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2 years ago

A circuit is supplied with a constant voltage. As the resistance of the circuit decreases, the power given to the circuit

1 answer:

7 0

Here, we are required to determine the reaction of the power given to a circuit supplied with a constant voltage as the distance of the circuit decreases.

In a circuit supposed with a constant voltage, as the resistance of the circuit decreases, the power given to the circuit Increases as evident in the formular: P = V²/R

The power given to a circuit is given by the formula;

Power, P = V × I

where, V = voltage supplied

and I = Current through the circuit.

However, from Ohm's law;

V = I × R

In essence, I = V/R

By substituting, I = V/R into the equation P = V × I

Therefore, we have;

P = V²/R.

Ultimately, if the circuit is supplied with a constant voltage, As the resistance of the circuit decreases, the power given to the circuit Increases.

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Answer:

Since it is falling freely, the only force on it is its weight, w.

w = m × g = 250 kg × 9.8 m/s^2 = 2450 Newton/N

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2 years ago

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A mass is attached to a vertical spring, which then goes into oscillation. At the high point of the oscillation, the spring is i

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Answer:

0.34 sec

Explanation:

Low point of spring ( length of stretched spring ) = 5.8 cm

midpoint of spring = 5.8 / 2 = 2.9 cm

Determine the oscillation period

at equilibrum condition

Kx = Mg

g= 9.8 m/s^2

x = 2.9 * 10^-2 m

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2 years ago

A helicopter carrying Dr. Evil takes off with a constant upward acceleration of 5.0 m/s2. Secret agent Austin Powers jumps on ju

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Answer:

a) $h=250\ m$

b) $\Delta h=0.0835\ m$

Explanation:

Given:

upward acceleration of the helicopter, $a=5\ m.s^{-2}$

time after the takeoff after which the engine is shut off, $t_a=10\ s$

Maximum height reached by the helicopter:

using the equation of motion,

$h=u.t+\frac{1}{2} a.t^2$

where:

u = initial velocity of the helicopter = 0 (took-off from ground)

t = time of observation

$h=0+0.5\times 5\times 10^2$

$h=250\ m$

time after which Austin Powers deploys parachute(time of free fall), $t_f=7\ s$

acceleration after deploying the parachute, $a_p=2\ m.s^{-2}$

height fallen freely by Austin:

$h_f=u.t_f+\frac{1}{2} g.t_f^2$

where:

$u=$ initial velocity of fall at the top = 0 (begins from the max height where the system is momentarily at rest)

$t_f=$ time of free fall

$h_f=0+0.5\times 9.8\times 7^2$

$h_f=240.1\ m$

Velocity just before opening the parachute:

$v_f=u+g.t_f$

$v_f=0+9.8\times 7$

$v_f=68.6\ m.s^{-1}$

Time taken by the helicopter to fall:

$h=u.t_h+\frac{1}{2} g.t_h^2$

where:

$u=$ initial velocity of the helicopter just before it begins falling freely = 0

$t_h=$ time taken by the helicopter to fall on ground

$h=$ height from where it falls = 250 m

now,

$250=0+0.5\times 9.8\times t_h^2$

$t_h=7.1429\ s$

From the above time 7 seconds are taken for free fall and the remaining time to fall with parachute.

remaining time,

$t'=t_h-t_f$

$t'=7.1428-7$

$t'=0.1428\ s$

Now the height fallen in the remaining time using parachute:

$h'=v_f.t'+\frac{1}{2} a_p.t'^2$

$h'=68.6\times 0.1428+0.5\times 2\times 0.1428^2$

$h'=9.8165\ m$

Now the height of Austin above the ground when the helicopter crashed on the ground:

$\Delta h=h-(h_f+h')$

$\Delta h=250-(240.1+9.8165)$

$\Delta h=0.0835\ m$

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2 years ago

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