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Sidana [21]
3 years ago
13

The stiffness of a particular spring is 42 N/m. One end of the spring is attached to a wall. A force of 2 N is required to hold

the spring elongated a total length 17 cm. What was the relaxed length of the spring? Answer in units of m.
Physics
1 answer:
Liula [17]3 years ago
6 0

Answer:

L_{o}=0.1224m

Explanation:

Given data

Force F=2 N

Length L=17 cm = 0.17 m

Spring Constant k=42 N/m

To find

Relaxed length of the spring

Solution

From Hooke's Law we know that

F_{spring}=k_{s}|s|\\F_{spring}=k_{s}(L-L_{o})\\ 2N=(42N/m)(0.17m-L_{o})\\2=7.14-42L_{o}\\-42L_{o}=2-7.14\\42L_{o}=5.14\\L_{o}=(5.14/42)\\L_{o}=0.1224m

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The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 23
Burka [1]

Answer:

a) v(2) = 3.9\,\frac{m}{s}, b) v(4) = -15.7\,\frac{m}{s}

Explanation:

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v(t) = 23.5 -9.8\cdot t

The velocities at given instants are, respectivelly:

v(2) = 3.9\,\frac{m}{s}

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3 years ago
The ground state energy of an oscillating electron is 1.23 eV. How much energy must be added to the electron to move it to the t
Vikki [24]

Answer:

  • The energy that must be added to the electron to move it to the third excited state is  -1.153 eV
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Explanation:

Given;

Energy of electron in ground state (n = 1 ) = 1.23 eV

E₁ = 1.23 eV

Eₙ = E₁ /n²

where;

E₁ is the energy of the electron in ground state

n is the energy level,

For third excited state, n = 4

E₄ = E₁ /4²

E₄ = (1.23 eV) / 16

E₄ = 0.077 eV

Change in energy level, = E₄ - E₁ = 0.077 eV - 1.23 eV = -1.153 eV

The energy that must be added to the electron to move it to the third excited state is  -1.153 eV

For fourth excited state, n = 5

E₅ = E₁ /5²

E₄ = (1.23 eV) / 25

E₄ = 0.049 eV

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The energy that must be added to the electron to move it to the fourth excited state is  -1.181 eV

5 0
3 years ago
A dragster takes off from rest and crosses the finish line 320 m away. If the dragster is able to accelerate at 10LaTeX: \frac{m
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Answer:

t = 8 s

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In order to find the time taken by the dragster we will use equations of motion. Here, we will use second equation of motion:

s = Vi t + (1/2)at²

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s = distance covered = 320 m

Vi = Initial Velocity = 0 m/s (Since, dragster starts from rest)

t = time taken = ?

a = acceleration of dragster = 10 m/s²

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320 m = (0 m/s)t + (1/2)(10 m/s²)t²

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3 0
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una bala de 20 g choca con un fango como se muestra en la figura y penetra una distancia de 6 cm antes de detenerse. calcule la
aleksklad [387]

Answer:

A 20g bullet collides with a mud as shown in the figure and penetrates a distance of 6cm before stopping. calculate the braking force f if the input speed was 80m/s

Explanation:

Given that,

A bullet of mass

M = 20g = 0.02kg

The bullet is fired into a mud initially at rest

The bullet penetrate a distance of 6cm in the mud

S = 6cm = 0.06m

The input velocity into the mud is 80m/s, this implies that, the initial velocity is 80m/s

U = 80m/s

So, the bullet stop after a distance of 6cm, so the final velocity is 0m/s

V = 0m/s

So, we need to find the braking force

Breaking force (F) is given as

F = ma

Where m is mass and 'a' is deceleration

So, we need to find the deceleration, using equation of motion

V² = U² + 2as

0² = 80² + 2 × a × 0.06

0 = 6400 + 0.12a

0.12a = -6400

a = -6400 / 0.12

a = - 53,333.33 m/s²

So, the negative sign shows that the bullet is decelerating

So, a = 53,333.33 m/s²

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In spanish

Dado que,

Una bala de masa

M = 20 g = 0.02 kg

La bala se dispara en un lodo inicialmente en reposo

La bala penetra una distancia de 6 cm en el barro.

S = 6 cm = 0.06 m

La velocidad de entrada al lodo es de 80 m / s, esto implica que la velocidad inicial es de 80 m / s

U = 80 m / s

Entonces, la bala se detiene después de una distancia de 6 cm, por lo que la velocidad final es de 0 m / s

V = 0 m / s

Entonces, necesitamos encontrar la fuerza de frenado

La fuerza de ruptura (F) se da como

F = ma

Donde m es masa y 'a' es desaceleración

Entonces, necesitamos encontrar la desaceleración, usando la ecuación de movimiento

V² = U² + 2as

0² = 80² + 2 × a × 0.06

0 = 6400 + 0.12a

0.12a = -6400

a = -6400 / 0.12

a = - 53,333.33 m / s²

Entonces, el signo negativo muestra que la bala se está desacelerando

Entonces, a = 53,333.33 m / s²

Entonces, fuerza de ruptura

F = ma

F = 0.02 × 53,333.33

F = 1066.67 N

La fuerza de ruptura es 1066.67 N

7 0
3 years ago
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