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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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A bathtub contains 65 gallons of water and the total weight of the tub and water is approximately 931.925 pounds. You pull the p
levacccp [35]

Answer:

Q = 8,345 * v

Explanation:

So, we are looking for an expression of the amount of water that has been drained from the tub. The expression is in terms of v that represent the number of gallons of water drained since the plug was pulled. Since we are interested in the pounds of water that has been drained from the tub we need to take into account that for every gallon of water drained, 8.345 pounds have left the tub. Therefore, the expression for the weight of water Q that has been drained from the tub in terms of v is simply :

Q = 8,345 * v

Where v is the amount of gallons that has been drained from the tub.

Have a nice day. let me know if I can help with anything else

8 0
4 years ago
Desde el balcón de un edificio se deja caer una manzana y llega a la planta baja en 5 s. ¿Desde qué piso se dejó caer, si cada p
miv72 [106K]

Answer: 42

Explanation:

I will answer this in English.

We know that the apple needs 5 seconds to reach the ground.

Each floor of the building has a height of 2.88m.

Now, when we drop something, the only force acting on the object is the gravitational one, so the acceleration of the apple is:

a = -g

for the velocity, we integrate the acceleration over time, and as the apple is dropped, we do not have any initial velocity, so we do not have a constant in the integration:

v = -g*t

for the position we integrate again, now we have an initial height H, so the position is:

p = (-g/2)*t^2 + H

now the apple hits the ground when p = 0, so we can solve this equation to find H.

i will use g = 9.8m/s^2

0 = (-4.9m/s^2)*(5s)^2 + H  

H = 122.5 m

now knowing H, we can divide it by the height of a floor in the building and get the number of the floor.

N = 122.5m/2.88m = 42.5

this means that the apple was dropped in the floor 42 (the 0.5 means that the apple was not right where the floor 42 starts, it was dropped around the middle of the floor 42)

5 0
3 years ago
Air bags are designed to deploy in 10 ms. Given that the air bags expand 20 cm as they deploy, estimate the acceleration of the
joja [24]

As it is given that the air bag deploy in time

t = 10 ms = 0.010 s

total distance moved by the front face of the bag

d = 20 cm = 0.20 m

Now we will use kinematics to find the acceleration

d = v_i*t + \frac{1}{2}at^2

0.20 = 0 + \frac{1}{2}a*0.010^2

0.20 = 5 * 10^{-5}* a

a = 4000 m/s^2

now as we know that

g = 10 m/s^2

so we have

a = 400g

so the acceleration is 400g for the front surface of balloon

3 0
3 years ago
•• CP Two blocks connected by a light horizontal rope sit at rest on a horizontal, frictionless surface. Block AA has mass 15.0
Firdavs [7]

Answer:

(a) T= 38.4 N

(b) m= 26.67 kg

Explanation:

We apply Newton's second law:

∑F = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Kinematics

d= v₀t+ (1/2)*a*t² (Formula 2)

d:displacement in meters (m)  

t : time in seconds (s)

v₀: initial speed in m/s  

vf: final speed in m/s  

a: acceleration in m/s²

v₀=0, d=18 m , t=5 s

We apply the formula 2 to calculate the accelerations of the blocks:

d= v₀t+ (1/2)*a*t²

18= 0+  (1/2)*a*(5)²

a= (2*18) / ( 25) = 1.44 m/s² to the right

We apply Newton's second law to the block A

∑Fx = m*ax

60-T = 15*1.44

60 - 15*1.44 = T

T = 38.4 N

We apply Newton's second law to the block B

∑Fx = m*ax

T = m*ax

38.4 = m*1.44

m= (38.4) / (1.44)

m = 26.67 kg

7 0
4 years ago
When measuring current, always place any type of meter in
Mars2501 [29]
     In series with the circuit, so for it pass the current to be mensured.

Letter A

If you notice any mistake in my english, please let me know, because i am not native.
5 0
3 years ago
Read 2 more answers
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