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Sav [38]
2 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]2 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]2 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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There are many ways to lose weight, but a sustainable plan generally involves behavior modification related to diet and exercise
Anna007 [38]

The answer is adequate nutrition, regular physical activity, and practical ways to reduce calories while retaining important nutrients.

Explanation:

Despite that adequate nutrition, regular physical activity, and practical ways to reduce calories while retaining important nutrients is one of the best strategy to reducing weight, most time it is very difficult for those that want to reduce or control their weight to discipline themselves enough to follow these routine. But one an individual that want to loose weight or live a healthy lifestyle is able to follow these procedures he/she will surely loose weight.

5 0
2 years ago
A particle with a charge of -1.24 x 10"° C is moving with instantaneous velocity (4.19 X 104 m/s)î + (-3.85 X 104 m/s)j. What is
astra-53 [7]

Answer:

(a) F= 6.68*10¹¹⁴ N (-k)

(b) F =( 6.68*10¹¹⁴ i  + 7.27*10¹¹⁴ j  ) N

Explanation

To find the magnetic force in terms of a fixed amount of charge q that moves at a constant speed v in a uniform magnetic field B we apply the following formula:

F=q* v X B Formula (1 )

q: charge (C)

v: velocity (m/s)

B: magnetic field (T)

vXB : cross product between the velocity vector and the magnetic field vector

Data

q= -1.24 * 10¹¹⁰ C

v= (4.19 * 10⁴ m/s)î + (-3.85 * 10⁴m/s)j

B  =(1.40 T)i  

B  =(1.40 T)k

Problem development

a) vXB = (4.19 * 10⁴ m/s)î + (-3.85* 10⁴m/s)j X (1.40 T)i =

            = - (-3.85*1.4) k = 5.39* 10⁴ m/s*T (k)

1T= 1 N/ C*m/s

We apply the formula (1)

  F= 1.24 * 10¹¹⁰ C*  5.39* 10⁴ m/s* N/ C*m/s (-k)

   F= 6.68*10¹¹⁴ N (-k)

a)  vXB = (4.19 * 10⁴ m/s)î + (-3.85* 10⁴m/s)j X (1.40 T)k =

             =( - 5.39* 10⁴i - 5.87* 10⁴j)m/s*T

1T= 1 N/ C*m/s

We apply the formula (1)

F= 1.24 * 10¹¹⁰ C*  (  5.39* 10⁴i + 5.87* 10⁴j) m/s* N/ C*m/s

F =( 6.68*10¹¹⁴  i  + 7.27*10¹¹⁴  j  ) N

8 0
2 years ago
Calculate the binding energy e of the boron nucleus 11 5b (1ev=1.602×10−19j). express your answer in millions of electron volts
nignag [31]
<span>Depends on the precision you're working to. proton mass ~ 1.00728 amu neutron mass ~ 1.00866 amu electron mass ~ electron mass = 0.000549 amu Binding mass is: mass of constituents - mass of atom Eg for nitrogen: (7*1.00728)-(7*1.00866)-(7*0.000549) -14.003074 = 0.11235amu Binding energy is: E=mc^2 where c is the speed of light. Nuclear physics is usually done in MeV[1] where 1 amu is about 931.5MeV/c^2. So: 0.11235 * 931.5 = 104.6MeV Binding energy per nucleon is total energy divided by number of nucleons. 104.6/14 = 7.47MeV This is probably about right; it sounds like the right size! Do the same thing for D/E/F and recheck using your numbers & you shouldn't go far wrong :) 1 - have you done this? MeV is Mega electron Volts, where one electronVolt (or eV) is the change in potential energy by moving one electron up a 1 volt potential. ie energy = charge * potential, so 1eV is about 1.6x10^-19J (the same number as the charge of an electron but in Joules). It's a measure of energy, but by E=mc^2 you can swap between energy and mass using the c^2 factor. Most nuclear physicists report mass in units of MeV/c^2 - so you know that its rest mass energy is that number in MeV.</span>
7 0
2 years ago
A Porsche challenges a Honda to a 200-m race.Because the Porsche's acceleration of 3.5 m/s2 is larger than the Honda's 3.0 m/s2,
Blizzard [7]

Answer:

Honda won by 0.14 s

Explanation:

We are given that

Distance =S=200 m

Initial velocity of Honda=u=0m/s

Initial velocity of Porsche=u'=0m/s

Acceleration of Honda=3.0m/s^2

Acceleration of Porsche's=3.5m/s^2

Time taken by Honda  to start=1 s

s=ut+\frac{1}{2}at^2

Substitute the values

200=0(t)+\frac{1}{2}(3)t^2

200=\frac{3}{2}t^2

t^2=\frac{200\times 2}{3}=\frac{400}{3}

t=\sqrt{\frac{400}{3}}=11.55s

Time taken by Honda=11.55 s

Now, time taken by  Porsche

200=\frac{1}{2}(3.5)t^2

t^2=\frac{200\times 2}{3.5}

t=\sqrt{\frac{400}{3.5}}=10.69 s

Total time taken by Porsche=10.69+1=11.69 s

Because it start 1 s late

Time taken by Honda is less than Porsche .Therefore, Honda won and

Time =11.69-11.55=0.14 s

Honda won by 0.14 s

3 0
2 years ago
At 5000-kg freight car runs into 10000-kg freight car at rest. they couple upon collision and move with a speed of 2 m/s. what w
aliina [53]

Solution for the problem is:

Total momentum before collision is always equal to total momentum after collision. So note that:
Momentum of car A = 5000 x Xm/s 
Momentum of car A + B = 15,000 x 2m/s 

So combining the two, will give us the equation:
15,000/5,000 = 3 
3 x 2 =6m/s

6 0
2 years ago
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