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2 years ago

Electromagnetic waves can behave like particles

1 answer:

3 0

Yes they can. Uh-huh. Indubitably. That's right, Vanessa. It's true.

A better way to say it might be:

Electromagnetic radiation can behave like waves AND like particles.

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A pulley system with a mechanical advantage of 15 is used to lift a 1750 N piano to a third floor balcony that is 7 m above the

devlian [24]

Answer:

The force is $E = 117 \ N$

Explanation:

From the question we are told that

The mechanical advantage is $ME = 15$

The load is $L = 1750 \ N$

The height is $h = 7 \ m$

Generally mechanical advantage is mathematically represented as

$ME = \frac{L }{E }$

Here E is the force that must be applied to move the piano

=> $E = \frac{L}{ ME }$

=> $E = \frac{1750 }{15 }$

=> $E = 117 \ N$

6 0

2 years ago

Consider a collection of charges in a given region and suppose all other charges are distant and have a negligible effect. Furth

AlladinOne [14]

Answer:

E. Some charges in the region are positive, and some are negative.

Explanation:

Electric potential is given as;

$V = \frac{W}{Q}$

where;

W is the work done in moving a charge between two points which have a difference in potential

Q is quantity of charge in the given region

If the electric potential at a given point in the region is zero, then sum of the charges in the given region must be equal to zero. For the charges to sum to zero, some will be positive while some will be negative,.

Therefore, the correct statement in the given options is "E"

E. Some charges in the region are positive, and some are negative.

3 0

3 years ago

The Hubble Space Telescope orbits the Earth at approximately 612,000m altitude. Its mass is 11,100 kg and the mass of earth is 5

nexus9112 [7]

Answer:

7.55 km/s

Explanation:

The force of gravity between the Earth and the Hubble Telescope corresponds to the centripetal force that keeps the telescope in uniform circular motion around the Earth:

$G\frac{mM}{R^2}=m\frac{v^2}{R}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2}$ is the gravitational constant

$m=11,100 kg$ is the mass of the telescope

$M=5.97\cdot 10^{24} kg$ is the mass of the Earth

$R=r+h=6.38\cdot 10^6 m+612,000 m=6.99\cdot 10^6 m$ is the distance between the telescope and the Earth's centre (given by the sum of the Earth's radius, r, and the telescope altitude, h)

v = ? is the orbital velocity of the Hubble telescope

Re-arranging the equation and substituting numbers, we find the orbital velocity:

$v=\sqrt{\frac{GM}{R}}=\sqrt{\frac{(6.67\cdot 10^{-11})(5.97\cdot 10^{24} kg)}{6.99\cdot 10^6 m}}=7548 m/s=7.55 km/s$

6 0

2 years ago

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

2 years ago

Which genotype has a 50% chance of being inherited

stepan [7]

Recessive genotype I hope this helps

8 0

2 years ago