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3 years ago

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The current through a certain heater wire is found to be fairly independent of its temperature. If the current through the heate

r wire is doubled, the amount of energy delivered by the heater in a given time interval will(a) increase by a factor of two. (b) decrease by a factor of two. (c) increase by a factor of four.(d) decrease by a factor of four. (e) increase by a factor of eight.

1 answer:

irina [24]3 years ago

5 0

Answer:

(c) increase by a factor of four

Explanation:

energy = power x time, and power = resistance x current ^2. 2^2 = 4.

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The force of gravity between two objects is given by:
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where
G is the gravitational constant
m1 and m2 are the masses of the two objects
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3 years ago

Read 2 more answers

Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are 4 MPa, 500◦C, and 80 m/s, and the exit

Cerrena [4.2K]

Answer:

a) ΔEC=-23.4kW

b)W=12106.2kW

c) $A=0.01297m^2$

Explanation:

A)

The kinetic energy is defined as:

$\frac{m*vel^2}{2}$ (vel is the velocity, to differentiate with v, specific volume).

The kinetic energy change will be: Δ ( $\frac{mvel^2}{2}$ )= $\frac{m*vel_2^2}{2}-\frac{m*vel_1^2}{2}$

Δ ( $\frac{mvel^2}{2}$ )= $\frac{m}{2}*(vel_2^2-vel_1^2)$

Where 1 and 2 subscripts mean initial and final state respectively.

Δ( $\frac{mvel^2}{2}$ )= $\frac{12\frac{kg}{s}}{2}*(50^2-80^2)\frac{m^2}{s^2}=-23400W=-23.4kW$

This amount is negative because the steam is losing that energy.

B)

Consider the energy balance, with a neglective height difference: The energy that enters to the turbine (which is in the steam) is the same that goes out (which is in the steam and in the work done).

$H_1+\frac{m*vel_1^2}{2}=H_2+\frac{m*vel_2^2}{2}+W\\W=m*(h_1-h_2)+\frac{m}{2} *(vel_1^2-vel_2^2)$

We already know the last quantity: $\frac{m}{2} *(vel_1^2-vel_2^2)$ = $-$ Δ ( $\frac{mvel^2}{2}$ )=23400W

For the steam enthalpies, review the steam tables (I attach the ones that I used); according to that, $h_1=h(T=500C,P=4MPa)=3445.3\frac{kJ}{kg}$

The exit state is a liquid-vapor mixture, so its enthalpy is:

$h_2=h_f+xh_{fg}=289.23+0.92*2366.1=2483.4\frac{kJ}{kg}$

Finally, the work can be obtained:

$W=12\frac{kg}{s}*(3445.3-2438.4)\frac{kJ}{kg} +23.400kW)=12106.2kW$

C) For the area, consider the equation of mass flow:

$m=p*vel*A$ where $p$ is the density, and A the area. The density is the inverse of the specific volume, so $m=\frac{vel*A}{v}$

The specific volume of the inlet steam can be read also from the steam tables, and its value is: $0.08643\frac{m^3}{kg}$ , so:

$A=\frac{m*v}{vel}=\frac{12\frac{kg}{s}*0.08643\frac{m^3}{kg}}{80\frac{m}{s}}=0.01297m^2$

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3 years ago

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Vsevolod [243]

The gravitational force experienced by Earth due to the Moon is <u>equal to </u>the gravitational force experienced by the Moon due to Earth.

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The force that attracts any two objects/bodies with mass towards each other is defined as gravitational force. Generally the gravitational force is attractive, as it always pulls the masses together and never pushes them apart.

The gravitational force can be calculated effectively using the following formula: F=GMmr^2

where “G” is the gravitational constant.

Though gravity has the ability to pull the masses together, it is the weakest force in the nature.

The mass of the Earth and moon varies, but still the gravitational force felt by the Earth and Moon are alike.

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3 years ago