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3 years ago

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A rectangular block of copper metal has a mass of 100 g. The dimensions of the block are 2 cm by 2 cm by 2 cm. From this data, w

hat is the density of copper?

2 answers:

lina2011 [118]3 years ago

6 0

Can I have some more detail

shusha [124]3 years ago

4 0

I think it is 4 copper

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A block of mass 14.9 kg is pulled to the right by an applied force of 39.4 N. If it moves with constant velocity, how much frict

lakkis [162]

The frictional force is 39.4 N

Explanation:

We can solve this problem by applying Newton's 2nd law of motion: in fact, the net force acting on the block is equal to the product between its mass and its acceleration. So we can write

$\sum F = ma$

where

$\sum F$ is the net force

m is the mass

a is the acceleration

Here we know that the box is moving with constant velocity, so its acceleration is zero:

$a=0$

This means that the net force is also zero:

$\sum F=0$

The net force on the block is given by the applied force, forward, and the frictional force, backward:

$\sum F = F_a-F_f=0$

where

$F_a=39.4 N$ is the applied force

$F_f$ is the frictional force

Therefore, solving for $F_f$ ,

$F_f=F_a=39.4 N$

Learn more about friction:

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#LearnwithBrainly

8 0

3 years ago

Steam enters an adiabatic turbine steadily at 7 MPa, 5008C, and 45 m/s, and leaves at 100 kPa and 75 m/s. If the power output of

marusya05 [52]

Answer:

a) $\dot m = 6.878\,\frac{kg}{s}$ , b) $T = 104.3^{\textdegree}C$ , c) $\dot S_{gen} = 11.8\,\frac{kW}{K}$

Explanation:

a) The turbine is modelled by means of the First Principle of Thermodynamics. Changes in kinetic and potential energy are negligible.

$-\dot W_{out} + \dot m \cdot (h_{in}-h_{out}) = 0$

The mass flow rate is:

$\dot m = \frac{\dot W_{out}}{h_{in}-h_{out}}$

According to property water tables, specific enthalpies and entropies are:

State 1 - Superheated steam

$P = 7000\,kPa$

$T = 500^{\textdegree}C$

$h = 3411.4\,\frac{kJ}{kg}$

$s = 6.8000\,\frac{kJ}{kg\cdot K}$

State 2s - Liquid-Vapor Mixture

$P = 100\,kPa$

$h = 2467.32\,\frac{kJ}{kg}$

$s = 6.8000\,\frac{kJ}{kg\cdot K}$

$x = 0.908$

The isentropic efficiency is given by the following expression:

$\eta_{s} = \frac{h_{1}-h_{2}}{h_{1}-h_{2s}}$

The real specific enthalpy at outlet is:

$h_{2} = h_{1} - \eta_{s}\cdot (h_{1}-h_{2s})$

$h_{2} = 3411.4\,\frac{kJ}{kg} - 0.77\cdot (3411.4\,\frac{kJ}{kg} - 2467.32\,\frac{kJ}{kg} )$

$h_{2} = 2684.46\,\frac{kJ}{kg}$

State 2 - Superheated Vapor

$P = 100\,kPa$

$T = 104.3^{\textdegree}C$

$h = 2684.46\,\frac{kJ}{kg}$

$s = 7.3829\,\frac{kJ}{kg\cdot K}$

The mass flow rate is:

$\dot m = \frac{5000\,kW}{3411.4\,\frac{kJ}{kg} -2684.46\,\frac{kJ}{kg}}$

$\dot m = 6.878\,\frac{kg}{s}$

b) The temperature at the turbine exit is:

$T = 104.3^{\textdegree}C$

c) The rate of entropy generation is determined by means of the Second Law of Thermodynamics:

$\dot m \cdot (s_{in}-s_{out}) + \dot S_{gen} = 0$

$\dot S_{gen}=\dot m \cdot (s_{out}-s_{in})$

$\dot S_{gen} = (6.878\,\frac{kg}{s})\cdot (7.3829\,\frac{kJ}{kg\cdot K} - 6.8000\,\frac{kJ}{kg\cdot K} )$

$\dot S_{gen} = 11.8\,\frac{kW}{K}$

4 0

3 years ago

When energy is changed from one form to another, ____.

vampirchik [111]

I think b is the answer but im not 100% sure

5 0

4 years ago

Read 2 more answers

During a test, a NATO surveillance radar system, operating at 37 GHz at 182 kW of power, attempts to detect an incoming stealth

Marta_Voda [28]

(a) $2.68\cdot 10^{-6} W/m^2$

The intensity of an electromagnetic wave is given by

$I=\frac{P}{A}$

where

P is the power

A is the area of the surface considered

For the waves in the problem,

$P=182 kW = 1.82\cdot 10^5 W$ is the power

The area is a hemisphere of radius

$r=104 km=1.04\cdot 10^5 m$

so

$A=2\pi r^2=2\pi (1.04\cdot 10^5 m)^2=6.8\cdot 10^{10} m^2$

So, the intensity is

$I=\frac{1.82\cdot 10^5 W}{6.8\cdot 10^{10}m^2}=2.68\cdot 10^{-6} W/m^2$

(b) $5.9\cdot 10^{-7} W$

In this case, the area of the reflection is

$A=0.22 m^2$

So, if we use the intensity of the wave that we found previously, we can calculate the power of the aircraft's reflection using the same formula:

$P=IA=(2.68\cdot 10^{-6} W/m^2)(0.22 m^2)=5.9\cdot 10^{-7} W$

(c) $8.7\cdot 10^{-18} W/m^2$

We said that the power of the waves reflected by the aircraft is

$P=5.9\cdot 10^{-7} W$

If we assume that the reflected waves also propagate over a hemisphere of radius

$r=104 km=1.04\cdot 10^5 m$

which has an area of

$A=2\pi r^2=2\pi (1.04\cdot 10^5 m)^2=6.8\cdot 10^{10} m^2$

Then the intensity of the reflected waves at the radar site will be

$I=\frac{P}{A}=\frac{5.9\cdot 10^{-7} W}{6.8\cdot 10^{10} m^2}=8.7\cdot 10^{-18} W/m^2$

(d) $8.1\cdot 10^{-8} V/m$

The intensity of a wave is related to the maximum value of the electric field by

$I=\frac{1}{2}c\epsilon_0 E_0^2$

where

c is the speed of light

$\epsilon_0$ is the vacuum permittivity

$E_0$ is the maximum value of the electric field vector

Solving the equation for $E_0$ ,

$E_0=\sqrt{\frac{2I}{c\epsilon_0}}=\sqrt{\frac{2(8.7\cdot 10^{-18} W/m^2)}{(3\cdot 10^8 m/s)(8.85\cdot 10^{-12} F/m)}}=8.1\cdot 10^{-8} V/m$

(e) $1.9\cdot 10^{-16} T$

The maximum value of the magnetic field vector is given by

$B_0 = \frac{E_0}{c}$

Substituting the values,

$B_0 = \frac{(8.1\cdot 10^{-8} V/m)}{3\cdot 10^8 m/s}=2.7\cdot 10^{-16} T$

And the rms value of the magnetic field is given by

$B_{rms} = \frac{B_0}{\sqrt{2}}=\frac{2.7\cdot 10^{-16} T}{\sqrt{2}}=1.9\cdot 10^{-16} T$

7 0

3 years ago

A battery connects to a resistor. The rate of energy dissipated by the resistor is 1.44 W. The rate of energy supplied by the ba

wel

Answer:

(C) Equal to 1.44 watt

Explanation:

We have given energy dissipated through the resistor = 1.44 watt

We have to find the energy supplied by the battery

From Telegence theorem we know that energy dissipated through resistor is equal to the energy supplied

As energy supplied is given as 1.44 watt

So energy supplied by the battery will be equal to 1.44 watt

So option (c) will be correct answer

3 0

3 years ago