20 points + Brainliest. Please help!

. A PNP transistor is connected in a circuit so that the collector-base junction remains reverse biased and the emitter-base jun

Elodia [21]

Explanation :

A power amplifier is used to amplify electric signals i.e. to increase the low power signal to higher powers.

A PNP transistor is connected in a circuit so that the collector-base junction remains reverse biased and the emitter-base junction is forward biased.

This transistor can be used as a power amplifier because it gives a much larger output current. The gain of an amplifier shows the amount of amplification. It is the difference between the input and the output signals.

6 0

3 years ago

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a ultrasonic wave at 8x10^4 Hz is emitted into a vien where the speed of sound in blood is 1570 m/s. the wave reflects off the r

Aneli [31]

Answer: 0.392 m/s

Explanation:

The Doppler shift equation is:

$f'=\frac{V+V_{o}}{V-V_{s}} f$

Where:

$f=8(10)^{4} Hz$ is the actual frequency of the sound wave

$f'=8.002(10)^{4} Hz$ is the "observed" frequency

$V=1570 m/s$ is the speed of sound

$V_{o}=0 m/s$ is the velocity of the observer, which is stationary

$V_{s}$ is the velocity of the source, which are the red blood cells

Isolating $V_{s}$ :

$V_{s}=\frac{V(f'-f)}{f'}$

$V_{s}=\frac{1570 m/s(8.002(10)^{4} Hz-8(10)^{4} Hz)}{8.002(10)^{4} Hz}$

Finally:

$V_{s}=0.392 m/s$

3 0

2 years ago

Friction is defined as *

den301095 [7]

Answer:

O a force that opposes motion

5 0

2 years ago

The diagram pictured shows a loop of wire being rotated at a constant rate about an axis in a uniform magnetic field.

faust18 [17]

The rotation is probably n

7 0

2 years ago

Consider a spring mass system (mass m1, spring constant k) with period T1. Now consider a spring mass system with the same sprin

tatuchka [14]

Answer:

Assuming that both mass here move horizontally on a frictionless surface, and that this spring follows Hooke's Law, then the mass of $m_2$ would be four times that of $m_1$ .

Explanation:

In general, if the mass in a spring-mass system moves horizontally on a frictionless surface, and that the spring follows Hooke's Law, then

$\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2$ .

Here's how this statement can be concluded from the equations for a simple harmonic motion (SHM.)

In an SHM, if the period is $T$ , then the angular velocity of the SHM would be

$\displaystyle \omega = \frac{2\pi}{T}$ .

Assume that the mass starts with a zero displacement and a positive velocity. If $A$ represent the amplitude of the SHM, then the displacement of the mass at time $t$ would be:

$\mathbf{x}(t) = A\sin(\omega\cdot t)$ .

The velocity of the mass at time $t$ would be:

$\mathbf{v}(t) = A\,\omega \, \cos(\omega\, t)$ .

The acceleration of the mass at time $t$ would be:

$\mathbf{a}(t) = -A\,\omega^2\, \sin(\omega \, t)$ .

Let $m$ represent the size of the mass attached to the spring. By Newton's Second Law, the net force on the mass at time $t$ would be:

$\mathbf{F}(t) = m\, \mathbf{a}(t) = -m\, A\, \omega^2 \, \cos(\omega\cdot t)$ ,

Since it is assumed that the mass here moves on a horizontal frictionless surface, only the spring could supply the net force on the mass. Therefore, the force that the spring exerts on the mass will be equal to the net force on the mass. If the spring satisfies Hooke's Law, then the spring constant $k$ will be equal to:

$\begin{aligned} k &= -\frac{\mathbf{F}(t)}{\mathbf{x}(t)} \\ &= \frac{m\, A\, \omega^2\, \cos(\omega\cdot t)}{A \cos(\omega \cdot t)} \\ &= m \, \omega^2\end{aligned}$ .

Since $\displaystyle \omega = \frac{2\pi}{T}$ , it can be concluded that:

$\begin{aligned} k &= m \, \omega^2 = m \left(\frac{2\pi}{T}\right)^2\end{aligned}$ .

For the first mass $m_1$ , if the time period is $T_1$ , then the spring constant would be:

$\displaystyle k = m_1\, \left(\frac{2\pi}{T_1}\right)^2$ .

Similarly, for the second mass $m_2$ , if the time period is $T_2$ , then the spring constant would be:

$\displaystyle k = m_2\, \left(\frac{2\pi}{T_2}\right)^2$ .

Since the two springs are the same, the two spring constants should be equal to each other. That is:

$\displaystyle m_1\, \left(\frac{2\pi}{T_1}\right)^2 = k = m_2\, \left(\frac{2\pi}{T_2}\right)^2$ .

Simplify to obtain:

$\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2$ .

6 0

3 years ago