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Reil [10]
3 years ago
13

Find the current i in:

Physics
1 answer:
lawyer [7]3 years ago
5 0

Answer:

I = 24 A

Explanation:

This is Parallel Circuit and it is the first principle of parallel circuit that voltage will be equal in all components in the circuit

It includes 10 resistors Therefore the voltage across,

R1 = R2 = R3 = R4 = R5 = R6 = R7 = R8 = R9 = R10 = voltage in battery

<h2>Ohm's Law</h2>

We will apply Ohm's Law to each resistor to find its current because we know the voltage across each resistor is 12 V  and the resistance of each resistor is 5Ω

I (R1) = E (R1) / R1

I (R1) = 12v / 5Ω

I (R1) = 2.4 A

The value resistance E of all resistors are same therefore by applying the formula above the value of current in all resistors will be 2.4 A

The Total current in the circuit will be

I (total) = I (1) + I (2) + I (3) + I (4) + I (5) + I (6) + I (7) + I (8) + I (9) + I (10)

I (total) = 2.4 + 2.4 + 2.4 + 2.4 + 2.4 + 2.4 + 2.4 + 2.4 + 2.4 + 2.4

I (total) = 24 A

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the idling engines of a landing turbojet produce forward thrust when operating in a normal manner, but they can produce reverse
Finger [1]

Forward thrust has positive values and reverse thrust has negative values.

Thrust is a sudden push or pull in a certain direction.

a)

Flight speed u = 150 km/h

1 km/h = \frac{1}{3.6} km/s

therefore, 150 km/h =  41.67 km / s

The thrust force represents the horizontal or x-component of momentum equation:

T = m_{exhaust} * U_{exhaust} - U_{flight}

T = 50 * (150 - 41.67)

T = 5416.67 N

Therefore, the value of forward thrust is 5416.67 N.

b)

Now the exhaust velocity is now vertical due to reverse thrust application, then it has a zero horizontal component,

thus thrust equation is:

T = m_{exhaust} * U_{exhaust} - U_{flight}

T = 50 * (0 - 41.67)

T = -2083.5 N

Therefore, the thrust force T is -2083.5 N in the reverse direction.

c)

Now the exhaust velocity and flight velocity is zero, then it has a zero horizontal component, thus thrust is also zero becauseU_{exhaust} = U_{flight} = 0\\

T = 0

Therefore, there is no difference in two velocities in x direction.

The given question is incomplete, the complete question is,

"The idling engines of a landing turbojet produce forward thrust when operating in a normal manner, but they can produce reverse thrust if the jet is properly deflected. Suppose that while the aircraft rolls down the runway at 150 km/h the idling engine consumes air at 50 kg/s and produces an exhaust velocity of 150 m/s.

a. What is the forward thrust of this engine?

b. What are the magnitude and direction (i.e., forward or reverse) if the exhaust is deflected 90 degree without affecting the mass flow?

c. What are the magnitude and direction of the thrust (forward or reverse) after the plane has come to a stop, with 90 degree exhaust deflection and an airflow of 40 kg/s?"

To know more about thrust,

brainly.com/question/14552836

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A) Determine the x and y-components of the ball's velocity at t = 0.0s, 2.0, 3.0 secs.
malfutka [58]

The kinematic relationships we can find the position, acceleration and launch angle of the body on the planet Exidor.

a) the position are

      time (s)  x (m)   y(m)

        0            0          0

        2.0         3.6        1.2

        3.0         5.4        0.9

b) The aceleration is  g = 0.6 m / s²

c) The launch angle      θ = 33.7º

given parameters

  • the initial velocity of the body vₓ = 1.8m / s and v_y = 1.2 m / s
  • the movement times t = 1.0s, 2.0s and 3.0 s

to find

    a) position

    b) acceleration

    c) launch angle

Projectile launch is an application of kinematics to the movement of the body in two dimensions where there is no acceleration on the x axis and the y axis has the planet's gravity acceleration

b) To calculate the acceleration of the plant acting on the y-axis, we use that the vertical velocity of the body at the highest point is zero.

         v_y = v_{oy} - g t

where v and v({oy}  are the velocities of the body, g the acceleration of the planet's gravity and t the time

          0 = v_{oy} - gt

           g = v_{oy} / t

from the graph we observe that the highest point occurs for t = 2.0 s

           g = 1.2 / 2.0

           g = 0.6 m / s²

 

a) The position is requested for several times

X axis

in this axis there is no acceleration so we can use the uniform motion relationships

          vₓ = x / t

          x = vₓ t

where x is the position, vx is the velocity and t is the time

we calculate for the time

t = 0.0 s

          x₀ = 0

           

t = 2.0 s

          x₂ = 1.8 2

          x₂ = 3.6 m

t = 3.0 s

          x₃ = 1.8 3

          x₃ = 5.4 m

Y axis

In this axis there is the acceleration of the planet, let us use for the position the relation

          y = v_{oy} t - ½ g t²

t = 0.0 s

          y₀ = 0

          y₀ = 0 m

t = 2.0 s

         y₂ = 1.2 2 - ½ 0.6 2²

         y₂ = 1.2 m

t = 3.0 s

        y₃ = 1.2  3 - ½  0.6  3²

        y₃ = 0.9 m

c) the launch angle use the trigonometry relation

        tan θ = \frac{v_y}{v_x}

        θ = tan⁻¹ \frac{v_y}{v_x}

        θ = tan⁻¹ \frac{1.2}{1.8}

        θ = 33.7º

measured counterclockwise from the positive side of the x-axis

With the kinematic relationships we can find the position, acceleration and launch angle of the body on the planet Exidor.

a) the position are

      time (s)  x (m)   y(m)

        0            0          0

        2.0         3.6        1.2

        3.0         5.4        0.9

b) The aceleration is  g = 0.6 m / s²

c) The launch angle      θ = 33.7ºto)

learn more about projectile launch here:

brainly.com/question/10903823

4 0
2 years ago
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