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Kobotan [32]
3 years ago
8

Of the three types of radiation listed below, which is the least penetrating and which is the most penetrating? -alpha -beta -ga

mma
Physics
1 answer:
Vikentia [17]3 years ago
3 0

Answer:

Alpha is the least penetrating radiation listed below.

Gamma is the most penetrating radiation in the given list.

These rays can be used to detect cancer, sterilizing the medical instrument, smoke detector, etc.

Radiation emitted from the radioactive elements also have very harmful. They affect the living being a lot. Waste from the nuclear reactor also contains harmful radiation which causes Land pollution if not disposed of properly.

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Which characterstics do venus and earth share
xxTIMURxx [149]

Answer:

Earth and Venus has similar size, mass and density. Both have a very similar composition, But atmosphere of Venus is composed only of carbon dioxide is not similar to Earth.

Explanation:

my mind

5 0
3 years ago
A rocket engine has a chamber pressure 4 MPa and a chamber temperature of 2000 K. Assuming isentropic expansion through the nozz
gladu [14]

This question is incomplete, the complete question is;

A rocket engine has a chamber pressure 4 MPa and a chamber temperature of 2000 K. Assuming isentropic expansion through the nozzle, and an exit Mach number of 3.2, what are the stagnation pressure and temperature in the exit plane of the nozzle?  Assume the specific heat ratio is 1.2.

Answer:

- stagnation pressure is 274.993 Mpa

- the stagnation temperature Tt is 4048 K

Explanation:

Given the data in the question;

To determine the stagnation pressure and temperature in the exit plane of the nozzle;

we us the expression;

Pt/P = (1 + (γ-1 / 2) M²)^(γ/γ -1) = ( Tt/T )^(γ/γ -1)

where Pt is stagnant pressure = ?

P is static pressure = 4 MPa = 4 × 10⁶ Pa  

Tt is stagnation temperature = ?

T is the static temperature  = 2000 K

γ is ratio of specific heats = 1.2

M is Mach number M = 3.2

we substitute

Pt/P = (1 + (γ-1 / 2) M²)^(γ/γ -1)

Pt = P(1 + (γ-1 / 2) M²)^(γ/γ -1)

Pt = 4 × 10⁶(1 + (1.2-1 / 2) 3.2²)^(1.2/1.2 -1)

Pt = 4 × 10⁶ × 68.7484

Pt = 274.993 × 10⁶ Pa

Pt = 274.993 Mpa

Therefore stagnation pressure is 274.993 Mpa

Now, to get our stagnation Temperature

Pt/P = ( Tt/T )^(γ/γ -1)

we substitute

274.993 × 10⁶ Pa / 4 × 10⁶ Pa =  ( Tt / 2000 )^(1.2/1.2 -1)

68.7484 =  Tt⁶ / 6.4 × 10¹⁹

Tt⁶ = 68.7484 × 6.4 × 10¹⁹

Tt⁶ = 4.3998976 × 10²¹

Tt = ⁶√(4.3998976 × 10²¹)

Tt = 4047.999 ≈ 4048 K

Therefore, the stagnation temperature Tt is 4048 K

6 0
3 years ago
An ocean thermal energy conversion system is being proposed for electric power generation. Such a system is based on the standar
defon

Answer:

Explanation:

Dear Student, this question is incomplete, and to attempt this question, we have attached the complete copy of the question in the image below. Please, Kindly refer to it when going through the solution to the question.

To objective is to find the:

(i) required heat exchanger area.

(ii) flow rate to be maintained in the evaporator.

Given that:

water temperature = 300 K

At a reasonable depth, the water is cold and its temperature = 280 K

The power output W = 2 MW

Efficiency \zeta = 3%

where;

\zeta = \dfrac{W_{out}}{Q_{supplied }}

Q_{supplied } = \dfrac{2}{0.03} \ MW

Q_{supplied } = 66.66 \ MW

However, from the evaporator, the heat transfer Q can be determined by using the formula:

Q = UA(L MTD)

where;

LMTD = \dfrac{\Delta T_1 - \Delta T_2}{In (\dfrac{\Delta T_1}{\Delta T_2} )}

Also;

\Delta T_1 = T_{h_{in}}- T_{c_{out}} \\ \\ \Delta T_1 = 300 -290 \\ \\ \Delta T_1 = 10 \ K

\Delta T_2 = T_{h_{in}}- T_{c_{out}} \\ \\ \Delta T_2 = 292 -290 \\ \\ \Delta T_2 = 2\ K

LMTD = \dfrac{10 -2}{In (\dfrac{10}{2} )}

LMTD = \dfrac{8}{In (5)}

LMTD = 4.97

Thus, the required heat exchanger area A is calculated by using the formula:

Q_H = UA (LMTD)

where;

U = overall heat coefficient given as 1200 W/m².K

66.667 \times 10^6 = 1200 \times A \times 4.97 \\ \\  A= \dfrac{66.667 \times 10^6}{1200 \times 4.97} \\ \\  \mathbf{A = 11178.236 \ m^2}

The mass flow rate:

Q_{H} = mC_p(T_{in} -T_{out} )  \\ \\  66.667 \times 10^6= m \times 4.18 (300 -292) \\ \\ m = \dfrac{  66.667 \times 10^6}{4.18 \times 8} \\ \\  \mathbf{m = 1993630.383 \ kg/s}

3 0
3 years ago
A 3.00-kg model airplane has velocity components of 5.00 m/s due east and 8.00 m/s due north. What is the plane’s kinetic energy
GalinKa [24]

Answer:

Kinetic energy, E = 133.38 Joules

Explanation:

It is given that,

Mass of the model airplane, m = 3 kg

Velocity component, v₁ = 5 m/s (due east)

Velocity component, v₂ = 8 m/s (due north)

Let v is the resultant of velocity. It is given by :

v=\sqrt{v_1^2+v_2^2}

v=\sqrt{5^2+8^2}=9.43\ m/s

Let E is the kinetic energy of the plane. It is given by :

E=\dfrac{1}{2}mv^2

E=\dfrac{1}{2}\times 3\ kg\times (9.43\ m/s)^2

E = 133.38 Joules

So, the kinetic energy of the plane is 133.38 Joules. Hence, this is the required solution.

5 0
3 years ago
Read 2 more answers
A bus travels north on some busy city streets for 2.5 km, and a trip
d1i1m1o1n [39]

Answer:

V = 4.63 m/s

V = 11.31 m/s

Explanation:

Given,

The distance traveled by the bus, towards north, d = 2.5 km

                                                                                     = 2500 m

The time taken by the trip is, t = 9 min

                                                  = 540 s

The velocity of the bus,

                                       V = d / t

                                           = 2500 / 540

                                          = 4.63 m/s

At another  point, the bus travels at a constant speed of v = 18 m/s

Therefore the velocity becomes

                                                V = (4.63 + 18)/2

                                                   = 11.31 m/s

Hence, the velocity of the  bus, V = 11.31 m/s

8 0
3 years ago
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