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3 years ago

When testing an PNP transistor with an ohmmeter, what are the high or low resistance values expected for a good transistor?

Physics

1 answer:

aksik [14]3 years ago

7 0

Answer:

0.45 V and 0.9 V.

Explanation:

To test a PNP transistor with an ohmmeter,

Plug the positive lead from the multimeter to the transistor EMITTER (E). Plug the negative meter into the transistor BASE (B). If you are using a PNP resistor you must watch OL that is over limit, the voltage decrease will indicate between 0.45V and 0.9V if you are measuring it.

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Why silicon have large forward current as compared to germanium?

Iteru [2.4K]

Answer:

The structure of Germanium crystals will be destroyed at higher temperature. However, Silicon crystals are not easily damaged by excess heat. Peak Inverse Voltage ratings of Silicon diodes are greater than Germanium diodes. Si is less expensive due to the greater abundance of element.

4 0

2 years ago

A body oscillates with simple harmonic motion along the x axis. Its displacement varies with time according to the equation x =

frutty [35]

Answer:

θ = (7π / 3) rad

Explanation:

given,

displacement of simple harmonic motion along x-axis

equation is given as

x = 5 sin (π t + π/3 )

general equation of simple harmonic motion

x = A sin θ

θ is the phase angle

θ = π t + π/3

at t = 2 s

$\theta =\pi \times 2 +\dfrac{\pi}{3}$

$\theta =\dfrac{7\pi}{3}\ rad$

Phase of the motion at t =2 s is θ = (7π / 3) rad

7 0

3 years ago

Before colliding, the momentum of block A is -100 kg*m/, and block B is -150 kg*m/s. After, block A has a momentum -200 kg*m/s.

rjkz [21]

Answer:

Momentum of block B after collision = $-50\ kg\ ms^{-1}$

Explanation:

Given

Before collision:

Momentum of block A = $p_{A1}$ = $-100\ kg\ ms^{-1}$

Momentum of block B = $p_{B1}$ = $-150\ kg\ ms^{-1}$

After collision:

Momentum of block A = $p_{A2}$ = $-200\ kg\ ms^{-1}$

Applying law of conservation of momentum to find momentum of block B after collision $p_{B2}$ .

$p_{A1}+p_{B1}=p_{A2}+p_{B2}$

Plugging in the given values and simplifying.

$-100-150=-200+p_{B2}$

$-250=-200+p_{B2}$

Adding 200 to both sides.

$200-250=-200+p_{B2}+200$

$-50=p_{B2}$

∴ $p_{B2}=-50\ kg\ ms^{-1}$

Momentum of block B after collision = $-50\ kg\ ms^{-1}$

6 0

2 years ago

Choose the situation below in which the force applied is the greatest.

Gnesinka [82]

Answer:

Explanation:

We know the formula for Work to be:

W = f * d

Where W is work done

f is force

d is the distance

Work = 50

Distance = 50

So, Force is:

Force = 50/50 = 1

Work = 400

Distance = 80

Force = 400/80 = 5

Work = 365

Distance = 73

Force = 365/73 = 5

Work = 144

Distance = 16

Force = 144/16 = 9

Hence, D is the situation in which the force applied is the greatest.

6 0

3 years ago

6. A car, 1110 kg, is traveling down a horizontal road at 20.0 m/s when it locks up its brakes. The coefficient of friction betw

USPshnik [31]

Answer:

$x=22.65m$

Explanation:

We have an uniformly accelerated motion, with a negative acceleration. Thus, we use the kinematic equations to calculate the distance will it take to bring the car to a stop:

$v_f^2=v_0^2-2ax\\\frac{0^2-v_0^2}{-2a}=x\\x=\frac{v_0^2}{2a}$

The acceleration can be calculated using Newton's second law:

$\sum F_x:F_f=ma\\\sum F_y:N=mg$

Recall that the maximum force of friction is defined as $F_f=\mu N$ . So, replacing this:

$\mu N=ma\\\mu mg=ma\\a=\mu g\\a=0.901(9.8\frac{m}{s^2})\\a=8.83\frac{m}{s^2}$

Now, we calculate the distance:

$x=\frac{v_0^2}{2a}\\x=\frac{(20\frac{m}{s})^2}{2(8.83\frac{m}{s^2})}\\x=22.65m$

7 0

3 years ago