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solmaris [256]
3 years ago
8

A 6.0 g marble is fired vertically upward using a spring gun. The spring must be compressed 9.4 cm if the marble is to just reac

h a target 22 m above the marble's position on the compressed spring. (a) What is the change ΔUg in the gravitational potential energy of the marble-Earth system during the 22 m ascent? (b) What is the change ΔUs in the elastic potential energy of the spring during its launch of the marble? (c) What is the spring constant of the spring?
Physics
1 answer:
RoseWind [281]3 years ago
4 0

Answer:

a) \Delta U_{g} = 12.945\,J, b) \Delta U_{k} = 12.945\,J, c) k = 2930.059\,\frac{N}{m}

Explanation:

a) The change in the gravitational potential energy of the marble-Earth system is:

\Delta U_{g} = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (22\,m)

\Delta U_{g} = 12.945\,J

b) The change in the elastic potential energy of the spring is equal to the change in the gravitational potential energy, then:

\Delta U_{k} = 12.945\,J

c) The spring constant of the gun is:

\Delta U_{k} = \frac{1}{2} \cdot k \cdot x^{2}

k = \frac{2\cdot \Delta U_{k}}{x^{2}}

k = \frac{2\cdot (12.945\,J)}{(0.094\,m)^{2}}

k = 2930.059\,\frac{N}{m}

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Answer:

1. Distance travelled = 12 km.

2. Displacement = 8.6 km

Explanation:

From the question given above, the following data were obtained:

Distance 1 (d₁) = 7 km

Distance 2 (d₂) = 5 km

Total distance =?

Displacement =?

1. Determination of the distance travelled.

Distance 1 (d₁) = 7 km

Distance 2 (d₂) = 5 km

Total distance (dₜ) =?

dₜ = d₁ + d₂

dₜ = 7 + 5

dₜ = 12 km

2. Determination of the displacement.

In the attached photo, R is the displacement.

We can obtain the value of R by using the pythagoras theory as illustrated below:

R² = 7² + 5²

R² = 49 + 25

R² = 74

Take the square root of both side

R = √74

R = 8.6 km

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2 years ago
Plz help This is for Rock formations...only 5 sentences if not more
dsp73

Answer:

The coastal zone is not a stable and constant environment, but a dynamic place that can change rapidly in response to natural processes such as seasonal weather patterns. Waves, winds, currents, tides and storms are the major forces on the coast.

Explanation:

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3 years ago
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lana66690 [7]

Answer:

a = (v2 - v1) / t

From A to B    (8 - 4) m/s / 1 s = 4 m / s^2

From A to D    ( 7 - 4) m/s / 5 s = .6 m / s^2

Note these equations hold for "uniform" values

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Three boards, each 50 mm thick, are nailed together to form a beam that is subjected to a 1200-N vertical shear. Knowing that th
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Answer:

Using equation

qs=F

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q=998.45 N

qs=F

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Two thin 80.0-cm rods are oriented at right angles to each other. Each rod has one end at the origin of the coordinates, and one
kogti [31]

Answer:

The net force on the electron is given as:

F = 1.35 x 10⁻¹³ N j - 1.35 x 10⁻¹³ N i

Explanation:

Given:

charge on rod along x-axis = Q₁ = -15 x 10⁻⁶ C

charge on rod along y-axis = Q₂ = 15 x 10⁻⁶ C

distance of electron from rod 1 = r₁ = 0.4 m

distance of electron from rod 1 = r₂ = 0.4 m

charge on electron = q = -1.6 x 10⁻¹⁹ C

ε° = 8.85 x 10⁻¹² C²/Nm²

Electric force on charge due to rod 1:

F₁ = qE = 1/4πε°(qQ₁/r₁²)

F₁ = (9 x 10⁹ x -1.6 x 10⁻¹⁹ x -15 x 10⁻⁶)/0.4²

F₁ = 1.35 x 10⁻¹³ N

Negative negative repels each other so the rod will Force the electron in positive y-direction.

F₁ = 1.35 x 10⁻¹³ N j

Electric force on charge due to rod 2:

F₂ = qE = 1/4πε°(qQ₂/r₂²)

F₂ = (9 x 10⁹ x -1.6 x 10⁻¹⁹ x 15 x 10⁻⁶)/0.4²

F₂ = - 1.35 x 10⁻¹³ N

Opposite charges attract each other so the rod will force the electron in negative x-direction.

F₂ =  - 1.35 x 10⁻¹³ N i

Net Force:

F = F₁ + F₂

F = 1.35 x 10⁻¹³ N j - 1.35 x 10⁻¹³ N i

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3 years ago
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