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3 years ago

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A series circuit that is connected to a 50 V, 60 Hz source is made up of 25 ohm resistor, capacite wieh X= 18 ohms, and inductor

with X 24 ohms, what is the circuit impedance? a. 25.70 b. 29.2Q 32.4Q c. d. 35.90 5

1 answer:

Allisa [31]3 years ago

6 0

Answer:

the impedance of the circuit is 25.7 ohms.

Explanation:

It is given that,

Voltage, V = 50 volts

Frequency, f = 60 Hz

Resistance, R = 25 ohms

Capacitive resistance, $X_C=18\ ohms$

Inductive resistance, $X_L=24\ ohms$

We need to find the impedance of the circuit. It is given by :

$Z=\sqrt{R^2+(X_L-X_C)^2}$

$Z=\sqrt{25^2+(24-18)^2}$

Z = 25.7 ohms

So, the impedance of the circuit is 25.7 ohms. Hence, this is the required solution.

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3 years ago

If he leaves the ramp with a speed of 31.0 m/s and has a speed of 29.5 m/s at the top of his trajectory, determine his maximum h

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Answer:

The maximum height reached is 4.63 m.

Explanation:

Given:

Initial speed of the man (u) = 31.0 m/s

Speed at the top of trajectory ( $u_x$ ) = 29.5 m/s

Acceleration due to gravity (g) = 9.8 m/s²

When the man reaches the top of the trajectory, the vertical component of velocity becomes zero and hence only horizontal component of velocity acts on him.

Also, since there is no net force acting in the horizontal direction, the acceleration is zero in the horizontal direction from Newton's second law. Thus, the horizontal component of velocity always remains the same.

So, speed at the top of trajectory is nothing but the horizontal component of initial velocity.

Now, initial velocity can be rewritten in terms of its components as:

$u^2=u_x^2+u_y^2$

Where, $u_x\ and\ u_y$ are the initial horizontal and vertical velocities of the man.

Now, plug in the given values and simplify. This gives,

$(31.0)^2=(29.5)^2+u_y^2\\\\961=870.25+u_y^2\\\\u_y^2=961-870.25\\\\u_y^2=90.75\ m^2/s^2--------1$

Now, we know that, for a projectile motion, the maximum height is given as:

$H=\frac{u_y^2}{2g}$

Plug in the value from equation (1) and 9.8 for 'g' to solve for 'H'. This gives,

$H=\frac{90.75}{2\times 9.8}\\\\H=4.63\ m$

Therefore, the maximum height reached is 4.63 m.

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4 years ago

Looking at the velocity vs time graph, between points b and c on the graph, the object was

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Between B and C, the object was going at a constant velocity; it is going 60 m/min consistently for that time frame.

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3 years ago

Assume you have a rocket in Earth orbit and want to go to Mars. The required change in velocity is ΔV≈9.6km/s . There are two op

Nostrana [21]

Answer: Part 1: Propellant Fraction (MR) = 8.76

Part 2: Propellant Fraction (MR) = 1.63

Explanation: The Ideal Rocket Equation is given by:

Δv = $v_{ex}.ln(\frac{m_{f}}{m_{e}} )$

Where:

$v_{ex}$ is relationship between exhaust velocity and specific impulse

$\frac{m_{f}}{m_{e}}$ is the porpellant fraction, also written as MR.

The relationship $v_{ex}$ is: $v_{ex} = g_{0}.Isp$

To determine the fraction:

Δv = $v_{ex}.ln(\frac{m_{f}}{m_{e}} )$

$ln(MR) = \frac{v}{v_{ex}}$

Knowing that change in velocity is Δv = 9.6km/s and $g_{0}$ = 9.81m/s²

<u>Note:</u> Velocity and gravity have different measures, so to cancel them out, transform km in m by multiplying velocity by 10³.

<u />

<u>Part 1</u>: Isp = 450s

$ln(MR) = \frac{v}{v_{ex}}$

ln(MR) = $\frac{9.6.10^{3}}{9.81.450}$

ln (MR) = 2.17

MR = $e^{2.17}$

MR = 8.76

<u>Part 2:</u> Isp = 2000s

$ln(MR) = \frac{v}{v_{ex}}$

ln (MR) = $\frac{9.6.10^{3}}{9.81.2.10^{3}}$

ln (MR) = 0.49

MR = $e^{0.49}$

MR = 1.63

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3 years ago

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Diagram A will result in the movement of the block to the left as a result of the forces.

<h3>What is Force?</h3>

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Read more about Force here brainly.com/question/25239010

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