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2 years ago

To span the distance between two switches, why is adding an intermediate switch not an attractive choice

1 answer:

3 0

Adding an intermediate switch is not an attractive choice to span the distance between two switches due to expense issues. It may have different ports.

<h3>What is a network switch?</h3>

A switch refers to a hardware tool capable of connecting different computer devices in a network.

A network switch can connect devices on a network by centralizing communications among many devices in a given local area network.

A switch may support different speeds for transmission of data (for example, one gigabit Gbps per second).

Learn more about network switches here:

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A toaster using a Nichrome heating element operates on 120 V. When it is switched on at 28 ∘С, the heating element carries an in

sukhopar [10]

Answer:

The final temperature of the element = 262.67°C

The power dissipated in the heating element initially = 163.21 W

The power dissipated in the heating element when the current reaches 1.23 A = 147.60 W

Explanation:

Our given parameters include;

A Nichrome heating element operates on 120 V.

Voltage (V) = 120V

Initial Current (I₁) = 1.36 A

Initial Temperature (T₁) = 28°C

Final Current (I₂) = 1.23 A

Final Temperature (T₂) = unknown ????

Temperature dependencies of resistance is given by:

$R_{T(2)}=R_1[1+\alpha (T_2-T_1)]$ ---------------------- (1)

in which R₁ is the resistance at temperature T₁

$R_{T(2)$ is the resistance at temperature T₂

Given that V= IR

R = $\frac{V}{I}$

Therefore, the resistance at temperature 28°C is;

$R_{28}= \frac{120V}{1.36A}$

= 88.24Ω

$R_{T(2)$ = $\frac{120V}{1.23A}$

= 97.56Ω

From (1) above;

$R_{T(2)}=R_1[1+\alpha (T_2-T_1)]$

97.56 = 88.24 [ 1 + 4.5×10⁻⁴(°C)⁻¹(T₂-28°C)]

$\frac{97.56}{88.24}= 1+(4.5*10^{-4})(T-28^0C)$

1.1056 - 1 = 4.5×10⁻⁴(°C)⁻¹(T₂-28°C)

0.1056 = 4.5×10⁻⁴(T₂-28°C)

$\frac{0.1056}{4.5*10^{-4}}= T-28^0C$

T - 28° C = 234.67

T = 234.67 + 28° C

T = 262.67 ° C

(b)

What is the power dissipated in the heating element initially and when the current reaches 1.23 A

The power dissipated in the heating element initially can be calculated as:

P = I²₁R₂₈

P = (1.36A)²(88.24Ω)

P = 163.209 W

P ≅ 163.21 W

The power dissipated in the heating element when the current reaches 1.23 A can be calculated as:

$P= I^2_2R_{T^0C$

P = (1.23)²(97.56Ω)

P = 147.598524

P ≅ 147.60 W

6 0

3 years ago

A 2 kg object with a weight of 20 N is being pulled up by a rope with a tension of 12N what is the acceleration of the object

son4ous [18]

Answer:

The object accelerates downward at 4 m/s² since the tension on the rope is less than weight of the object.

Explanation:

Given;

mass of the object, m = 2 kg

weigh of the object, W = 20 N

tension on the rope, T = 12 N

The acceleration of the object is calculated by applying Newton's second law of motion as follows;

T = F + W

T = ma + W

ma = T - W

$a = \frac{T-W}{m} \\\\a = \frac{12 - 20}{2} \\\\a = -4 \ m/s^2$ (the negative sign indicates deceleration of the object)

The object accelerates downward at 4 m/s² since the tension on the rope is less than weight of the object.

7 0

3 years ago

HEY CAN ANYONE PLS HELP ME OUT IN DIS PLS

Zanzabum

Answer:

Its the second choice

Explanation:

4 0

3 years ago

Read 2 more answers

An airplane is flying through a thundercloud at a height of 2000 m (This is a very dangerous thing to do because of updrafts, tu

Doss [256]

Answer:

$400000\ \text{N/C}$

Explanation:

$q_1$ = Charge at 3000 m = 40 C

$q_2$ = Charge at 1000 m = -40 C

$r_1$ = 3000 m

$r_2$ = 1000 m

k = Coulomb constant = $9\times10^9\ \text{Nm}^2/\text{C}^2$

Electric field due to the charge at 3000 m

$E_1=\dfrac{k|q_1|}{r_1^2}\\\Rightarrow E_1=\dfrac{9\times 10^9\times 40}{3000^2}\\\Rightarrow E_1=40000\ \text{N/C}$

Electric field due to the charge at 1000 m

$E_2=\dfrac{k|q_2|}{r_2^2}\\\Rightarrow E_2=\dfrac{9\times 10^9\times 40}{1000^2}\\\Rightarrow E_2=360000\ \text{N/C}$

Electric field at the aircraft is $E_1+E_2=40000+360000=400000\ \text{N/C}$ .

7 0

3 years ago

Elaborate on the reason why heating water up will increase the rate of dissolving for a solute.

Harrizon [31]

<span> D) <span> <span> <span> The average kinetic energy of the molecules that make up the solution also increases with increasing temperature. This increase in kinetic energy allows the solvent molecules to more effectively break apart the solute molecules that are held together by intermolecular attractions. </span></span></span></span>

5 0

3 years ago

Read 2 more answers