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3 years ago

15

If you were to walk toward a Van der Graaff generator, the effect of its electrical field would get weaker as you walked towards

it.
True or False

1 answer:

gavmur [86]3 years ago

7 0

This is False.

If you walk towards a Vander Graaff generator and touch it, your body will be electrically charged with the same sign as the generator (that’s why your hair stands up, because it is charged with static electricity, and…electrical charges of the same sign repel each other)

Note that the Vander Graaff generator is isolated from the ground, so, if the electrical charge is increased, electrical arcs will form and fly off it to any nearby grounded object or person.

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Which are properties of a liquid? Check all that apply.

Strike441 [17]

Answer:

the property of liquid are

1 they can flow from one place to another if surface is slanted

2 it cannot be compressed

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2 years ago

What is the frequency of an ocean wave that is traveling at a speed of 45 m/s if it has a wavelength of 3 meters.

vaieri [72.5K]

Answer:

Frequency, f = 15 Hz

Explanation:

We have,

Speed of an ocean wave is 45 m/s

Wavelength of a wave is 3 m

It is required to find the frequency of an ocean wave.

Speed of a wave, $v=f\lambda$ , f = frequency of ocean wave

$f=\dfrac{v}{\lambda}\\\\f=\dfrac{45}{3}\\\\f=15\ Hz$

So, the frequency of an ocean wave is 15 Hz.

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3 years ago

An electron accelerated from rest through a voltage of 780 v enters a region of constant magnetic field. part a part complete if

maxonik [38]

The electron is accelerated through a potential difference of $\Delta V=780 V$ , so the kinetic energy gained by the electron is equal to its variation of electrical potential energy:
$\frac{1}{2}mv^2 = e \Delta V$
where
m is the electron mass
v is the final speed of the electron
e is the electron charge
$\Delta V$ is the potential difference

Re-arranging this equation, we can find the speed of the electron before entering the magnetic field:
$v= \sqrt{ \frac{2 e \Delta V}{m} } = \sqrt{ \frac{2(1.6 \cdot 10^{-19}C)(780 V)}{9.1 \cdot 10^{-31} kg} }=1.66 \cdot 10^7 m/s$

Now the electron enters the magnetic field. The Lorentz force provides the centripetal force that keeps the electron in circular orbit:
$evB=m \frac{v^2}{r}$
where B is the intensity of the magnetic field and r is the orbital radius. Since the radius is r=25 cm=0.25 m, we can re-arrange this equation to find B:
$B= \frac{mv}{er}= \frac{(9.1 \cdot 10^{-31}kg)(1.66 \cdot 10^7 m/s)}{(1.6 \cdot 10^{-19}C)(0.25 m)} =3.8 \cdot 10^{-4} T$

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3 years ago

"relate the fertile phase of the menstrual cycle to the process of fertilisation"

DerKrebs [107]

Explanation:

During your menstrual cycle , Harmones make the eggs in your Ovaries mature -

• when an egg is mature , That means it's ready to be fertilized by a sperm cell .

• These hormones also make the lining of your uterus thick and spongy . So if your egg does get Fertilised , It has a nice cushy place to land and start a pregnancy .

<h3>Hope this helps </h3>

8 0

2 years ago

Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bu

vlada-n [284]

Answer:

$\frac{1}{10}$ M

Explanation:

To apply the concept of <u>angular momentum conservation</u>, there should be no external torque before and after

As the <u>asteroid is travelling directly towards the center of the Earth</u>, after impact ,it <u>does not impose any torque on earth's rotation,</u> So angular momentum of earth is conserved

⇒ $I_{1} \times W_{1} =I_{2} \times W_{2}$

$I_{1}$ is the moment of interia of earth before impact
$W_{1}$ is the angular velocity of earth about an axis passing through the center of earth before impact
$I_{2}$ is moment of interia of earth and asteroid system
$W_{2}$ is the angular velocity of earth and asteroid system about the same axis

let $W_{1}=W$

since $\text{Time period of rotation}∝$ $\frac{1}{\text{Angular velocity}}$

⇒ if time period is to increase by 25%, which is $\frac{5}{4}$ times, the angular velocity decreases 25% which is $\frac{4}{5}$ times

therefore $W_{1}$ $=$ $\frac{4}{5} \times W_{1}$

$I_{1}=\frac{2}{5} \times M\times R^{2}$ (moment of inertia of solid sphere)

where M is mass of earth

R is radius of earth

$I_{2}=\frac{2}{5} \times M\times R^{2}+M_{1}\times R^{2}$

(As given asteroid is very small compared to earth, we assume it be a particle compared to earth, therefore by parallel axis theorem we find its moment of inertia with respect to axis)

where $M_{1}$ is mass of asteroid

⇒ $\frac{2}{5} \times M\times R^{2} \times W_{1}=}(\frac{2}{5} \times M\times R^{2}$ $+$ $M_{1}\times R^{2})\times(\frac{4}{5} \times W_{1})$

$\frac{1}{2} \times M\times R^{2}$ = $(\frac{2}{5} \times M\times R^{2}$ + $M_{1}\times R^{2})$

$M_{1}\times R^{2}=$ $\frac{1}{10} \times M\times R^{2}$

⇒ $M_{1}=}$ $\frac{1}{10} \times M$

3 0

3 years ago