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julia-pushkina [17]

4 years ago

7

What mass of a material with density rho is required to make a hollow spherical shell having inner radius r1 and outer radius r2

? (Use any variable or symbol stated above as necessary.)

1 answer:

satela [25.4K]4 years ago

8 0

Answer:

$\frac{4}{3}\rho\pi(r_2^3 - r_1^3)$

Explanation:

The volume of the shell is the difference between the volume of the outer sphere and the volume of the inner sphere

$V = V_o - V_i = \frac{4}{3}\pi r_2^3 - \frac{4}{3}\pi r_1^3$

$V = \frac{4}{3}\pi(r_2^3 - r_1^3)$

The mass of the shell is the product of the volume and its density

$m = V\rho = \frac{4}{3}\rho\pi(r_2^3 - r_1^3)$

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7. 13 a turbine receives steam at 6 mpa, 600°c with an exit pressure of 600 kpa. Assume the turbine is adiabatic and neglect kin

AnnyKZ [126]

The work done by the turbine will be 708.2 kJ/kg. The work done by the turbine is the difference of the enthalpy at inlet and exit.

<h3 /><h3>What is temperature?</h3>

Temperature directs the hotness or coldness of a body. In clear terms, it is the method of finding the kinetic energy of particles within an entity. Faster the motion of particles, more the temperature.

If the given turbine is assumed to be reversible;

$\rm P_I$ (Initial pressure)=60 mpa = 60 bar

$\rm T_i$ (Initial temperature)=600° C

$\rm P_e$ (Exit pressure)=600 kpa=6 bar

The heat balance equation is;

$\rm q-W_t=h_e-h_i\\\\ q=0\\\\\ W_t=h_i-h_e$

The change in the entropy is;

$\rm S_2-S_1=\frac{\delta q}{dt}$

The work done by the turbine is;

$\rm W_t = h_i-h_e\\\\ W_t =3658.4 -29.5019\\\\ W_t =708.2 \ kJ/kg$

Hence,the work done by the turbine will be 708.2 kJ/kg.

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6 0

2 years ago

which force field can accelerate an electron, but never change its speed? a) electric field b) magnetic field c) both of these d

Drupady [299]

Option a; Electric field can accelerate an electron, but never change its speed

An electric field (also known as an E-field) is a physical field that surrounds electrically charged particles and exerts force on all other charged particles in the field, either attracting or repelling them. It can also refer to the physical field of a charged particle system. Electric fields are created by electric charges and time-varying electric currents. Electric and magnetic fields are both aspects of the electromagnetic field, one of nature's four fundamental interactions (also known as forces). Electric fields are significant in many areas of physics and are used in electrical technology. In atomic physics and chemistry, for example, the electric field is the attractive force that holds the atomic nucleus and electrons together in atoms. It is also the driving force behind chemical bonds between atoms.

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4 0

1 year ago

(a) What is the ionization energy of a hydrogen atom that is in the n = 6 excited state? (b) For a hydrogen atom, determine the

crimeas [40]

Answer:

(a) 0.3778 eV

(b) Ratio = 0.0278

Explanation:

The Bohr's formula for the calculation of the energy of the electron in nth orbit is:

$E=\frac {-13.6}{n^2}\ eV$

(a) The energy of the electron in n= 6 excited state is:

$E=\frac {-13.6}{6^2}\ eV$

$E=-0.3778\ eV$

Ionisation energy is the amount of this energy required to remove the electron. Thus, |E| = 0.3778 eV

(b) For first orbit energy is:

$E=\frac {-13.6}{1^2}\ eV$

$E=-13.6\ eV$

$Ratio=\frac {E_6}{E_1}$

$Ratio=\frac {-0.3778}{-13.6}$

Ratio = 0.0278

7 0

4 years ago

While working on her science fair project Venus connected a battery to a circuit that contained a light bulb. Venus decided to c

Dmitriy789 [7]

<h2>Answer:</h2>

<u>A) Increase the voltage by adding a bigger battery </u>

<h2>Explanation:</h2>

According to Ohm's law

V = IR

where V is voltage, I is current and R is the resistance. If we write the equation for resistance we would get

R= V / I

Here we can see that Voltage is directly proportional to Resistance so in order to keep the balance if we increase the resistance then we must increase the voltage to keep the current constant.

4 0

3 years ago

Read 2 more answers

A torsional pendulum consists of a disk of mass 450 g and radius 3.5 cm, hanging from a wire. If the disk is given an initial an

Montano1993 [528]

To solve this problem we will use the kinematic equations of angular motion, starting from the definition of angular velocity in terms of frequency, to verify the angular displacement and its respective derivative, let's start:

$\omega = 2\pi f$

$\omega = 2\pi (2.5)$

$\omega = 5\pi rad/s$

The angular displacement is given as the form:

$\theta (t) = \theta_0 cos(\omega t)$

In the equlibrium we have to $t=0, \theta(t) = \theta_0$ and in the given position we have to

$\theta(t) = \theta_0 cos(5\pi t)$

Derived the expression we will have the equivalent to angular velocity

$\frac{d\theta}{dt} = 2.7rad/s$

Replacing,

$\theta_0(sin(5\pi t))5\pi = 2.7$

Finally

$\theta_0 = \frac{2.7}{5\pi}rad = 9.848\°$

Therefore the maximum angular displacement is 9.848°

6 0

3 years ago