Lauren is comparing rocks from two different locations. She has the following

The total energy of a block—spring system is 0.18 J. The amplitude is 14.0 cm and the maximum speed is 1.25 m/s. Find: (a) the m

algol13

a) The mass is 0.23 kg

b) The spring constant is 1.25 N/m

c) The frequency is 1.42 Hz

d) The speed of the block is 1.08 m/s

Explanation:

a)

We can find the mass of the block by applying the law of conservation of energy: in fact, the total mechanical energy of the system (which is sum of elastic potential energy, PE, and kinetic energy, KE) is constant:

$E=PE+KE=const.$

The potential energy is given by

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

where k is the spring constant and x is the displacement. When the block is crossing the position of equilibrium, x = 0, so all the energy is kinetic energy:

$E=KE_{max}=\frac{1}{2}mv_{max}^2$ (1)

where

m is the mass of the block

$v_{max}=1.25 m/s$ is the maximum speed

We also know that the total energy is

$E=0.18 J$

Re-arranging eq.(1), we can find the mass:

$m=\frac{2E}{v_{max}^2}=\frac{2(0.18)}{(1.25)^2}=0.23 kg$

b)

The maximum speed in a spring-mass system is also given by

$v_{max} =\sqrt{\frac{k}{m}} A$

where

k is the spring constant

m is the mass

A is the amplitude

Here we have:

$v_{max}=1.25 m/s$ is the maximum speed

m = 0.23 kg is the mass

A = 14.0 cm = 0.14 m is the amplitude

Solving for k, we find the spring constant

$k=\frac{v_{max}^2}{A^2}m=\frac{1.25^2}{0.14^2}(0.23)=18.3 N/m$

c)

The frequency in a spring-mass system is given by

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$

where

k is the spring constant

m is the mass

In this problem, we have:

k = 18.3 N/m is the spring constant (found in part b)

m = 0.23 kg is the mass (found in part a)

Substituting and solving for f, we find the frequency of the system:

$f=\frac{1}{2\pi}\sqrt{\frac{18.3}{0.23}}=1.42 Hz$

d)

We can solve this part by using the law of conservation of energy; in fact, we have

$E=PE+KE=\frac{1}{2}kx^2 + \frac{1}{2}mv^2$

Where v is the speed of the system when the displacement is equal to x.

We know that the total energy of the system is

E = 0.18 J

Also we know that

k = 18.3 N/m is the spring constant

m = 0.23 kg is the mass

Substituting

x = 7.00 cm = 0.07 m

We can solve the equation to find the corresponding speed v:

$v=\sqrt{\frac{2E-kx^2}{m}}=\sqrt{\frac{2(0.18)-(18.3)(0.07)^2}{0.23}}=1.08 m/s$

#LearnwithBrainly

3 0

2 years ago

(a) As a soap bubble thins it becomes dark, because the path length difference becomes small compared with the wavelength of lig

OLga [1]

Answer:

t< 75 nm

Explanation:

A soap bubble is a thin film where when the beam enters the film it has a 180º phase change due to the refractive index and the wavelength changes between

λ = λ₀ / n

In the case of constructive interference in the curve of the spherical film it is

2 nt = (m + ½) λ₀

Where t is the thickness of the film and n the refractive index that does not indicate that we use that of water n = 1.33, m is an integer. The thickness of the film for the first interference (m = 0) is

t = λ₀ / 4 n

A thickness less than this gives destructive interference.

Let's look for the thickness for the visible spectrum

Violet light λ₀ = 400 nm = 400 10⁻⁹ m

t₁ = 400 10⁻⁹ / 4 1.33

t₁ = 75.2 10-9 m

Red light λ₀ = 700 nm = 700 10⁻⁹ m

t₂ = 700 10⁻⁹ / 4 1.33

t₂ = 131.6 10⁻⁹ m

Therefore, for all wavelengths to have destructive interference, the thickness must be less than 75 10⁻⁹ m = 75 nm

b) a film like eta is very thin, it is achieved when gravity thins the pomp, but any movement or burst of air breaks it,

3 0

3 years ago

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timama [110]

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4 0

2 years ago

Read 2 more answers

The options are stationary, running, and walking

Sunny_sXe [5.5K]

Answer:

A is walking

B is stationary

C is running

D is walking

E is stationary

4 0

2 years ago

How much heat (in kJ) is required to warm 13.0 g of ice, initially at -12.0 ∘C, to steam at 109.0 ∘C?

Sunny_sXe [5.5K]

<h2>Answer:</h2>

39.699 kJ

<h2>Explanation:</h2>

In this situation, there are a few transformations as follows;

(i) Heat required to warm the ice from -12°C to its melting point.

(ii) Heat required to melt the ice.

(iii) Heat required to boil the melted ice to boiling point (i.e to steam)

(iv) Heat required to vapourize the water

(v) Heat required to heat the steam from 100°C to 109.0°C

The sum of all the heat processes gives the heat required to warm the ice to steam;

<h3><em>Calculate each of these heat processes</em></h3>

Let the heat required to warm the ice from -12.0°C to its melting point (0°C) be Q₁.

Q₁ = m x c x ΔT -----------------------(i)

Where;

m = mass of ice = 13.0g

c = specific heat capacity of ice = 2.09 J/g°C

ΔT = final temperature - initial temperature = 0°C - (-12°C) = 12°C

Substitute these values into equation (i) as follows;

Q₁ = 13.0 x 2.09 x 12 = 326.04 J

Let the heat required to melt the ice be Q₂. This heat is called the heat of fusion and it is given by;

Q₂ = m x L -----------------------(ii)

Where;

m = mass of ice = 13.0g

L = latent heat of fusion of ice = 333.6 J/g

Substitute these values into equation (ii) as follows;

Q₂ = 13.0 x 333.6

Q₂ = 4336.8 J

Let the heat required to boil the melted ice from 0°C to boiling point of 100°C be Q₃.

Q₃ = m x c x ΔT -----------------------(i)

Where;

m = mass of melted ice (water) which is still 13.0g

c = specific heat capacity of melted ice (water) = 4.2 J/g°C

ΔT = final temperature - initial temperature = 100°C - 0°C = 100°C

Substitute these values into equation (i) as follows;

Q₁ = 13.0 x 4.2 x 100 = 5460 J

Let the heat required to vaporize the water (melted ice) be Q₄. This heat is called the heat of vaporization and it is given by;

Q₄ = m x L -----------------------(iv)

Where;

m = mass of ice = 13.0g

L = latent heat of vaporization of water = 2257 J/g

Substitute these values into equation (iv) as follows;

Q₄ = 13.0 x 2257

Q₄ = 29341 J

Let the heat required to heat the steam from 100°C to 109°C be Q₅.

Q₅ = m x c x ΔT -----------------------(i)

Where;

m = mass of steam which is still 13.0g

c = specific heat capacity of steam = 2.01 J/g°C

ΔT = final temperature - initial temperature = 109.0°C - 100°C = 9°C

Substitute these values into equation (i) as follows;

Q₅ = 13.0 x 2.01 x 9 = 235.17J

<em>Finally:</em>

<em>Sum all the heat values together;</em>

Q = Q₁ + Q₂ + Q₃ + Q₄ + Q₅

Q = 326.04 + 4336.8 + 5460 + 29341 + 235.17

Q = 39699.01 J

Q = 39.699 kJ

Therefore, the amount of heat (in kJ) required is 39.699

7 0

2 years ago