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3 years ago

Rotational dynamics physics question please help.

Physics

1 answer:

soldier1979 [14.2K]3 years ago

8 0

Looking at the force-time graph, wouldn't the force be integral fdt between 0 and 10s, a sort of "smoothed out pulse" ? And it looks like a familiar bell shaped curve. doesn't that produce a turning effect/torque and isn't there something about a circular analogue to F=ma in newtonian linear mechs ???
Looks like a difficult question for 5 points

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When frequency division multiplexing (FDM) is used to aggregate several modulated channels together, no separation between chann

zaharov [31]

Answer:

False.

Separation between channel is required when frequency division multiplexing (FDM) is used to aggregate several modulated channels together.

Explanation:

In Frequency Division Multiplexing (FDM), the total bandwidth is divided to a set of frequency bands that do not overlap. Each of these bands is a carrier of a different signal that is generated and modulated by one of the sending devices.

The frequency bands are separated from one another by strips of unused frequencies called the guard bands, to prevent overlapping of signals.

The modulated signals are combined together using a multiplexer (MUX) in the sending end. The combined signal is transmitted over the communication channel, thus allowing multiple independent data streams to be transmitted simultaneously. At the receiving end, the individual signals are extracted from the combined signal by the process of demultiplexing (DEMUX).

6 0

3 years ago

Which of these organs is an endocrine structure?

Gre4nikov [31]

Answer: Actually three of them are. The ovaries, the uterus and fallopian tubes.

6 0

2 years ago

A 2.0 kg wood block is launched up a wooden ramp that is inclined at a 30˚ angle. The block’s initial speed is 10 m/s. What vert

WARRIOR [948]

Answer:

The wood block reaches a height of 4.249 meters above its starting point.

Explanation:

The block represents a non-conservative system, since friction between wood block and the ramp is dissipating energy. The final height that block can reach is determined by Principle of Energy Conservation and Work-Energy Theorem. Let suppose that initial height has a value of zero and please notice that maximum height reached by the block is when its speed is zero.

$\frac{1}{2}\cdot m \cdot v^{2} = m \cdot g\cdot h + \mu_{k}\cdot m\cdot g\cdot s \cdot \sin \theta$

$\frac{1}{2}\cdot v^{2} = g\cdot h + \mu_{k}\cdot g\cdot \left(\frac{h}{\sin \theta} \right)\cdot \sin \theta$

$\frac{1}{2}\cdot v^{2} = g\cdot h +\mu_{k}\cdot g\cdot h$

$\frac{1}{2}\cdot v^{2} = (1 +\mu_{k})\cdot g\cdot h$

$h = \frac{v^{2}}{2\cdot (1 + \mu_{k})\cdot g}$ (1)

Where:

$h$ - Maximum height of the wood block, in meters.

$v$ - Initial speed of the block, in meters per second.

$\mu_{k}$ - Kinetic coefficient of friction, no unit.

$g$ - Gravitational acceleration, in meters per square second.

$m$ - Mass, in kilograms.

$s$ - Distance travelled by the wood block along the wooden ramp, in meters.

$\theta$ - Inclination of the wooden ramp, in sexagesimal degrees.

If we know that $v = 10\,\frac{m}{s}$ , $\mu_{k} = 0.20$ and $g = 9.807\,\frac{m}{s^{2}}$ , then the height reached by the block above its starting point is:

$h = \frac{\left(10\,\frac{m}{s} \right)^{2}}{2\cdot (1+0.20)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}$

$h = 4.249\,m$

The wood block reaches a height of 4.249 meters above its starting point.

5 0

3 years ago

4. Suppose the observed motion of a moving airplane, illustrated by a distance vs. time graph, is nearly a straight line with ze

jekas [21]

It shows that the airplane covers equal distance in equal time interval, that's it has a straight line from the origin.

The plane is moving at uniform speed.

7 0

3 years ago

A parallel combination of a 1.13-μF capacitor and a 2.85-μF one is connected in series to a 4.25-μF capacitor. This three-capaci

Nata [24]

Answer:

(a) Charge of 4.25 μF capacitor is 35.46 μC.

(b) Charge of 1.13 μF capacitor is 10.05 μC.

(c) Charge of 2.85 μF capacitor is 25.36 μC.

Explanation:

Let C₁ , C₂ and C₃ are the capacitor which are connected to the battery having voltage V. According to the problem, C₁ and C₂ are connected in parallel. There equivalent capacitance is:

C₄ = C₁ + C₂

Substitute 1.13 μF for C₁ and 2.85 μF for C₂ in the above equation.

C₄ = ( 1.13 + 2.85 ) μF = 3.98 μF

Since, C₄ and C₃ are connected in series, there equivalent capacitance is:

C₅ = $\frac{C_{3}C_{4} }{C_{3} + C_{4} }$

Substitute 4.25 μF for C₃ and 3.98 μF for C₄ in the above equation.

C₅ = $\frac{4.25\times3.98 }{4.25 + 3.98 }$

C₅ = 2.05 μF

The charge on the equivalent capacitance is determine by the relation :

Q = C₅ V

Substitute 2.05 μF for C₅ and 17.3 volts for V in the above equation.

Q = 2.05 μF x 17.3 = 35.46 μC

Since, the capacitors C₃ and C₄ are connected in series, so the charge on these capacitors are equal to the charge on the equivalent capacitor C₅.

Charge on the capacitor, C₃ = 35.46 μC

Charge on the capacitor, C₄ = 35.46 μC

Voltage on the capacitor C₄ = $\frac{Q}{C_{4} }$ = $\frac{35.46\times10^{-6} }{3.98\times10^{-6}}$ = 8.90 volts

Since, C₁ and C₂ are connected in parallel, the voltage drop on both the capacitors are same, that is equal to 8.90 volts.

Charge on the capacitor, C₁ = C₁ V = 1.13 μF x 8.90 = 10.05 μC

Charge on the capacitor, C₂ = C₂ V = 2.85 μF x 8.90 = 25.36 μC

5 0

3 years ago