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Anton [14]
2 years ago
7

A non-conducting sphere of radius R = 3.0 cm carries a charge Q = 2.0 mC distributed uniformly throughout its volume. At what di

stance, measured from the center of the sphere, does the electric field reach a value equal to half its maximum value?
Physics
1 answer:
BlackZzzverrR [31]2 years ago
4 0

Answer:

r =3 *\sqrt{2} = 4.24 cm

Explanation:

given data

Radius of sphere 3.0 cm

charge Q = 2.0 m C

We know that maximum electric field is given as

E_{MAX}= \frac{KQ}{r^{2}}

electric field inside the sphere can be determine by using below relation

\frac{KQ}{r^{2}}= \frac{1}{2}*\frac{KQ}{R^{2}}

r = \sqrt{2}R

r =3 *\sqrt{2} = 4.24 cm

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Pions have a half-life of 1.8 x 10^-8 s. A pion beam leaves an accelerator at a speed of 0.8c. What is the expected distance ove
Nuetrik [128]

Answer:

the expected distance is 4.32 m

Explanation:

given data

half life time = 1.8 × 10^{-8} s

speed = 0.8 c = 0.8 × 3 × 10^{8}

to find out

expected distance over

solution

we know c is speed of light in air is 3 × 10^{8} m/s

we calculate expected distance by given formula that is

expected distance = half life time × speed   .........1

put here all these value

expected distance = half life time × speed

expected distance = 1.8 × 10^{-8} ×  0.8 × 3 × 10^{8}

expected distance = 4.32

so the expected distance is 4.32 m

5 0
2 years ago
In at least 150 words, discuss how Chang's use of personification, metaphor, or connotation express one his themes in "Garden of
Kay [80]
<span>Poet Kuangchi Chang did not remain in China long enough to be "re-educated." Following the Communist takeover he fled to the United States. His poem "Garden of My Childhood" describes China before the revolution as a peaceful, idyllic garden with a violent horde rapidly approaching. A vine, the wind, and the sea are each personified, and each beckons for him to run. It is not until "eons later," when he is "worlds away," that his "running is all done," and he finds himself at his destination: another garden, just like the one he had left behind.</span>
6 0
2 years ago
You have negotiated with the Omicronians for a base on the planet Omicron Persei 7. The architects working with you to plan the
steposvetlana [31]

Answer:

5.724 meters / second^2

Explanation:

We are given two pieces of information, 5.24 flurg = 1 meter, 1 grom = 0.493 second. If that is so, we can say that there are two possible conversion units,  5.25 flurg / meter, and 0.493 second / grom.

_____

We want to convert 7.29 flurg / grom^2 ( I believe? ) to the units meters / second^2. But, let's break this down into bits. It would be convenient to first convert 7.29 flurg / grom^2 to the units meters / grom^2, by dividing the conversion factors as to cancel out the appropriate things, which we will go into detail on a bit later ( using the first conversion factor ). Respectively we can convert meters / grom^2 to meters / grom * s, canceling out the flurg ( through the second conversion factor ). And now we would need to get rid of the grom, dividing similarly.

_____

( 1 ) ( flurg / grom^2 ) / ( flurg / meters  ) - first conversion unit

= flurg / grom^2 * meters /flurg

= ( meters * flurg ) / ( grom^2 * flurg )  

= meters /grom^2,

7.29 flurg / grom^2 / 5.24 flurg / meter = ( About ) 1.39 meter / grom^2

( 2 ) ( meter / grom^2 ) / ( second / grom  ) - second conversion unit

= meter / grom^2 * grom / second

= ( meter * grom ) / ( grom^2 * second )

= meter / ( grom * second ),

( 1.39 meter / grom^2 ) / 0.493 second / grom = ( About ) 2.82195 meter /  grom * second

( 3 ) ( 2.82195 meter / ( grom * second ) ) / 0.493 second / grom = 5.724 meter / second^2

( And thus, the value of gOP7 in the units the architects will use should be about 5.724 meters / second^2 )

8 0
2 years ago
An object has a mass of 8.00kg. What is the gravitational force on the object by the earth
Firdavs [7]
78.4 F because you do 8.00 muliplyed by 9.8
6 0
2 years ago
José is pinned against the walls of the Rotor, a ride with a radius of 3.00 meters that spins so fast that the floor can be remo
zaharov [31]

r = radius of the circle of the ride = 3.00 meters

v = linear speed of the person during the ride = 17.0 m/s

m = mass of the person in angular motion in the ride

L = angular momentum of the person in the ride = 3570 kg m²/s

Angular momentum is given as

L = m v r

inserting the values

3570 kg m²/s = m (17 m/s) (3.00 m)

m = 3570 kg m²/s/(51 m²/s)

m = 7 kg

hence the mass comes out to be 7 kg


8 0
2 years ago
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