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3 years ago

Does a thicker core make an electromagnet stronger?

Engineering

2 answers:

mel-nik [20]3 years ago

6 0

Answer:

Yes

Explanation:

The core of an electromagnet serves to stabilize the magnetic field created by the wire. The thicker the core, the more metal there is to amplify the current. Therefore, a thicker core does make an electromagnet stronger. Hope this helps!

iogann1982 [59]3 years ago

3 0

My conclusion is that the electromagnet with the thick core is stronger than the electromagnet with the thin core. Wire size did not make a significant difference in electromagnet strength.

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Which allows a user to run applications on a computing device? Group of answer choices Application software CSS Operating system

sveticcg [70]

Answer:

The operating system

Explanation:

The job of the operating system is to manage system resources allowing the abstraction of the hardware, providing a simple user interface for the user. The operating system is also responsible for handling application's access to system resources.

For this purpose, the operating system allows a user to run applications on their computing device.

Cheers.

4 0

3 years ago

A heat engine receives 6 kW from a 250oC source and rejects heat at 30oC. Examine each of three cases with respect to the inequa

taurus [48]

Answer:

Explanation:

Given

$T_h=250^{\circ}C\approx 523\ K$

$T_L=30^{\circ}C\approx 303\ K$

$Q_1=6 kW$

From Clausius inequality

$\oint \frac{dQ}{T}=0$ =Reversible cycle

$\oint \frac{dQ}{T}$ =Irreversible cycle

$\oint \frac{dQ}{T}>0$ =Impossible

(a)For $P_{out}=3 kW$

Rejected heat $Q_2=6-3=3\ kW$

$\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}$

$=\frac{6}{523}-\frac{3}{303}=1.57\times 10^{-3} kW/K$

thus it is Impossible cycle

(b) $P_{out}=2 kW$

$Q_2=6-2=4 kW$

$\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}$

$=\frac{6}{523}-\frac{4}{303}=-1.73\times 10^{-3} kW/K$

Possible

(c)Carnot cycle

$\frac{Q_2}{Q_1}=\frac{T_1}{T_2}$

$Q_2=3.47\ kW$

$\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}$

$=\frac{6}{523}-\frac{3.47}{303}=0$

and maximum Work is obtained for reversible cycle when operate between same temperature limits

$P_{out}=Q_1-Q_2=6-3.47=2.53\ kW$

Thus it is possible

6 0

4 years ago

The 5-kg collar has a velocity of 5 m>s to the right when it is at A. It then travels along the smooth guide. Determine its s

Gnoma [55]

Answer:

The speed at point B is 5.33 m/s

The normal force at point B is 694 N

Explanation:

The length of the spring when the collar is in point A is equal to:

$lA=\sqrt{0.2^{2}+0.2^{2} }=0.2\sqrt{2}m$

The length in point B is:

lB=0.2+0.2=0.4 m

The equation of conservation of energy is:

$(Tc+Ts+Vc+Vs)_{A}=(Tc+Ts+Vc+Vs)_{B}$ (eq. 1)

Where in point A: Tc = 1/2 mcVA^2, Ts=0, Vc=mcghA, Vs=1/2k(lA-lul)^2

in point B: Ts=0, Vc=0, Tc = 1/2 mcVB^2, Vs=1/2k(lB-lul)^2

Replacing in eq. 1:

$\frac{1}{2}m_{c}v_{A}^{2}+0+m_{c}gh_{A}+ \frac{1}{2}k(l_{A}-l_{ul}) ^{2}=\frac{1}{2}m_{c}v_{B}^{2}+0+0+\frac{1}{2}k(l_{B}-l_{ul}) ^{2}$

Replacing values and clearing vB:

vB = 5.33 m/s

The balance forces acting in point B is:

Fc-NB-Fs=0

$\frac{m_{C}v_{B}^{2} }{R}-N_{B}-k(l_{B}-l_{ul})=0$

Replacing values and clearing NB:

NB = 694 N

6 0

4 years ago

Read 2 more answers

Jim starts walking from a bus stop. He walks north for 10 km. Next, he walks east for 5 km. He then walks south for 10 km. Next,

Luda [366]

Answer:

Explanation:

8 0

3 years ago

1 kg of saturated steam at 1000 kPa is in a piston-cylinder and the massless cylinder is held in place by pins. The pins are rem

BARSIC [14]

Answer:

The final specific internal energy of the system is 1509.91 kJ/kg

Explanation:

The parameters given are;

Mass of steam = 1 kg

Initial pressure of saturated steam p₁ = 1000 kPa

Initial volume of steam, = V₁

Final volume of steam = 5 × V₁

Where condition of steam = saturated at 1000 kPa

Initial temperature, T₁ = 179.866 °C = 453.016 K

External pressure = Atmospheric = 60 kPa

Thermodynamic process = Adiabatic expansion

The specific heat ratio for steam = 1.33

Therefore, we have;

$\dfrac{p_1}{p_2} = \left (\dfrac{V_2}{V_1} \right )^k = \left [\dfrac{T_1}{T_2} \right ]^{\dfrac{k}{k-1}}$

Adding the effect of the atmospheric pressure, we have;

p = 1000 + 60 = 1060

We therefore have;

$\dfrac{1060}{p_2} = \left (\dfrac{5\cdot V_1}{V_1} \right )^{1.33}$

$P_2= \dfrac{1060}{5^{1.33}} = 124.65 \ kPa$

$\left [\dfrac{V_2}{V_1} \right ]^k = \left [\dfrac{T_1}{T_2} \right ]^{\dfrac{k}{k-1}}$

$\left [\dfrac{V_2}{V_1} \right ]^{k-1} = \left \dfrac{T_1}{T_2} \right$

$5^{0.33} = \left \dfrac{T_1}{T_2} \right$

T₁/T₂ = 1.70083

T₁ = 1.70083·T₂

T₂ - T₁ = T₂ - 1.70083·T₂

Whereby the temperature of saturation T₁ = 179.866 °C = 453.016 K, we have;

T₂ = 453.016/1.70083 = 266.35 K

ΔU = 3× $c_v$ ×(T₂ - T₁)

$c_v$ = cv for steam at 453.016 K = 1.926 + (453.016 -450)/(500-450)*(1.954-1.926) = 1.93 kJ/(kg·K)

cv for steam at 266.35 K = 1.86 kJ/(kg·K)

We use cv given by (1.93 + 1.86)/2 = 1.895 kJ/(kg·K)

ΔU = 3× $c_v$ ×(T₂ - T₁) = 3*1.895 *(266.35 -453.016) = -1061.2 kJ/kg

The internal energy for steam = $U_g = h_g -pV_g$

$h_g$ = 2777.12 kJ/kg

$V_g$ = 0.194349 m³/kg

p = 1000 kPa

$U_{g1}$ = 2777.12 - 0.194349 * 1060 = 2571.11 kJ/kg

The final specific internal energy of the system is therefore, $U_{g1}$ + ΔU = 2571.11 - 1061.2 = 1509.91 kJ/kg.

3 0

3 years ago