A spinning turbine can generate electricity only in the form of a/an ______

Five-gram samples of copper and aluminum are at room temperature. Both receive equal amounts of energy due to heat flow. The spe

hammer [34]

Answer:

c. The copper will get hotter than the aluminum.

Explanation:

The energy due to heat flow, i.e. the heat energy, is proportional to the product of the mass, the specific heat capacity, and the change in temperature:

$Heat=mass\times \text{specific heat capacity }\times\Delta T$

From which you obtain:

$\Delta T=\dfrac{Heat}{mass\times \text{specific heat capacity}}$

That equation tells that the change in temperature is inversely related to the product of the mass and the specific heat.

In the problem, the masses of the samples of copper and aluminum are equal (5.00g) and both samples receive equal amounts of energy due to heat flow, thus the only difference is the specific heat capacity of each sample.

From the above stated relationship between the change in temperature, the heat, the mass, and the specific heat capacity, under the assumption of all the other conditions equal (heat energy and mass), the higher the specific heat capacity the lower the change in temperature, and the lower the specific heat capacity the greater the change in temperature.

The heat capacity of copper (0.09 cal/g°C) is lower than the specific heat capacity of aluminum is (0.22 cal/g°C), thus the increase in temeperature of the copper sample will be greater than that of the aluminum sample. This means that the copper will get hotter than the aluminum (option c.)

5 0

2 years ago

. The inner and outer surfaces of a 4-m × 7-m brick wall of thickness 30 cm and thermal conductivity 0.69 W/m-K are maintained a

Anni [7]

Answer:

$\frac{dQ}{dt} = 966 W$

Explanation:

As we know that the rate of heat transfer due to temperature difference is given by the formula

$\frac{dQ}{dt} = \frac{KA(\Delta T)}{L}$

here we know that

$K = 0.69 W/m-K$

A = 4 m x 7 m

thickness = 30 cm

temperature difference is given as

$\Delta T = 20 - 5 = 15 ^oC$

now we have

$\frac{dQ}{dt} = \frac{(0.69W/m-K)(28 m^2)(15)}{0.30}$

$\frac{dQ}{dt} = 966 W$

4 0

2 years ago

A 10.0 N package of whole wheat flour is suddenly placed on the pan of a scale such as you find in grocery stores. The pan is su

inna [77]

Answer:

0.025 m

Explanation:

From the question,

Applying Hook's law

F = ke................... Equation 1

Where F = Force, k = spring constant of the scale, e = maximum distance at which the spring will compress.

make e the subject of the equation

e = F/k....................... Equation 2

Given: F = 10 N, e = 395 N/m

Substitute these values into equation 2

e = 10/395

e = 0.025 m

7 0

3 years ago

A mass of 0.5 kg hangs motionless from a vertical spring whose length is 1.10 m and whose unstretched length is 0.50 m. Next the

ser-zykov [4K]

Answer:

The maximum length during the motion is $L_{max} = 1.45m$

Explanation:

From the question we are told that

The mass is $m =0.5 kg$

The vertical spring length is $L = 1.10m$

The unstretched length is $L_{un} = 1.30m$

The initial speed is $v_i = 1.3m/s$

The new length of the spring $L_{new} = 1.30 m$

The spring constant k is mathematically represented as

$k = -\frac{F}{y}$

Where F is the force applied $= m * g = 0.5 * 9.8=4.9N$

y is the difference in weight which is $=1.10-0.50=0.6m$

The negative sign is because the displacement of the spring (i.e its extension occurs against the force F)

Now substituting values accordingly

$k = \frac{4.9}{0.6}$

$= 8.17 N/m$

The elastic potential energy is given as $E_{PE} = \frac{1}{2} k D^2$

where D is this the is the displacement

Since Energy is conserved the total elastic potential energy would be

$E_T = initial \ elastic\ potential \ energy + kinetic \ energy$

$E_T = \frac{1}{2} k D_{max}^2 = \frac{1}{2} k D^2 + \frac{1}{2} mv^2$

Substituting value accordingly

$\frac{1}{2} *8.17 *D_{max}^2 =\frac{1}{2} * 8.17*(1.30 - 0.50)^2 + \frac{1}{2} * 0.5 *1.30^2$

$4.085 * D_{max}^2 = 3.69$

$D^2_{max} = 0.9033$

$D_{max} = 0.950m$

So to obtain total length we would add the unstretched length

So we have

$L_{max} = 0.950 + 0.5 = 1.45m$

5 0

2 years ago

Read 2 more answers

A geneticist looks through a microscope to determine the phenotype of a fruit fly. The microscope is set to an overall magnifica

guajiro [1.7K]

Answer:

$f_{e}$ = 1.7 cm

Explanation:

The magnification of the compound microscope is given by the product of the magnification of each lens

M = M₀ $m_{e}$

M = - L/f₀ 25/ $f_{e}$

Where f₀ and $f_{e}$ are the focal lengths of the lens and eyepiece, respectively, all values in centimeters

In this exercise they give us the magnification (M = 400X), the focal length of the lens (f₀ = 0.6 cm), the distance of the tube (L = 16 cm), let's look for the focal length of the eyepiece ( $f_{e}$ )

$f_{e}$ = - L / f₀ 25 / M

Let's calculate

$f_{e}$ = - 16 / 0.6 25 / (-400)

$f_{e}$ = 1.67 cm

The minus sign in the magnification is because the image is inverted.

$f_{e}$ = 1.7 cm

6 0

2 years ago

A spinning turbine can generate electricity only in the form of a/an _______ current.