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3 years ago

A spring has a spring constant of 115 N/m. How much energy is stored in the

Physics

2 answers:

noname [10]3 years ago

8 0

Explanation:

Elastic energy stored in a spring is:

EE = ½ kx²

EE = ½ (115 N/m) (0.15 m)²

EE = 1.3 J

andrey2020 [161]3 years ago

6 0

<u>Answer:</u> The energy stored in the spring is 1.33 J

<u>Explanation:</u>

To calculate the energy stored in the spring, we use the equation:

$E=\frac{1}{2}kx^2$

where,

E = energy stored in the spring

k = spring constant = 115 N/m

x = compression of the spring = 0.15 m

Putting values in above equation, we get:

$E=\frac{1}{2}\times 115\times (0.15)^2\\\\E=1.33J$

Hence, the energy stored in the spring is 1.33 J

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You are raising up a big bucket of water from a 25.9 m deep well. The combined mass of the water and the bucket is 13.9 kg. The

Mars2501 [29]

The total work done is 5980 Joules and the power expended is 57 Watts.

The work done is the work done in the gravitational field as the bucket is raised up Thus work required to remove the bucket Wb;

Wb = 13.9 kg * 25.9 m * 9.8 m/s^2 = 3530 Joules

Height of the center of mass of chain = 25.9 / 2 = 12.95 m

Work done by the chain Wc;

Wc = 12.95 * 19.3 * 9.8 = 2450 Joules

Total work = 3530 + 2450 = 5980 Joules

Power expended = W / t = 5980 J / 105 sec = 57 J/s = 57 Watts

Learn more about work done:brainly.com/question/13662169

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3 0

2 years ago

A block of concrete has a mass of 48kg a crane lifts the block to a height of 12m above the ground calculate the gravitational p

Afina-wow [57]

Answer:

5760 J

Explanation:

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

Mass of block = 48 kg

Height (h) = 12 m

Gravitational field strength (g) = 10 N/Kg

Gravitational potential energy (PE) =?

The gravitational potential energy stored by the block can simply be obtained as follow:

PE = mgh

PE = 48 × 10 × 12

PE = 5760 J

Therefore, the gravitational potential energy stored by the block is 5760 J

3 0

3 years ago

If a proton and an electron are released when they are 7.00×10−10 m apart (typical atomic distances), find the initial accelerat

Ugo [173]

Answer:

The acceleration of the proton is 2.823 x 10¹⁷ m/s²

The acceleration of the electron is 5.175 x 10²⁰ m/s²

Explanation:

Given;

distance between the electron and proton, r = 7 x 10⁻¹⁰ m

mass of proton, $m_p$ = 1.67 x 10⁻²⁷ kg

mass of electron, $m_e$ = 9.11 x 10⁻³¹ kg

The attractive force between the two charges is given by Coulomb's law;

$F = \frac{k(q_p)(q_e)}{r^2}$

where;

k is Coulomb's constant = 9 x 10⁹ Nm²/c²

$F = \frac{k(q_p)(q_e)}{r^2} \\\\F = \frac{(9*10^9)(1.602*10^{-19})(1.602*10^{-19})}{(7*10^{-10})^2} \\\\F = 4.714 *10^{-10} \ N$

Acceleration of proton is given by;

F = ma

$F = m_pa_p\\\\a_p = \frac{F}{m_p}\\\\a_p = \frac{4.714*10^{-10}}{1.67*10^{-27}}\\\\a_p = 2.823 *10^{17} \ m/s^2$

Acceleration of the electron is given by;

$F = m_ea_e\\\\a_e = \frac{F}{m_e}\\\\a_e = \frac{4.714*10^{-10}}{9.11*10^{-31}}\\\\a_e = 5.175 *10^{20} \ m/s^2$

7 0

3 years ago

A weightlifter works out at the gym each day. Part of her routine is to lie on her back and lift a 43 kg barbell straight up fro

Tasya [4]

Answer:

A. 231.77 J

B. 5330.71 J

C. 46 donuts

Explanation:

A. To lift the barbell once, she will have to extend it the full length of her arm. The work done will then be:

W = F * d

Where the force is the weight of the barbell.

F = m * g

Hence, the work done in lifting the barbell is:

W = m * g * d

W = 43 * 9.8 * 0.55

W = 231.77 J

B. If she does 23 repetitions, the total energy she expend will be equal to the Potential energy when the barbell is lifted multiplied by 23:

E = 23 * m * g * d

E = 23 * 231.77

E = 5330.71 J

C. 1 Joule = 4.184 calories

5330.71 Joules = 5330.71 * 4.184 = 22303.69

If 1 donut contains 490 calories, the number of donuts she will need will be:

N = 22303.69/490 = 45.5 donuts or 46 donuts

5 0

4 years ago

An electron in the n = 7 level of the hydrogen atom relaxes to a lower energy level, emitting light of 397 nm. what is the value

Dimas [21]

Answer:

$n_f=2$

Explanation:

It is given that,

Initially, the electron is in n = 7 energy level. When it relaxes to a lower energy level, emitting light of 397 nm. We need to find the value of n for the level to which the electron relaxed. It can be calculate using the formula as :

$\dfrac{1}{\lambda}=R(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2})$

$\dfrac{1}{397\times 10^{-9}\ m}=R(\dfrac{1}{n_f^2}-\dfrac{1}{(7)^2})$

R = Rydberg constant, $R=1.097\times 10^7\ m^{-1}$

$\dfrac{1}{397\times 10^{-9}\ m}=1.097\times 10^7\ m^{-1}\times (\dfrac{1}{n_f^2}-\dfrac{1}{(7)^2})$

Solving above equation we get the value of final n is,

$n_f=2.04$

$n_f=2$

So, it will relax in the n = 2. Hence, this is the required solution.

6 0

3 years ago

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