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3 years ago

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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Which can be observed both on Earth and in an accelerating ship in space that is free from the effect of any gravitational field

Reil [10]

Answer:

apples falling from trees
people's feet touching the ground
sky divers moving toward the ground
balls bending downward after being thrown

Explanation:

When a space ship is accelerating in space, there is a force known as Inertia that kicks in. Inertia will mirror the effects of gravity on the ship even if there is no gravitational field effect such that anything that would happen where there is gravity, would continue to happen.

This means that apples will fall from trees, people's feet will touch the ground, sky divers will be pulled downwards and balls will bend downwards when thrown as well. These are the same effects expected on earth where gravity pulls things towards the earth's core.

8 0

2 years ago

Read 2 more answers

A train 471 m long is moving on a straight track with a speed of 75.1 km/h. The engineer applies the brakes at a crossing, and l

hram777 [196]

Answer:

$t = 37.6 s$

Explanation:

As we know that train is initially moving with the speed

$v_i = 75.1 km/h$

now we know that

$v_i = 20.86 m/s$

now the final speed of the train when it crossed the crossing

$v_f = 15 km/h$

$v_f = 4.17 m/s$

now we can use kinematics here

$v_f^2 - v_i^2 = 2 a d$

$4.17^2 - 20.86^2 = 2 a(471)$

$a = -0.44 m/s^2$

Now the time to cross that junction is given as

$v_f - v_i = at$

$4.17 - 20.86 = a(-0.44)$

$t = 37.6 s$

4 0

2 years ago

0.125 C of charge flow out of a

zimovet [89]

Answer:

Resistance = 252.53 Ohms

Explanation:

Given the following data;

Charge = 0.125 C

Voltage = 5 V

Time = 6.3 seconds

To find the resistance;

First of all, we would determine the current flowing through the battery;

Quantity of charge, Q = current * time

0.125 = current * 6.3

Current = 0.125/6.3

Current = 0.0198 A

Next, we find the resistance;

Resistance = voltage/current

Resistance = 5/0.0198

Resistance = 252.53 Ohms

3 0

2 years ago

Solar evaporation ponds are shallow so that the sun can heat them up, causing the water to evaporate, which increases the salt c

ehidna [41]

Answer:

B. stearothermophilus and S. ruber

Explanation:

B. stearothermophilus and S. ruber

In solar evaporation ponds the temperature is higher and the salt concentration is also higher because of the water evaporated so sunder such extreme conditions this hybrid bacteria is capable of surviving. B. stearothermophilus is thermophilus bacteria which grows at high temperature and S. ruber is halophilic bacteria which grows in saline environment. So, these two bacteria best suited for the above hybrid condition.

6 0

3 years ago

An electron is placed on a line connecting two fixed point charges of equal charge but the opposite sign. The distance between t

viktelen [127]

Answer:

a) F_net = 6.48 10⁻¹⁸ ( $\frac{1}{x^2} + \frac{1}{(0.300-x)^2}$ ), b) x = 0.15 m

Explanation:

a) In this problem we use that the electric force is a vector, that charges of different signs attract and charges of the same sign repel.

The electric force is given by Coulomb's law

F = $k \frac{q_2q_2}{r^2}$

Since when we have the two negative charges they repel each other and when we fear one negative and the other positive attract each other, the forces point towards the same side, which is why they must be added.

F_net= ∑ F = F₁ + F₂

let's locate a reference system in the load that is on the left side, the distances are

left side - electron r₁ = x

right side -electron r₂ = d-x

let's call the charge of the electron (q) and the fixed charge that has equal magnitude Q

we substitute

F_net = k q Q ( $\frac{1}{r_1^2}+ \frac{1}{r_2^2}$ )

F _net = kqQ ( $\frac{1}{x^2} + \frac{1}{(d-x)^2}$ )

let's substitute the values

F_net = 9 10⁹ 1.6 10⁻¹⁹ 4.50 10⁻⁹ ( $\frac{1}{x^2} + \frac{1}{(0.30-x)^2}$ )

F_net = 6.48 10⁻¹⁸ ( $\frac{1}{x^2} + \frac{1}{(0.300-x)^2}$ )

now we can substitute the value of x from 0.05 m to 0.25 m, the easiest way to do this is in a spreadsheet, in the table the values of the distance (x) and the net force are given

x (m) F (N)

0.05 27.0 10-16

0.10 8.10 10-16

0.15 5.76 10-16

0.20 8.10 10-16

0.25 27.0 10-16

b) in the adjoint we can see a graph of the force against the distance, it can be seen that it has the shape of a parabola with a minimum close to x = 0.15 m

4 0

3 years ago