Please answer the question number 2 in this pic in detail

If a voltmeter has a less than ideal resistance, say 1 MΩ, and is used to measure the voltage across a resistor of a comparable

Naddik [55]

Answer:

As the difference between the resistance of voltmeter and the resistance being measured gets reduced the error in the reading of the voltmeter gets increased.

Explanation:

An ideal voltmeter has infinite parallel resistance and because of this it doesn't draw any current from the circuit of measurement which means it will measure the exact voltage across the elements.

But practically speaking, a real voltmeter doesn't has infinite resistance therefore, all the practical voltmeters face loading effect to some extent.

As the difference between the resistance of voltmeter and the resistance being measured gets reduced the error in the reading of the voltmeter gets increased. This is why we want to have a greater value of voltmeter resistance, ideally infinite so that the corresponding error is minimized.

Lets consider the given scenario,

A voltmeter has 1 MΩ parallel resistance and the resistance of of measuring element is 500 kΩ or 0.5 MΩ

lets suppose the supplied voltage is 1 V.

First lets assume that the voltmeter is ideal and it has infinite resistance, so in this case voltmeter will measure a voltage of 1 V across the 0.5 MΩ resistor.

Now consider the loading effect, when we connect the voltmeter across the 0.5 MΩ resistor they both become parallel so the resistance is

R = (1*0.5)/(1+0.5)

R = 0.33 MΩ

As you can see the voltmeter will see a reduced resistance and the corresponding voltage also reduces because resistance and voltage are directly proportional.

Therefore, it is preferred to have a very high parallel resistance of the voltmeter.

8 0

3 years ago

Neutron stars are extremely dense objects that are formed from the remnants of supernova explosions. Many rotate very rapidly. S

Vlada [557]

Answer:

The required angular speed the neutron star is 10992.32 rad/s

Explanation:

Given the data in the question;

mass of the sun M $_S$ = 1.99 × 10³⁰ kg

Mass of the neutron star

M $_N$ = 2( M $_S$ )

M $_N$ = 2( 1.99 × 10³⁰ kg )

M $_N$ = ( 3.98 × 10³⁰ kg )

Radius of neutron star R $_N$ = 13.0 km = 13 × 10³ m

Now, let mass of a small object on the neutron star be m

angular speed be ω $_N$ .

During rotational motion, the gravitational force on the object supplies the necessary centripetal force.

GmM $_N$ = / R $_N$ ² = mR $_N$ ω $_N$ ²

ω $_N$ ² = GM $_N$ = / R $_N$ ³

ω $_N$ = √(GM $_N$ = / R $_N$ ³)

we know that gravitational G = 6.67 × 10⁻¹¹ Nm²/kg²

we substitute

ω $_N$ = √( ( 6.67 × 10⁻¹¹ )( 3.98 × 10³⁰ ) ) / (13 × 10³ )³)

ω $_N$ = √( 2.65466 × 10²⁰ / 2.197 × 10¹²

ω $_N$ = √ 120831133.3636777

ω $_N$ = 10992.32 rad/s

Therefore, The required angular speed the neutron star is 10992.32 rad/s

5 0

3 years ago

Add these two velocity vectors to find the magnitude of their resultant vector.

hammer [34]

The magnitude of their resultant vector is 4.6 meters/seconds

Since we are to add the velocity vectors in order to find the magnitude of their resultant vector.

Hence:

Resultant vector magnitude=5.8 meters/seconds + (1.2 meters/seconds)

Resultant vector magnitude=5.8 meters/seconds-1.2 meters/seconds

Resultant vector magnitude 4.6 meters/seconds

Inconclusion The magnitude of their resultant vector is 4.6 meters/seconds

Learn more here:

brainly.com/question/11134601

6 0

2 years ago

Ice floats on water.

OleMash [197]

Answer:

Yeah ice floats on water.

Observation

Example in those areas were ice is found like Antarctica ice is found on top of water.

5 0

3 years ago

the gravitational force between two objects is 1600 and what will be the gravitational force between the objects if the distance

Xelga [282]

I believe this is what you have to do:

The force between a mass M and a point mass m is represented by

$F = G\frac{Mm}{r^{2} }$

So lets compare it to the original force before it doubles, it would just be the exact formula so lets call that F₁

So F₁ = G(Mm/r^2)

Now the distance has doubled so lets account for this in F₂:

F₂ = G(Mm/(2r)^2)

Now square the 2 that gives you four and we can pull that out in front to give

F₂ = $\frac{1}{4}$ G(Mm/r^2)

Now we can replace G(Mm/r^2) with F₁ as that is the value of the force before alterations

now we see that:

F₂ = $\frac{1}{4}$ F₁

So the second force will be 0.25 (1/4) x 1600 or 400 N.

6 0

3 years ago