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2 years ago

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An electron is to be accelerated from a velocity of 3.00×106m/s to a velocity of 8.00×106m/s . Through what potential difference

must the electron pass to accomplish this?

1 answer:

Jobisdone [24]2 years ago

8 0

mmmm I don't really understand

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Stars combine Hydrogen to make Helium during nuclear fusion. Living things are made of heavier elements like Carbon, Oxygen, Iro

alex41 [277]

They were formed in the nuclear<span> fusion reaction inside older </span><span>stars.

As a star burns, fusion reactions inside its core create heavier elements. Those materials are released when the star dies of old age in an explosion.</span>

8 0

2 years ago

# A cheetah can start from rest and attain the velocity 72km/h in 2 seconds. Calculate the acceleration of cheetah

yan [13]

Answer:

<h2> $10 \:$ m/s ^2</h2><h2 />

Explanation:

Solution,

When a certain object comes in motion from rest, in the case, initial velocity = 0 m/s

Initial velocity ( u ) = 0 m/s

Final velocity ( v ) = 72 km/h ( Given)

We have to convert 72 km /h in m/s

$72 \: km \: per \: hour$

$= 72 \times \frac{1000}{60 \times 60}$

$= 20$ m/s

Final velocity ( v ) = 20 m/s

Time taken ( t ) = 2 seconds

Acceleration (a) = ?

Now,

we have,

$a = \frac{v - u}{t}$

$a = \frac{20 - 0}{2}$

$a = \frac{20}{2}$

$a = 10$ m/s ^2

Hope this helps...

Good luck on your assignment..

7 0

2 years ago

You use 8x binoculars were used on a warbler (14cm long) in a tree 18cm away. What angle (in degrees) does the image of the warb

mafiozo [28]

Answer:

The angle it subtend on the retina is $\theta_z = 0.44586^o$

Explanation:

From the question we are told that

The length of the warbler is $L = 14cm = \frac{14}{100} = 0.14m$

The distance from the binoculars is $d = 18cm = \frac{18}{100} = 0.18m$

The magnification of the binoculars is $M =8$

Without the 8 X binoculars the angle made with the angular size of the object is mathematically represented as

$\theta = \frac{L}{d}$

$\theta = \frac{0.14}{0.18}$

$= 0.007778 rad$

Now magnification can be represented mathematically as

$M = \frac{\theta _z}{\theta}$

Where $\theta_z$ is the angle the image of the warbler subtend on your retina when the binoculars i.e the binoculars zoom.

So

$\theta_z = M * \theta$

=> $\theta_z =8 * 0.007778$

$= 0.0622222224$

Generally the conversion to degrees can be mathematically evaluated as

$\theta_z = 0.062222224 * (\frac{360 }{2 \pi rad} )$

$\theta_z = 0.44586^o$

7 0

3 years ago

A bar magnet is dropped toward a conducting ring lying on the floor. As the magnet falls toward the ring, does it move as a free

Maslowich

Consider that the bar magnet has a magnetic field that is acting around it, which will imply that there is a change in the magnetic flux through the loop whenever it moves towards the conducting loop. This could be described as an induction of the electromotive Force in the circuit from Faraday's law.

In turn by Lenz's law, said electromotive force opposes the change in the magnetic flux of the circuit. Therefore, there is a force that opposes the movement of the bar magnet through the conductor loop. Therefore, the bar magnet does not suffer free fall motion.

The bar magnet does not move as a freely falling object.

3 0

2 years ago

Can somebody please help?

DedPeter [7]

Answer:

Im not sure

Explanation:

I don't take physics cuz im in 9th grade. so. idk but I will find out and come back with an answer.

7 0

2 years ago