Discuss the relationship between amperage, voltage, and power.

A small sphere of reference-grade iron with a specific heat of 447 J/kg K and a mass of 0.515 kg is suddenly immersed in a water

elena-14-01-66 [18.8K]

Answer:

The specific heat of the unknown material is 131.750 joules per kilogram-degree Celsius.

Explanation:

Let suppose that sphere is cooled down at steady state, then we can estimate the rate of heat transfer ( $\dot Q$ ), measured in watts, that is, joules per second, by the following formula:

$\dot Q = m\cdot c\cdot \frac{T_{f}-T_{o}}{\Delta t}$ (1)

Where:

$m$ - Mass of the sphere, measured in kilograms.

$c$ - Specific heat of the material, measured in joules per kilogram-degree Celsius.

$T_{o}$ , $T_{f}$ - Initial and final temperatures of the sphere, measured in degrees Celsius.

$\Delta t$ - Time, measured in seconds.

In addition, we assume that both spheres experiment the same heat transfer rate, then we have the following identity:

$\frac{m_{I}\cdot c_{I}}{\Delta t_{I}} = \frac{m_{X}\cdot c_{X}}{\Delta t_{X}}$ (2)

Where:

$m_{I}$ , $m_{X}$ - Masses of the iron and unknown spheres, measured in kilograms.

$\Delta t_{I}$ , $\Delta t_{X}$ - Times of the iron and unknown spheres, measured in seconds.

$c_{I}$ , $c_{X}$ - Specific heats of the iron and unknown materials, measured in joules per kilogram-degree Celsius.

$c_{X} = \left(\frac{\Delta t_{X}}{\Delta t_{I}}\right)\cdot \left(\frac{m_{I}}{m_{X}} \right) \cdot c_{I}$

If we know that $\Delta t_{I} = 6.35\,s$ , $\Delta t_{X} = 4.59\,s$ , $m_{I} = 0.515\,kg$ , $m_{X} = 1.263\,kg$ and $c_{I} = 447\,\frac{J}{kg\cdot ^{\circ}C}$ , then the specific heat of the unknown material is:

$c_{X} = \left(\frac{4.59\,s}{6.35\,s} \right)\cdot \left(\frac{0.515\,kg}{1.263\,kg} \right)\cdot \left(447\,\frac{J}{kg\cdot ^{\circ}C} \right)$

$c_{X} = 131.750\,\frac{J}{kg\cdot ^{\circ}C}$

Then, the specific heat of the unknown material is 131.750 joules per kilogram-degree Celsius.

3 0

3 years ago

How much work is required to pull a wagon 6 meters if you use 70 N of force?

yKpoI14uk [10]

Answer:

..

Explanation:

7 0

4 years ago

100%

xxMikexx [17]

Answer:

1. The elastic potential energy is 0.0176 Joules

2. The kinetic energy of the pinball the instant it leaves the spring is 0.0176 Joules

3. The speed of the pinball the instant it leaves the spring is approximately 2.42212 m/s

4. The height of the part where the pinball is located on the machine above the ground is approximately 0.213 meters

Explanation:

The spring constant of the pinball machine's plunger, k = 22 N/m

The amount by which the pinball machine's plunger is compressed, x = 0.04 m

The mass of the pinball ball, m = 0.006 kg

1. The elastic potential energy, P.E. = 1/2·k·x²

By substitution, we get;

P.E. = 1/2 × 22 N/m × (0.04 m)² = 0.0176 J

The elastic potential energy, P.E. = 0.0176 J

2. At the instant the pinball leaves the spring, the plunger and therefore the force of the plunger no longer acts on the pinball

Since there are no external forces acting on the pinball to increase the speed of the pinball after it leaves the spring, the velocity reached is its maximum velocity, and therefore, the kinetic energy, K.E. is the maximum kinetic energy which by the conservation of energy, is equal to the initial potential energy

Therefore;

K.E. = P.E. = 0.0176 J

The kinetic energy of the pinball the instant it leaves the spring, K.E.= 0.0176 J

3. The kinetic energy, K.E., is given by the following formula;

K.E. = 1/2·m·v²

Where;

v = The speed or velocity of the object having kinetic energy K.E.

Therefore, from K.E. = 0.0176 J, and by plugging in the values of the variables, we have;

K.E. = 0.0176 J = 1/2 × 0.006 kg × v²

v² = 0.0176 J/(1/2 × 0.006 kg) = 88/15 m²/s²

v = √(88/15 m²/s²) ≈ (2·√330)/15 m/s ≈ 2.42212 m/s

The speed of the pinball the instant it leaves the spring, v ≈ 2.42212 m/s

4. The height of the pinball is given by the following kinematic equation of motion;

$v_h$ ² = u² - 2·g·h

Where;

$v_h$ = The velocity of the pinball at the given height = 1.3 m/s

u = v ≈ 2.42212 m/s (The initial velocity of the pinball as it the spring)

g = The acceleration due to gravity ≈ 9.8 m/s²

h = The height of the pinball above the ground

We get;

$v_h$ ² = 1.3² = 2.42212² - 2 × 9.8 × h

∴ h = (2.42212² - 1.3²)/(2 × 9.8) ≈ 0.213

The height of the part where the pinball is located on the machine above the ground, h ≈ 0.213 m

5 0

3 years ago

25. Describe how the atomic mass unit (amu) was derived and how the mass of the electron relates to the

Gwar [14]

Answer:

ok

Explanation:

7 0

4 years ago

Read 2 more answers

A diverging lens has a focal length of -30.0 cm. An object is placed 18.0 cm in front of this lens.

yulyashka [42]

Answer:

A) Calculate the distance

8 0

4 years ago