} s

now the deflection is given as

d = \frac{1}{2}at^2

0.310 \times 10^{-3} = \frac{1}{2}(7.75 \times 10^{15} q)(1.11 \times 10^{-3})^2

0.310 \times 10^{-3} = 4.78 \times 10^9 q

q = 6.48 \times 10^{-14} C

5 0
3 years ago
X rays of wavelength 0.0169 nm are directed in the positive direction of an x axis onto a target containing loosely bound electr
mamaluj [8]

Answer:

a) 4.04*10^-12m

b) 0.0209nm

c) 0.253MeV

Explanation:

The formula for Compton's scattering is given by:

\Delta \lambda=\lambda_f-\lambda_i=\frac{h}{m_oc}(1-cos\theta)

where h is the Planck's constant, m is the mass of the electron and c is the speed of light.

a) by replacing in the formula you obtain the Compton shift:

\Delta \lambda=\frac{6.62*10^{-34}Js}{(9.1*10^{-31}kg)(3*10^8m/s)}(1-cos132\°)=4.04*10^{-12}m

b) The change in photon energy is given by:

\Delta E=E_f-E_i=h\frac{c}{\lambda_f}-h\frac{c}{\lambda_i}=hc(\frac{1}{\lambda_f}-\frac{1}{\lambda_i})\\\\\lambda_f=4.04*10^{-12}m +\lambda_i=4.04*10^{-12}m+(0.0169*10^{-9}m)=2.09*10^{-11}m=0.0209nm

c) The electron Compton wavelength is 2.43 × 10-12 m. Hence you can use the Broglie's relation to compute the momentum of the electron and then the kinetic energy.

P=\frac{h}{\lambda_e}=\frac{6.62*10^{-34}Js}{2.43*10^{-12}m}=2.72*10^{-22}kgm\\

E_e=\frac{p^2}{2m_e}=\frac{(2.72*10^{-22}kgm)^2}{2(9.1*10^{-31}kg)}=4.06*10^{-14}J\\\\1J=6.242*10^{18}eV\\\\E_e=4.06*10^{-14}(6.242*10^{18}eV)=0.253MeV

5 0
4 years ago
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