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3 years ago

11

At what distance of separation, or, must two 5.87*10^-8 coulomb charges be positioned in order for the repulsive force between t

hem to be 64.9 N?
Answer with work shown in GUESS format

G (givens)
U (unknown)
S (substitute)
S (solve)

50 POINTS AND BRAINLIEST ANSWER!

1 answer:

kupik [55]3 years ago

3 0

Explanation:

r^2= 9×10^9 × 5.87×5.87×10^-16 / 64.9

=47.7× 10^-8

so taking sq. root

r = 6.9 ×10^-4 m

or

r= 6.9×10^ -2 cm

this gives the required seperation

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Any four difference between speed and acceleration

Mumz [18]

<u>Speed:</u>

Measured in distance / time
It tells the speed of a body
It is a scalar quantity
Speed is always positive
It's unit is meter per second

<u>Acceleration:</u>

Measured in velocity / time
It tells the change in the velocity
Implied on change in a velocity
It is a vector quantity
It's unit is meter per second square

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3 years ago

Alpha particles, each having a charge of +2e and a mass of 6.64 ×10-27 kg, are accelerated in a uniform 0.50 T magnetic field to

sergij07 [2.7K]

Answer:

$KE=1.2036\times 10^{-12}\ J$

Explanation:

Given:

charge on the alpha particle, $q=2e=3.2\times 10^{-19}\ C$

mass of the alpha particle, $m=6.64\times 10^{-27}\ kg$

strength of a uniform magnetic field, $B=0.5\ T$

radius of the final orbit, $r=0.5\ m$

<u>During the motion of a charge the magnetic force and the centripetal forces are balanced:</u>

$q.v.B=m.\frac{v^2}{r}$

$m.v=q.B.r$

where:

v = velocity of the alpha particle

$v=\frac{q.B.r}{m}$

$v=\frac{3.2\times 10^{-19}\times 0.5\times 0.5}{6.64\times 10^{-27}}$

$v=1.2048\times 10^{7}\ m.s^{-1}$

Here we observe that the velocity of the aprticle is close to the velocity of light. So the kinetic energy will be relativistic.

<u>We firstly find the relativistic mass as:</u>

$m'=\frac{1}{\sqrt{1-\frac{v^2}{c^2} } } \times m$

$m'=\frac{6.64\times 10^{-27}}{\sqrt{1-\frac{(1.2048\times 10^7)^2}{(3\times 10^8)^2} } }$

$m'=6.6533\times10^{-27}\ kg$

now kinetic energy:

$KE=m'.c-m.c$

$KE=6.6533\times 10^{-27}\times (3\times 10^8)^2-6.64\times 10^{-27}\times (3\times 10^8)^2$

$KE=1.2036\times 10^{-12}\ J$

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3 years ago

A pulse of light takes 3.00 ns to travel through air from an emitter to a detector. When a piece of transparent material with a

cupoosta [38]

Answer:

(a) $1.75\times10^8\text{ m/s}$

(b) 1.71

Explanation:

(a) The difference in the times of travel in the two case = $3.20 - 3.00 = 0.20\text{ ns} = 2.0\times10^{-9}\text{ s}$

This difference is the time in the transparent material. With a thickness of 35.0 cm, the speed in the material is

$v = \dfrac{35 cm}{2.0\times10^{-9}\text{ s}}=\dfrac{0.35 m}{2.0\times10^{-9}\text{ s}} = 1.75\times10^8\text{ m/s}$

(b) The refractive index of the material is the ratio of the velocity of light in vacuum to its velocity in the material. Using speed of light in vacuum as $c = 3.00\times10^8\text{ m/s}$ , the refractive index, $n$ , is

$n=\dfrac{c}{v} = \dfrac{ 3.00\times10^8\text{ m/s}}{1.75\times10^8\text{ m/s}}= 1.71$

4 0

3 years ago

The way matter moves in a longitudinal wave

serg [7]

Answer:

A longitudinal wave is a type of mechanical wave, or wave that travels through matter, called the medium. In a longitudinal wave, particles of the medium vibrate in a direction that is parallel to the direction that the wave travels. Places where particles of the medium crowd closer together are called compression's.

Explanation:

6 0

4 years ago

(a) What magnitude point charge creates a 10,000 N/C electric field at a distance of 0.250 m? (b) How large is the field at 10.0

Sergio039 [100]

<h2>Answer:</h2>

(a) 6.95 x 10⁻⁸ C

(b) 6.25N/C

<h2>Explanation:</h2>

The electric field (E) on a point charge, Q, is given by;

E = k x Q / r² ---------------(i)

Where;

k = constant = 8.99 x 10⁹ N m²/C²

r = distance of the charge from a reference point.

Given from the question;

E = 10000N/C

r = 0.250m

Substitute these values into equation(i) as follows;

10000 = 8.99 x 10⁹ x Q / (0.25)²

10000 = 8.99 x 10⁹ x Q / (0.0625)

10000 = 143.84 x 10⁹ x Q

Solve for Q;

Q = 10000/(143.84 x 10⁹)

Q = 0.00695 x 10⁻⁵C

Q = 6.95 x 10⁻⁸ C

The magnitude of the charge is 6.95 x 10⁻⁸ C

(b) To get how large the field (E) will be at r = 10.0m, substitute these values including Q = 6.95 x 10⁻⁸ C into equation (i) as follows;

E = k x Q / r²

E = 8.99 x 10⁹ x 6.95 x 10⁻⁸ / 10²

E = 8.99 x 10⁹ x 6.95 x 10⁻⁸ / 100

E = 6.25N/C

Therefore, at 10.0m, the electric field will be just 6.25N/C

3 0

3 years ago