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4 years ago

What charge does a “neutron” carry in an atom? Help

Engineering

1 answer:

Mademuasel [1]4 years ago

4 0

A neutron carries no charge in an atom

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The outer surface of a spacecraft in space has an emissivity of 0.6 and an absorptivity of 0.2 for solar radiation. If solar rad

kondaur [170]

Answer:

T surface = 3.9°C

Explanation:

given data

emissivity 0.6

absorptivity = 0.2

solar radiation is incident rate = 1200 W/m²

solution

we get here surface temperature by equality of emitted and absorbed heat rate that is

Q (absorbed) = Q (heat ) .................1

α Qinc = $\epsilon * \sigma *A*T^4(surface)$

T surface = $\sqrt[4]{\frac{\alpha Qinc}{\epsilon *\sigma * A} }$ ..........................2

put here value and we get

T surface = $\sqrt[4]{\frac{0.2*1000}{0.6*5.67**10^{-8}} }$

T surface = 276.9 K

T surface = 3.9°C

8 0

3 years ago

19/32 reduced to its lowest form

d1i1m1o1n [39]

Answer:

19/32

Explanation:

19/32 is already in its lowest form

8 0

3 years ago

Read 2 more answers

Steam enters a nozzle at 400°C and 800 kPa with a velocity of 10 m/s, and leaves at 300°C and 200 kPa while losing heat at a rat

jeyben [28]

Answer:606 m/s

Explanation:

Given

Steam Inlet temperature $400^{\circ} C$ and pressure 800 KPa

$h_1=3267.7 kJ/kg$

initial Velocity 10 m/s

Steam Outlet Velocity is $300^{\circ} C$ and pressure is 200 KPa

$h_2=3072.1 kJ/kg$

$\nu _2=1.31632$

From steam table

Heat loss 25 KW

inlet area 800 cm^2

applying Steady Flow Energy Equation

$h_1+\frac{1}{2}v_1^2++Q=h_2+\frac{1}{2} v_2^2+W$

$3267.7+\frac{1}{2000}10^2-25=3072.1+\frac{1}{2000}v_2^2$

$3267.7-3072.1+0.05-25=\frac{1}{2}v_2^2$

$v_2=\sqrt{195.65\times 2}=606 m/s$

and volume flow rate is $m=\dot{m}\mu _2=2.082\times 1.31623=2.74 m^3/s$

5 0

3 years ago

Read 2 more answers

Madison how do you do this

Pani-rosa [81]

Answer:

Do what

Explanation:

Nothing is there

8 0

3 years ago

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Technicians have to determine the flow rate of water in a pipe with the aid of a venturi installation and a mercury au.oaeter. T

Schach [20]

Answer:

Manometric difference x=142.85 mm.

Explanation:

Given :

Pipe diameter $d_1=40 mm$

venturi meter $d_2=20 mm$

We can know that discharge through venturi meter is given as

$Q=C_d\dfrac{A_1A_2\sqrt{2gh}}{\sqrt {A_1^2-A_2^2}}$

$A_1=1.24\times 10^{-3},A_2=3.12\times 10^{-4}$

$Q=A_1V_1$

$Q=1.24\times 10^{-3}\times 1.5=0.00186 m^3/s$

$0.00186=0.97\dfrac{1.24\times 10^{-3}\times 3.12\times 10^{-4} \sqrt{2gh}}{\sqrt {(1.24\times 10^{-3})^2-(3.12\times 10^{-4})^2}}$

h=1.8 m

We know that $h=x\left (\dfrac{\rho_{hg}}{\rho_w}-1\right )$

Where x is the manometric deflection

⇒ $1.8=x\left (\dfrac{13600}{1000}-1\right )$

So x=14.28 mm

Manometric difference x=142.85 mm.

7 0

3 years ago