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grandymaker [24]

4 years ago

7

If an amplitude modulation (AM) has an amplitude modulating signal of 0.5 V and carrier amplitude of 1.3 V, what is the modulati

on index?

1 answer:

Olenka [21]4 years ago

3 0

Answer:

The modulation index in the amplitude modulation will be 0.384

Explanation:

We have given amplitude of adulating signal $A_m=0.5volt$

Amplitude of carrier signal $A_C=1.3volt$

We have to find modulation index

Modulation index is the ratio of amplitude of modulating signal and amplitude of carrier signal

So modulation index $m=\frac{A_m}{A_c}=\frac{0.5}{1.3}=0.384$

So the modulation index in the amplitude modulation will be 0.384

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A kettle transfers 6,000j of energy electrically. 1,500j of this is wasted. what is the efficiency of this kettle?

Mashcka [7]

Answer:75 percent

Explanation:so in order tro fin d thge efficiency i used the forumla ,efficency=useful output energy/input energyx100%,in order to use this formula i needed the output,which i found by subtracting the input energy with wasted energy,that gave me the output,and after founding the output,i put that into the formula,

output energy=input energy - wasted energy

output energy=6000j-1500j

output energy=4500

put that into the formula

efficiency =output energy/input enrgy x100%

efficiency=4500/6000 multiplied by 100%

efficiency=0.75x100%

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7 0

2 years ago

13.An airplane starts from rest at the end of a runway and accelerates at a constant rate. In the first second, the airplane tra

Murrr4er [49]

Answer:

The answer is 3.33m

Explanation:

The acceleration "a" is constant.

Acceleration is the variation of velocity over time,

$\frac{dv}{dt} = a$ .

solving the last equation

$\int_{v_0}^v dv = a\int_0^t dt \rightarrow v-v_0 = at$ ,

where $v_0=0$ because the airplane starts from rest.

Once again, velocity is the variation of distance over time.

$\frac{dx}{dt} = at \rightarrow \int_{x_0}^x dx = a\int_0^t t\ dt$

then

$x- x_0 = \frac{1}{2}at^2$

where $x_0=0$ if we consider the end of the runway as the initial point (this step is for simplicity but you can let it expressed, it's going to cancel anyway).

If $x=1.11\ m$ at $t=1s$ , then

$a = \frac{2x}{t^2} = 2.22\ m/s^2$

and the final expression for the distance is

$x = 1.11 t^2$ .

If t = 2s, x = 4.44 m. Which means thad the additional distance is

$x(2s) - x(1s) = 4.44 - 1.11 = 3.33\ m$

8 0

4 years ago

How do scientists know the true environmental lapse rate in a column of air?

stepan [7]

With weather ballons carry instruments to measure the temperature ,pressure,wind direction and wind speed

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4 years ago

Very far from earth (at R- oo), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitational force

Margaret [11]

Answer:

Speed of the spacecraft right before the collision: $\displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}$ .

Assumption: the earth is exactly spherical with a uniform density.

Explanation:

This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

the kinetic energy of this spacecraft, and
the (gravitational) potential energy of this spacecraft.

Let $m$ denote the mass of this spacecraft. At a distance of $R$ from the center of the earth (with mass $M_\text{e}$ ), the gravitational potential energy ( $\mathrm{GPE}$ ) of this spacecraft would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}$ .

Initially, $R$ (the denominator of this fraction) is infinitely large. Therefore, the initial value of $\mathrm{GPE}$ will be infinitely close to zero.

On the other hand, the question states that the initial kinetic energy ( $\rm KE$ ) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.

Right before the collision, the spacecraft would be very close to the surface of the earth. The distance $R$ between the spacecraft and the center of the earth would be approximately equal to $R_\text{e}$ , the radius of the earth.

The $\mathrm{GPE}$ of the spacecraft at that moment would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ .

Subtract this value from zero to find the loss in the $\rm GPE$ of this spacecraft:

$\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}$

Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the $\rm GPE$ of this spacecraft would be equal to the size of the gain in its $\rm KE$ .

Therefore, right before collision, the $\rm KE$ of this spacecraft would be:

$\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}$ .

On the other hand, let $v$ denote the speed of this spacecraft. The following equation that relates $v\!$ and $m$ to $\rm KE$ :

$\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^2$ .

Rearrange this equation to find an equation for $v$ :

$\displaystyle v = \sqrt{\frac{2\, \text{KE}}{m}}$ .

It is already found that right before the collision, $\displaystyle \text{KE} = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ . Make use of this equation to find $v$ at that moment:

$\begin{aligned}v &= \sqrt{\frac{2\, \text{KE}}{m}} \\ &= \sqrt{\frac{2\, G\cdot M_\text{e} \cdot m}{R_\text{e}\cdot m}} = \sqrt{\frac{2\, G\cdot M_\text{e}}{R_\text{e}}}\end{aligned}$ .

6 0

3 years ago

Suppose that you have a rod that is positively charged, a rod that is negatively charged and a stand that allows rotation. if yo

andreev551 [17]

Yes, you can. Remember that in electrophysics, opposite charges attract. Since there is a stand of rotation, it allows the rod with both ends having opposite charges to rotate. Now, if you place the unknown charge near the set-up, it could turn clockwise or counterclockwise. When the set-up turns such that the negative end is approaching the unknown charge, that charge must be positive. Otherwise, the unknown charge is negative.

6 0

3 years ago

Read 2 more answers