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Oksana_A [137]
3 years ago
15

A stone is dropped from rest from the top of a cliff into a pond below. If its initial height is 10 m, what is its speed when it

hits the water?is it 14 m/s, 9.9 m/s, or 4.8 m/s
Physics
1 answer:
Jobisdone [24]3 years ago
4 0

Answer:

14 m/s

Explanation:

The problem can be solved by using the law of conservation of energy. In fact:

- The mechanical energy of the stone at the top of the cliff is just gravitational potential energy (because the stone is at rest, so its kinetic energy is zero), and it is given by

E=U=mgh (1)

where m is the mass of the stone, g=9.8 m/s^2 is the acceleration due to gravity, and h=10 m is the height of the cliff.

- The mechanical energy of the stone when it hits the water is just kinetic energy (because the height of the stone has become zero, so the gravitational potential energy is zero), so it is

E=K=\frac{1}{2}mv^2 (2)

where v is the speed of the stone when it hits the water.

Since the mechanical energy is conserved, we can equalize (1) and (2), and solving for v we find:

mgh=\frac{1}{2}mv^2\\v=\sqrt{2gh}=\sqrt{2(9.8 m/s^2)(10 m)}=14 m/s

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A 2.00 kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0300 m . The spring has
iogann1982 [59]

Answer:

v = 0.41 m/s

Explanation:

  • In this case, the change in the mechanical energy, is equal to the work done by the fricition force on the block.
  • At any point, the total mechanical energy is the sum of the kinetic energy plus the elastic potential energy.
  • So, we can write the following general equation, taking the initial and final values of the energies:

       \Delta K + \Delta U = W_{ffr}  (1)

  • Since the block and spring start at rest, the change in the kinetic energy is just the final kinetic energy value, Kf.
  • ⇒ Kf = 1/2*m*vf²  (2)
  • The change in the potential energy, can be written as follows:

       \Delta U = U_{f}  - U_{o}  = \frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (3)

       where k = force constant = 815 N/m

       xf = final displacement of the block = 0.01 m (taking as x=0 the position

      for the spring at equilibrium)

      x₀ = initial displacement of  the block = 0.03 m

  • Regarding the work done by the force of friction, it can be written as follows:

       W_{ffr} = - \mu_{k}* F_{n} * \Delta x  (4)

       where μk = coefficient of kinettic friction, Fn = normal force, and Δx =

       horizontal displacement.

  • Since the surface is horizontal, and no acceleration is present in the vertical direction, the normal force must be equal and opposite to the force due to gravity, Fg:
  • Fn = Fg= m*g (5)
  • Replacing (5) in (4), and (3) and (4) in (1), and rearranging, we get:

        \frac{1}{2} * m* v^{2} = W_{ffr} - \Delta U = W_{ffr} - (U_{f} -U_{o})  (6)

        \frac{1}{2} * m* v^{2} = (- \mu_{k}* m*g* \Delta x)  -\frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (7)

  • Replacing by the values of m, k, g, xf and x₀, in (7) and solving for v, we finally get:

    \frac{1}{2} * 2.00 kg* v^{2}  = (-0.4*2.00 kg*9.8m/s2*0.02m) +( (\frac{1}{2} *815 N/m)* (0.03m)^{2} - (0.01m)^{2}) = -0.1568 J + 0.326 J (8)

  • v =\sqrt{(0.326-0.1568}  =  0.41 m/s  (9)
7 0
3 years ago
A penny rides on top of a piston as it undergoes vertical simple harmonic motion with an amplitude of 4.0cm. If the frequency is
aniked [119]

1) At the moment of being at the top, the piston will not only tend to push the penny up but will also descend at a faster rate at which the penny can reach in 'free fall', in that short distance. Therefore, at the highest point, the penny will lose contact with the piston. Therefore the correct answer is C.

2) To solve this problem we will apply the equations related to the simple harmonic movement, hence we have that the acceleration can be defined as

a = -\omega^2 A

Where,

a = Acceleration

A = Amplitude

\omega= Angular velocity

From a reference system in which the downward acceleration is negative due to the force of gravity we will have to

a = -g

-\omega^2 A = -g

\omega = \sqrt{\frac{g}{A}}

From the definition of frequency and angular velocity we have to

\omega = 2\pi f

f = \frac{1}{2\pi} \sqrt{\frac{g}{A}}

f = \frac{1}{2\pi} \sqrt{\frac{9.8}{4*10^{-2}}}

f = 2.5Hz

Therefore the maximum frequency for which the penny just barely remains in place for the full cycle is 2.5Hz

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Answer:

$4.2

Explanation:

Given data

Power= 700W

time= 10 hours

Cost per kilowatt hours is cents $0.20

Let us find the number of hours in a month

=24*30

=720 hours

Energ= power*time

Energy= 700/1000*30

Energy= 7*3

Enery= 21 kwh

1 kwh= 0.2

21kwh= x

cross multiply

x=21*0.2

x= $4.2

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