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4 years ago

"what factors determine the duration of the quasi-stable output pulse from a one-shot multivibrator?"

1 answer:

6 0

<span>Answer: R-C network,connected externally to the 555 timer determine the duration of the quasi-stable output pulse from a one-shot multivibrator.</span>

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A series circuit contains a 100.0-Ω resistor, a 0.450-H inductor, a 0.360-µF capacitor, and a time-varying source of emf providi

eduard

Answer:

A) The resonance angular frequency in an L-R-C circuit is

$\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{0.45\times0.36\times 10^{-6}}}\\\omega_0 = 2.48\times10^3 rad/s$

B) The current can be found by calculating the impedance of the system.

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

Since $X_L = X_C$ at resonance.

So, $I = V/Z = V/R = 30 / 100 = 0.3A$

5 0

3 years ago

Exit

Mila [183]

The gravitation field strength of Planet X is 10 times that of Earth's, as the object weight has increased 10 times. This is only possible if the planet is more massive or more dense (smaller diameter).

Therefore, the answer is C.

3 0

3 years ago

Object A is moving due east, while object B is moving due north. They collide and stick together in a completely inelastic colli

serious [3.7K]

Answer:

pf = 198.8 kg*m/s

θ = 46.8º N of E.

Explanation:

Since total momentum is conserved, and momentum is a vector, the components of the momentum along two axes perpendicular each other must be conserved too.
If we call the positive x- axis to the W-E direction, and the positive y-axis to the S-N direction, we can write the following equation for the initial momentum along the x-axis:

$p_{ox} = p_{oAx} + p_{oBx} (1)$

We can do exactly the same for the initial momentum along the y-axis:

$p_{oy} = p_{oAy} + p_{oBy} (2)$

The final momentum along the x-axis, since the collision is inelastic and both objects stick together after the collision, can be written as follows:

$p_{fx} = (m_{A} + m_{B} ) * v_{fx} (3)$

We can repeat the process for the y-axis, as follows:

$p_{fy} = (m_{A} + m_{B} ) * v_{fy} (4)$

Since (1) is equal to (3), replacing for the givens, and since p₀Bₓ = 0, we can solve for vfₓ as follows:

$v_{fx} = \frac{p_{oAx}}{(m_{A}+ m_{B)}} = \frac{m_{A}*v_{oAx} }{(m_{A}+ m_{B)}} =\frac{17.0kg*8.00m/s}{46.0kg} = 2.96 m/s (5)$

In the same way, we can find the component of the final momentum along the y-axis, as follows:

$v_{fy} = \frac{p_{oBy}}{(m_{A}+ m_{B)}} = \frac{m_{B}*v_{oBy} }{(m_{A}+ m_{B)}} =\frac{29.0kg*5.00m/s}{46.0kg} = 3.15 m/s (6)$

With the values of vfx and vfy, we can find the magnitude of the final speed of the two-object system, applying the Pythagorean Theorem, as follows:

$v_{f} = \sqrt{v_{fx} ^{2} + v_{fy} ^{2}} = \sqrt{(2.96m/s)^{2} + (3.15m/s)^{2}} = 4.32 m/s (7)$

The magnitude of the final total momentum is just the product of the combined mass of both objects times the magnitude of the final speed:

$p_{f} = (m_{A} + m_{B})* v_{f} = 46 kg * 4.32 m/s = 198.8 kg*m/s (8)$

Finally, the angle that the final momentum vector makes with the positive x-axis, is the same that the final velocity vector makes with it.
We can find this angle applying the definition of tangent of an angle, as follows:

$tg \theta = \frac{v_{fy}}{v_{fx}} = \frac{3.15 m/s}{2.96m/s} = 1.06 (9)$

⇒ θ = tg⁻¹ (1.06) = 46.8º N of E

8 0

3 years ago

A force AB of length 100m and weight 600N has its centre of gravity 4.0 m from the end, and lies on horizontal ground calculate

Stells [14]

Answer:

F = 24 N

Explanation:

In this exercise we have a bar l = 100 m with a center of gravity x = 4 m, which force is needed to lift it from the other end

Let's use the rotational equilibrium relationship, where we consider the counterclockwise rotations as positive and fix the reference system at the point closest to the center of gravity

∑ τ = 0

F l -x W = 0

F = $\frac{x}{l} \ W$