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3 years ago

Which one has greater potential energy? An 8 kg rock sitting on a 2.2 m cliff or a 6 kg rock sitting on a 3.2 m cliff.

Physics

1 answer:

Evgesh-ka [11]3 years ago

6 0

Answer: The 6 kg rock sitting on a 3.2 m cliff.

Explanation:

The potential energy of an object of mass M that is at a height H above the ground us:

U = M*H*g

where g is the gravitational acceleration:

g = 9.8m/s^2

Then:

"An 8 kg rock sitting on a 2.2 m cliff"

M = 8kg

H = 2.2m

U = 8kg*2.2m*9.8 m/s^2 = 172.48 J

"a 6 kg rock sitting on a 3.2 m cliff"

M = 6kg

H = 3.2m

U = 6kg*3.2m*9.8m/s^2 = 188.16 J

You can see that the 6kg rock on a 3.2m cliff has a larger potential energy.

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MariettaO [177]

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3 years ago

A 2.20-kg object is attached to a spring and placed on frictionless, horizontal surface. A horizontal force of 29.0 N is require

g100num [7]

Answer:

a. 145 N/m b. 1.29 Hz c. 1.62 m/s d. 0 m e. 13.2 m/s² f. ± 0.2 m g. 2.9 J h. 0.54 m/s i. 4.39 m/s²

Explanation:

a. The force constant of the spring

The spring force F = kx and k = F/x where k is the spring constant. F = 29.0 N and x = 0.200 m

k = 29.0 N/0.200 m = 145 N/m

b. The frequency of oscillations, f

f = 1/2π√(k/m) m = mass = 2.20 kg

f = 1/2π√(145 N/m/2.20 kg) = 1.29 Hz

c. maximum speed of the object

The maximum elastic potential energy of the spring = maximum kinetic energy of the object

1/2kx² = 1/2mv²

v = (√k/m)x where v is the maximum speed of the object

v = (√145/2.2)0.2 = 1.62 m/s

d Where does the maximum speed occur?

The maximum speed occurs at 0 m

e. The maximum acceleration

a = kx/m = 145 × 0.2/2.2 = 13.2 m/s²

f. The maximum acceleration occurs at x = ± 0.2 m

g. The total energy of the system is the maximum elestic potential energy of the system

E = 1/2kx² = 1/2 × 145 × 0.2² = 2.9 J

h. When x = x₀/3

1/2k(x₀/3)² = 1/2mv²

kx₀²/9 = mv²

v = 1/3(√k/m)x₀ = 1/3(√145/2.2)0.2 = 0.54 m/s

i When x = x₀/3

a = kx₀/3m = 145 × 0.2/(2.2 × 3)= 4.39 m/s²

8 0

3 years ago

A ball is thrown vertically downwards with a speed 7.3 m/s from the top of a 51 m tall building. With what speed will it hit the

ddd [48]

Answer:

32.46m/s

Explanation:

Hello,

To solve this exercise we must be clear that the ball moves with constant acceleration with the value of gravity = 9.81m / S ^ 2

A body that moves with constant acceleration means that it moves in "a uniformly accelerated motion", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are the follow

$\frac {Vf^{2}-Vo^2}{2.a} =X$