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4 years ago

Describe how this instrument works. Operation:

Engineering

1 answer:

LenKa [72]4 years ago

5 0

Answer:

An instrument takes the image that you are looking at and reflects it off the aluminum coating. The image then bounces off the coating to the glass. The glass is angle toward the eyepiece so that you have a full visual of the image you are trying to see. The telescope allows us to see further; they are able to collect and focus more light from distant objects than our eyes can alone. This is achieved by refracting or reflecting the light using lenses or mirrors.

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A 50 Ibm block is moving at 4 ft/s. Find Its total kinetic energy? Also find Its kinetic energy per unit mass is

Kamila [148]

Answer:

Total kinetic energy= 400 puondal foot

Kinetic energy per unit mass= 8 (poundaul foot/lb)

Explanation:

To determine the total kinetic energy of the block we use the equation:

Kinetic energy = (1/2) × m × (v^{2}) = (1/2) × (50lb) × (4 foot/s)^{2}) =

= 400 poundal foot

To establish the kinetic energy per unit of mass we must simply divide the value of the total kinetic energy obtained previously by the mass of the block obtaining:

Kinetic energy per unit mass= (400 poundal foot) / (50lb) =

= 8 (poundaul foot/lb)

3 0

3 years ago

What is Applied Science?

andreev551 [17]

Answer:

Applied science is the application of existing scientific knowledge to practical applications, like technology or inventions. Within natural science, disciplines that are basic science develop basic information to predict and perhaps explain and understand phenomena in the natural world.

6 0

3 years ago

A 3.7 g mass is released from rest at C which has a height of 1.1 m above the base of a loop-the-loop and a radius of 0.2 m . Th

Dmitrij [34]

Answer:

Normal force = 0.326N

Explanation:

Given that:

mass released from rest at C = 3.7 g = 3.7 × 10⁻³ kg

height of the mass = 1.1 m

radius = 0.2 m

acceleration due to gravity = 9.8 m/s²

We are to determine the normal force pressing on the track at A.

To to that;

Let consider the conservation of energy relation; which says:

mgh = mgr + 1/2 mv²

gh = gr + 1/2 v²

gh - gr = 1/2v²

g(h-r) = 1/2v²

v² = 2g(h-r)

However; the normal force will result to a centripetal force; as such, using the relation

N =mv²/r

replacing the value for v² = 2g(h-r) in the above relation; we have:

Normal force = 2mg(h-r)/r

Normal force = 2 × 3.7 × 10⁻³ × 9.8 ( 1.1 - 0.2 )/ 0.2

Normal force = 0.065268/0.2

Normal force = 0.32634 N

Normal force = 0.326N

6 0

3 years ago

For better thermal control it is common to make catalytic reactors that have many tubes packed with catalysts inside a larger sh

Paul [167]

Answer:

the pressure drop is 0.21159 atm

Explanation:

Given that:

length of the reactor L = 2.5 m

inside diameter of the reactor d= 0.025 m

diameter of alumina sphere $dp$ = 0.003 m

particle density = 1300 kg/m³

the bed void fraction $\in =$ 0.38

superficial mass flux m = 4684 kg/m²hr

The Feed is methane with pressure P = 5 bar and temperature T = 400 K

Density of the methane gas $\rho$ = 0.15 mol/dm ⁻³

viscosity of methane gas $\mu$ = 1.429 x 10⁻⁵ Pas

The objective is to determine the pressure drop.

Let first convert the Density of the methane gas from 0.15 mol/dm ⁻³ to kg/m³

SO; we have :

Density = 0.15 mol/dm ⁻³

Molar mass of methane gas (CH₄) = (12 + (1×4) ) = 16 mol

Density = $0.1 5 *\dfrac{16}{0.1^3}$

Density = 2400

Density $\rho_f$ = 2.4 kg/m³

Density = mass /volume

Thus;

Volume = mass/density

Volume of the methane gas = 4684 kg/m²hr / 2.4 kg/m³

Volume of the methane gas = 1951.666 m/hr

To m/sec; we have :

Volume of the methane gas = 1951.666 * 1/3600 m/sec = 0.542130 m/sec

$Re = \dfrac{dV \rho}{\mu}$

$Re = \dfrac{0.025*0.5421430*2.4}{1.429*10^5}$

$Re=2276.317705$

For Re > 1000

$\dfrac{\Delta P}{L}=\dfrac{1.75 \rho_f(1- \in)v_o}{\phi_sdp \in^3}$

$\dfrac{\Delta P}{2.5}=\dfrac{(1.75 *2.4)(1- 0.38)*0.542130}{1*0.003 (0.38)^3}$

$\Delta P=8575.755212*2.5$

$\Delta = 21439.38803 \ Pa$

To atm ; we have

$\Delta P = \dfrac{21439.38803 }{101325}$

$\Delta P =0.2115903087 \ atm$

ΔP ≅ 0.21159 atm

Thus; the pressure drop is 0.21159 atm

4 0

3 years ago

In the idealized Otto cycle, heat is added during: a. Isentropic Compression b. Constant (minimum) volume c. Constant (maximum)

docker41 [41]

Answer:

(b) Constant (minimum) volume

Explanation:

In the idealized Otto cycle there are 4 process that are

Reversible adiabatic compression
Addition of heat at constant volume
Reversible adiabatic expansion
Rejection of constant volume

So from above discussion we can see that heat is added when there is constant (minimum) volume which is given in option (b) so option (b) will be the correct answer

3 0

3 years ago