Ask question

Ask question

All categories

3 years ago

8

Point charges of 28.0 µC and 42.0 µC are placed 0.500 m apart. (a) At what point (in m) along the line connecting them is the el

ectric field zero? m (from the smaller charge) (b) What is the magnitude (in N/C) and direction of the electric field halfway between them? magnitude N/C direction

1 answer:

olga_2 [115]3 years ago

6 0

Answer: a) electric field will be zero at zero meters apart

b) for smaller charge q, E = 6.048X10^6N/m towards away from the charge,

for bigger charge Q, E = 4.032X10^6N/m

Explanation:

Detailed explanation and calculation is shown in the image below.

You might be interested in

A small block of mass 20.0 grams is moving to the right on a horizontal frictionless surface with a speed of 0.68 m/s. The block

Usimov [2.4K]

Answer:

a) $v'=-0.227\ m.s^{-1}$

b) $v=1.36\ m.s^{-1}$

Explanation:

Given:

mass of the lighter block, $m'=0.02\kg$

velocity of the lighter block, $u'=0.68\ m.s^{-1}$

mass of the heavier block, $m=0.04\ kg$

velocity of the heavier block, $u=0\ m.s^{-1}$

a)

Using conservation of linear momentum:

$m'.u'+m.u=m'.v'+m.v$

where:

$v'=$ final velocity of the lighter block

$v=$ final velocity of the heavier block

$m'.u'=m'.v'+m.v$

$m'(u'-v')=m.v$ ........................(1)

Since kinetic energy is conserved in elastic collision:

$\frac{1}{2}m'.u'^2=\frac{1}{2}m'.v'^2+\frac{1}{2}m.v^2$

$m'(u'^2-v'^2)=m.v^2$

$m'(u'-v')(u'+v')= m.v^2$

divide the above equation by eq. (1)

$v=u'+v'$ .............................(2)

now we substitute the value of v from eq. (2) in eq. (1)

$m'(u'-v')=m(u'+v')$

$\frac{m'+m}{m'-m} =\frac{u'}{v'}$

$\frac{0.02+0.04}{0.02-0.04} =\frac{0.68}{v'}$

$v'=-0.227\ m.s^{-1}$ (negative sign denotes that the direction is towards left)

b)

now we substitute the value of v' from eq. (2) in eq. (1)

$m'(u'-v+u')=m.v$

$2m'.u'=(m-m')v$

$2\times 0.02\times 0.68=(0.04-0.02)\times v$

$v=1.36\ m.s^{-1}$

6 0

3 years ago

A neutron star is moving in outer space at 4,000 km/hr. What happened to the star that set it in motion on it's current course?

astra-53 [7]

Answer:

d

a balanced force acted on it and propelled it to 4,000 km/hr

Explanation:

For the neutrons star which is moving in outer space at 4,000 km/hr, it could only be possible as a result of the balanced force which had already acted on it. <em>This is based on newton's law of motion which states that 'To every action, there is equal and opposite reaction'. </em>

4 0

3 years ago

A light beam is traveling through an unknown substance. When it exits that substance and enters into air, the angle of reflectio

bogdanovich [222]

Answer:

0.79

Explanation:

Using Snell's law, we have that:

n(1) * sin θ1 = n(2) * sinθ2

Where n(1) = refractive index of air = 1.0003

θ1 = angle of incidence

n(2) = refractive index of second substance

θ2 = angle of refraction

The angle of reflection through the unknown substance is the same as the angle of incidence of air. This means that θ1 = 32°

=> 1.0003 * sin32 = n(2) * sin42

n(2) = (1.0003 * sin32) / sin42

n(2) = 0.79

3 0

3 years ago

Read 2 more answers

2. What is the gravitational force on a 70.0 kg object that is 3.18 x 10 ^6m on the

Vlada [557]

Can you send a pic of the question

6 0

3 years ago

an increase in the total energy of the particles in an object results in an increase in the blank energy of the object

____ [38]

no it would in fact decrease your blank energy

3 0

3 years ago