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tangare [24]
3 years ago
12

A mass of 4.8 kg is dropped from a height of 4.84 meters above a vertical spring anchored at its lower end to the floor. If the

spring has a height of 82 cm and a constant of 24 N/cm, how far, to the nearest tenth of a cm, is the spring compressed?
Physics
1 answer:
krek1111 [17]3 years ago
5 0

Answer:

45.6 cm

Explanation:

Let x (m) be the length that the spring is compressed. We know that when we drop the mass from 4.84 m above and compress the springi, ts gravitational energy shall be converted to spring potential energy due to the law of energy conservation

E_g = E_p

mgh = 0.5kx^2

where h = 4.84 + x is the distance from the dropping point the the compressed point, and k = 24N/cm = 2400N/m is the spring constant, g = 9.81 m/s2 is the gravitational acceleration constant. And m = 4.8 kg is the object mass.

4.8*9.81(4.84 + x) = 0.5 * 2400 * x^2

47.088x + 227.906 = 1200x^2

1200x^2 - 47.088x - 227.906 = 0

x = 0.456m or 45.6 cm

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2 years ago
A stone was dropped off a cliff and hit the ground with a speed of 88 ft/s. What is the height (in feet) of the cliff
igor_vitrenko [27]

Answer:

the height (in feet) of the cliff is 121 ft

Explanation:

A stone hit the cliff with

speed, v = 88 ft/s

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initial speed, u = 0 ft/s

height is h.

To solve this problem we will apply the linear motion kinematic equations, Equation of motion describes change in velocity, depending on the acceleration and the distance traveled

so, writing the formula of Equation of motion:

v^2 - u^2 = 2*a*h

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(88)^2 - 0 = 2*32* h

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learn more about height of the cliff here:

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3 0
1 year ago
An electron in a vacuum is first accelerated by a voltage of 51400 V and then enters a region in which there is a uniform magnet
zimovet [89]

Answer:

       F = 8.6 10⁻¹² N

Explanation:

For this exercise we use the law of conservation of energy

Initial. Field energy with the electron at rest

         Em₀ = U = q ΔV

Final. Electron with velocity, just out of the electric field

         Emf = K = ½ m v²

          Em₀ = Emf

          e ΔV = ½ m v²

          v =√ 2 e ΔV / m

          v = √(2 1.6 10⁻¹⁹ 51400 / 9.1 10⁻³¹)

           v = √(1.8075 10¹⁶)

           v = 1,344 10⁸ m / s

Now we can use the equation of the magnetic force

         F = q v x B

Since the speed and the magnetic field are perpendicular the force that

        F = e v B

        F = 1.6 10⁻¹⁹  1.344 10⁸ 0.4

       For this exercise we use the law of conservation of energy

Initial. Field energy with the electron at rest

         Emo = U = q DV

Final. Electron with velocity, just out of the electric field

         Emf = K = ½ m v2

          Emo = Emf

          .e DV = ½ m v2

          .v = RA 2 e DV / m

          .v = RA (2 1.6 10-19 51400 / 9.1 10-31)

           .v = RA (1.8075 10 16)

           .v = 1,344 108 m / s

Now we can use the equation of the magnetic force

         F = q v x B

Since the speed and the magnetic field are perpendicular the force that

        F = e v B

       F = 1.6 10-19 1,344 108 0.4

       F = 8.6 10-12 N

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