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4 years ago

When moving cylinders always remove and make

1 answer:

5 0

Unless cylinders are firmly secured on a special carrier intended for this purpose, regulators shall be removed and valve protection caps put in place before cylinders are moved. A suitable cylinder truck, chain, or other steadying device shall be used to keep cylinders from being knocked over while in use.

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A closely wound, circular coil with radius 2.80 cm has 720 turns. Part APart complete What must the current in the coil be if th

lara [203]

Answer:

I = 4.642 Ampere.

x = 2.145 cm

Explanation:

a) As we know, the magnetic field on the axis of the loop is given as

$B = \frac{ \mu NIa^{2} }{2(x^{2} + a^{2})^{\frac{3}{2} } }$

where a = radius of loop

x = point on the axis of loop

N = No of turns of coil

Current in the loop for which the magnetic field at the center is 0.0750 Tesla is given as x = 0

Therefore, the above equation can be rewritten as

$B_{x} = \frac{ \mu NI}{2a}$

I = $\frac{2aB}{μN}$

by putting values

I = $\frac{2 X 0.0750 T X 0.028 m}{4\pi X 10^{-7} T.\frac{m}{A} X 720}$

Current in the loop for which the magnetic field at the center is 0.0750 Tesla = I = 4.642 Ampere

Now for part b the magnetic field at a distance x from the center is given as

$B = \frac{ \mu N I a^{2} }{2(x^{2} + a^{2})^{\frac{3}{2} } }$

multiply and divide by a on both sides we get

$B = \frac{ \mu NI}{2a} X \frac{a^{3} }{(x^{2} + a^{2})^{\frac{3}{2} } }$

As we know, according to Biot sivorts law,the Magnetic Field at the Center of a circular loop is given as

$B = \frac{ \mu NI}{2a}$ ( Magnetic field at center) = $B_{c}$

So we got magnetic field at any point x as

$B_{x}$ = $B_{c}$ X $\frac{a^{3} }{(x^{2} + a^{2})^{\frac{3}{2} } }$

For magnetic field at x is half of the B at center

$B_{x}$ = $\frac{1}{2}$ $B_{c}$

from the above two equations

$\frac{a^{3} }{(x^{2} + a^{2})^{\frac{3}{2} } }$ = $\frac{1}{2}$

$(x^{2} + a^{2})^{3} = 4a^{6}$

$x = \sqrt{4^{\frac{1}{3}} -1 } X a^{}$

Putting a = 2.80 cm

We have x = 2.145 cm Ans

6 0

3 years ago

What is the line called that has the red arrow pointing to it in the attached picture?

Simora [160]

Answer:

csvadbvns egv,ekrhvybge e yrbge ngeeeeerhjyk4 r5y erhyniner mbrltjhnprihmb fghurijmb fm nbjrkfb

Explanation:

8 0

3 years ago

The free convection heat transfer coefficient on a thin hot vertical plate suspended in still air can be determined from observa

damaskus [11]

Answer:

h = 6.35 W/m².k

Explanation:

In order to solve this problem, we will use energy balance, taking the thin hot plate as a system. According to energy balance, the rate of heat transfer to surrounding through convection must be equal to the energy stored in the plate:

Rate of Heat Transfer Through Convection = Energy Stored in Plate

- h A (Ts - T₀) = m C dT/dt

where,

h = convection heat transfer coefficient = ?

A = Surface area of plate through which heat transfer takes place = 2 x 0.3 m x 0.3 m (2 is multiplied for two sides of thin plate) = 0.18 m²

Ts = Surface Temperature of hot thin plate = 225⁰C

T₀ = Ambient Temperature = 25°C

m = mass of plate = 3.75 kg

C = Specific Heat = 2770 J/kg. k

dT/dt = rate of change in plate temperature = - 0.022 K/s

Therefore,

- h (0.18 m²)(225 - 25) k = (3.75 kg)(2770 J/kg.k)(- 0.022 k/s)

h = (- 228.525 W)/(- 36 m².k)

4 0

3 years ago

Air is heated from 50 F to 200 F in a rigid container with a heat transfer of 500 Btu. Assume that the air behaves as an ideal g

marissa [1.9K]

Answer:

$V=68.86ft^3$

Explanation:

$T_1$ =10°C, $T_2$ =93.33°C

Q=500 btu=527.58 KJ

$P_1= 2atm$

If we assume that air is ideal gas PV=mRT, ΔU= $mC_v(T_2-T_1)$

Actually this is closed system so work will be zero.

Now fro first law

Q=ΔU= $mC_v(T_2-T_1)$ +W

⇒Q= $mC_v(T_2-T_1)$

527.58 = $m\times 0.71(200-50)$

m=4.9kg

PV=mRT

$200V=4.9\times 0.287\times (10+273)$

$V=1.95m^3$ ( $V=1m^3=35.31ft^3$ )

$V=68.86ft^3$

6 0

4 years ago

It is possible to maintain a pressure of 10 kPa in a condenser that is being cooled by river water entering at 20 °C?

vovangra [49]

Answer:

Yes, it is possible to maintain a pressure of 10 kPa in a condenser that is being cooled by river water that is entering at 20 °C because this temperature (20 °C) of the external cooling water is less than the saturation temperature of steam which is which is 45.81 °C, and heated by a boiler; as a result of this condition, coupled with the assumption that the turbine, pump, and interconnecting tube are adiabatic, and the condenser exchanges its heat with the external cooling river water, it possible to maintain a pressure of 10 kPa.

5 0

4 years ago