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SVEN [57.7K]
3 years ago
8

A voltage amplifier needs a high input resistance and a low output resistance. Select one: True False

Physics
1 answer:
cupoosta [38]3 years ago
6 0

Answer:

Explanation:

In electric circuit , the potential difference is always developed across the resistance .

Now is we are to amplify the voltage , that means low input voltage is converted into high voltage output .

Therefore we require low resistance at input for low voltage and high out put resistance for high output

Thus the statement given is wrong .

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Boyle's Law mainly involves _______.
goblinko [34]
Your answer is B, gases
6 0
4 years ago
A glass tube 1 mm in diameter is dipped into glycerin. The density of the glycerin is 1260 kg/m3, surface tension is 6.3x10-2 N/
Sunny_sXe [5.5K]

Answer:

The capillary rise of the glycerin is most nearly  y  =  0.0204 \ m

Explanation:

From the question we are told that

  The diameter of the glass tube is  d =  1 \ mm =  0.001 \ m

   The density of glycerin is  \rho =  1260 \ kg /m^3

   The surface tension of the glycerin is \sigma   =  6.3 *10^{-2} \ N /m

The capillary rise of the glycerin is mathematically represented as

       y  =  \frac{4 * \sigma  *  cos (\theta )}{ \rho * g *  d}

substituting value  

       y  =  \frac{4 * 6.3 *10^{-2}  *  cos (0 )}{ 1260 * 9.8 *  0.001}

      y  =  0.0204 \ m

Therefore the height  of the glass tube  the glycerin was able to cover is

y  =  0.0204 \ m  

4 0
3 years ago
In a photoelectric experiment, you shine light onto an electrode and record a current of 25 μA. When you apply +500 mV to the el
kkurt [141]

Answer:

2.083 V.

Explanation:

Stopping potential is the potential that is required to stop the current to zero . This potential is applied externally to oppose the potential created by the photoelectric effect . It gives the measure the photoelectric potential being generated .

Here current drops to 25 μA to 19 μA by a potential of 500mV

Change in current

= 25 - 19 = 6 μA

Voltage requirement for unit reduction in current

= 500 / 6 μA

To reduce current 0f 25 μA

requirement of V = (500 / 6 )  x 25 =   2083.33 mV = 2.083 V.

7 0
3 years ago
Which of the following statements are true? Select all correct responses. Choose all that apply. Choose all that apply. The tota
Ne4ueva [31]
<h3><u>Answer</u>;</h3>

-The total momentum of an isolated system is constant.

-The total momentum of any number of particles is equal to the vector sum of the momenta of the individual particles.

-The vector sum of forces acting on a particle equals the rate of change of momentum of the particle with respect to time.

<h3><u>Explanation</u>;</h3>
  • Momentum is a vector quantity, and therefore we need to use vector addition when summing together the momenta of the multiple bodies which make up a system.
  • The vector sum of forces acting on a particle is equivalent to the rate of change of momentum of the particle with respect to time. This is according to the Newton's second Law of motion. In mathematical terms, ֿF = d ֿp/dt, that is F= ma.
  • According to the Law of conservation of Momentum, or a collision occurring between object 1 and object 2 in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision.
3 0
3 years ago
A Hall probe, consisting of a rectangular slab of current-carrying material, is calibrated by placing it in a known magnetic fie
Citrus2011 [14]

Answer:

(a) 0.345 T

(b) 0.389 T

Solution:

As per the question:

Hall emf, V_{Hall} = 20\ mV = 0.02\ V

Magnetic Field, B = 0.10 T

Hall emf, V'_{Hall} = 69\ mV = 0.069\ V

Now,

Drift velocity, v_{d} = \frac{V_{Hall}}{B}

v_{d} = \frac{0.02}{0.10} = 0.2\ m/s

Now, the expression for the electric field is given by:

E_{Hall} = Bv_{d}sin\theta                            (1)

And

E_{Hall} = V_{Hall}d

Thus eqn (1) becomes

V_{Hall}d = dBv_{d}sin\theta

where

d = distance

B = \frac{V_{Hall}}{v_{d}sin\theta}                      (2)

(a) When \theta = 90^{\circ}

B = \frac{0.069}{0.2\times sin90} = 0.345\ T

(b) When \theta = 60^{\circ}

B = \frac{0.069}{0.2\times sin60} = 0.398\ T

5 0
3 years ago
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