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Which of the following takes place in the combustion chamber of a gas turbine engine?A. A glow plug is used to add enough heat t

erica [24]

Answer:

Fuel oil is mixed with a proper portion of compressed air

Explanation:

A gas turbine has three main part, which are

combustion chamber
air compressor
power turbine

The combustion chamber is responsible for mixing fuel with a proper portion of compressed air.

The air compressor supplies air in sufficient quantity to satisfy the requirements of the combustion chamber

The power turbine produces the power to drive the air compressor.

6 0

3 years ago

A slender rod is 90.0 cm long and has mass 0.120 kg. A small 0.0200 kg sphere is welded to one end of the rod, and a small 0.080

deff fn [24]

Answer:

Speed of 0.08 kg mass when it will reach to the bottom position is 1.94 m/s

Explanation:

When rod is released from rest then due to unbalanced torque about the hinge the system will rotate

Now moment of inertia of the system is given as

$I = \frac{ML^2}{12} + \frac{m_1L^2}{4} + \frac{m_2L^2}{4}$

now we have

$M = 0.120 kg$

$m_1 = 0.02 kg$

$m_3 = 0.08 kg$

now we have

$I = \frac{0.120(0.90)^2}{12} + \frac{0.02(0.90)^2}{4} + \frac{0.08(0.90)^2}{4}$

so we have

$I = 8.1 \times 10^[-3} + 4.05 \times 10^[-3} + 0.0162$

$I = 0.02835$

now by energy conservation we can say work done by gravity must be equal to change in kinetic energy

so we have

$\frac{1}{2}I\omega^2 = m_1g \frac{L}{2} - m_2 g\frac{L}{2}$

$\frac{1}{2}(0.02835)\omega^2 = (0.08 - 0.02)(9.81)(0.45)$

$\omega = 4.32 rad/s$

Now speed of 0.08 kg mass when it reaches to bottom point is given as

$v = \omega \frac{L}{2}$

$v = 4.32 (0.45)$

$v = 1.94 m/s$

3 0

3 years ago

A plant dying after being exposed to poison represents a physical change. True False

nasty-shy [4]

Answer:

It does not represent a physical change

5 0

3 years ago

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If you were to drop a rock from a tall building, assuming that it had not yet hit the ground, and neglecting air resistance, aft

8_murik_8 [283]

Answer:

Explanation:

Using the equation of motion v = u + at to get the speed at which the object would be travelling.

v is the final speed (in m/s)

u is the initial velocity (in m/s)

a is the acceleration (in m/s²)

t is the time taken (in secs)

Given parameters

u = 0m/s

t = 10s

a = g = 9.8m/s²

Substituting this values into the formula;

v = 0+9.8(10)

v = 0+ 98

v = 98m/s

Hence the rock will be travelling at a speed of 98m/s.

6 0

3 years ago

What is the shortest-wavelength x-ray radiation in m that can be generated in an x-ray tube with an applied voltage of 93.3 kV?

VikaD [51]

(a) $1.33\cdot 10^{-11} m$

The x-rays in the tube are emitted as a result of the collisions of electrons (accelerated through the potential difference applied) on the metal target. Therefore, all the energy of the accelerated electron is converted into energy of the emitted photon:

$e \Delta V = \frac{hc}{\lambda}$

where the term on the left is the electric potential energy given by the electron, and the term on the right is the energy of the emitted photon, and where:

$e=1.6\cdot 10^{-19}C$ is the electron's charge

$\Delta V = 93.3 kV = 93300 V$ is the potential difference

$h=6.63\cdot 10^{-34} Js$ is the Planck constant

$c=3.00\cdot 10^8 m/s$ is the speed of light

$\lambda$ is the wavelength of the emitted photon

Solving the formula for $\lambda$ , we find:

$\lambda=\frac{hc}{e\Delta V}=\frac{(6.63\cdot 10^{-34})(3\cdot 10^8)}{(1.6\cdot 10^{-19})(93300)}=1.33\cdot 10^{-11} m$

(b) 93300 eV (93.3 keV)

The energy of the emitted photon is given by:

$E=\frac{hc}{\lambda}$

where

h is Planck constant

c is the speed of light

$\lambda=1.33\cdot 10^{-11} m$ is the wavelength of the photon, calculated previously

Substituting,

$E=\frac{(6.63\cdot 10^{-34})(3\cdot 10^8)}{1.33\cdot 10^{-11}}=1.50\cdot 10^{-14} J$

Now if we want to convert into electronvolts, we have to divide by the charge of the electron:

$E=\frac{1.50\cdot 10^{-14} J}{1.6\cdot 10^{-19} J/eV}=93300 eV$

(c) The following statements are correct:

The maximum photon energy is just the applied voltage times the electron charge. (1)

The value of the voltage in volts equals the value of the maximum photon energy in electron volts.

In fact, we see that statement (1) corresponds to the equation that we wrote in part (a):

$e \Delta V = \frac{hc}{\lambda}$

While statement (2) is also true, since in part (b) we found that the photon energy is 93.3 keV, while the voltage was 93.3 kV.

3 0

3 years ago

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