Ask question

Ask question

All categories

Ksenya-84 [330]

3 years ago

11

Three point charges are placed at distances of d , 2 d , and 3 d from a point P. The particle that is 2 d away from P has a char

ge of 6 q . From the options given, assign charges to the remaining two point charges such that the potential at point P is zero.

1 answer:

mel-nik [20]3 years ago

7 0

Answer:

at d the charge will be 3q and at 3d it will be 9q

Explanation:

for V=Vp-V2d

V=KQ/d=K*6q/2d=3kq/d for potential to 2d at 6q be zero the Vp will equal 3kq/d; hence at d, Q=3q and at 3d, Q=9q

You might be interested in

The energy of microwaves is less than the energy of ultraviolet light. Which comparison of the energies of

adell [148]

Answer:

A. Gamma rays have higher energy than microwaves because gamma rays have shorter wavelengths.

Explanation:

Electromagnetic waves are waves produced by the interaction between both magnetic and electric fields. These waves have some properties that make them to be arranged in a definite form producing an electromagnetic spectrum.

The spectrum has a general property of which as the wavelength of the waves increases, the frequency decreases. And vice versa.

Thus, the energy of the waves increases as the frequency increases.

Gamma rays have higher frequency, but shorter wavelength. While microwaves has lower frequency, but higher wavelength.

8 0

2 years ago

Read 2 more answers

One of the foci for the moon's orbit would be the

suter [353]

The question is oversimplified, and pretty sloppy.

Relative to the Earth . . .
The Moon is in an elliptical orbit around us, with a period of
27.32... days, and with the Earth at one focus of the ellipse.

Relative to the Sun . . .
The Moon is in an elliptical orbit around the Sun, with a period
of 365.24... days, and with the Sun at one focus of the ellipse,
and the Moon itself makes little dimples or squiggles in its orbit
on account of the gravitational influence of the nearby Earth.

I'm sorry if that seems complicated. You know that motion is
always relative to something, and the solar system is not simple.

5 0

2 years ago

Read 2 more answers

A 100 kg roller coaster comes over the first hill at 2 m/sec (vo). The height of the first hill (h) is 20 meters. See roller dia

aleksandr82 [10.1K]

For the 100 kg roller coaster that comes over the first hill of height 20 meters at 2 m/s, we have:

1) The total energy for the roller coaster at the <u>initial point</u> is 19820 J

2) The potential energy at <u>point A</u> is 19620 J

3) The kinetic energy at <u>point B</u> is 10010 J

4) The potential energy at <u>point C</u> is zero

5) The kinetic energy at <u>point C</u> is 19820 J

6) The velocity of the roller coaster at <u>point C</u> is 19.91 m/s

1) The total energy for the roller coaster at the <u>initial point</u> can be found as follows:

$E_{t} = KE_{i} + PE_{i}$

Where:

KE: is the kinetic energy = (1/2)mv₀²

m: is the mass of the roller coaster = 100 kg

v₀: is the initial velocity = 2 m/s

PE: is the potential energy = mgh

g: is the acceleration due to gravity = 9.81 m/s²

h: is the height = 20 m

The<em> total energy</em> is:

$E_{t} = KE_{i} + PE_{i} = \frac{1}{2}mv_{0}^{2} + mgh = \frac{1}{2}*100 kg*(2 m/s)^{2} + 100 kg*9.81 m/s^{2}*20 m = 19820 J$

Hence, the total energy for the roller coaster at the <u>initial point</u> is 19820 J.

2) The <em>potential energy</em> at point A is:

$PE_{A} = mgh_{A} = 100 kg*9.81 m/s^{2}*20 m = 19620 J$

Then, the potential energy at <u>point A</u> is 19620 J.

3) The <em>kinetic energy</em> at point B is the following:

$KE_{A} + PE_{A} = KE_{B} + PE_{B}$

$KE_{B} = KE_{A} + PE_{A} - PE_{B}$

Since

$KE_{A} + PE_{A} = KE_{i} + PE_{i}$

we have:

$KE_{B} = KE_{i} + PE_{i} - PE_{B} = 19820 J - mgh_{B} = 19820 J - 100kg*9.81m/s^{2}*10 m = 10010 J$

Hence, the kinetic energy at <u>point B</u> is 10010 J.

4) The <em>potential energy</em> at <u>point C</u> is zero because h = 0 meters.

$PE_{C} = mgh = 100 kg*9.81 m/s^{2}*0 m = 0 J$

5) The <em>kinetic energy</em> of the roller coaster at point C is:

$KE_{i} + PE_{i} = KE_{C} + PE_{C}$

$KE_{C} = KE_{i} + PE_{i} = 19820 J$

Therefore, the kinetic energy at <u>point C</u> is 19820 J.

6) The <em>velocity</em> of the roller coaster at point C is given by:

$KE_{C} = \frac{1}{2}mv_{C}^{2}$

$v_{C} = \sqrt{\frac{2KE_{C}}{m}} = \sqrt{\frac{2*19820 J}{100 kg}} = 19.91 m/s$

Hence, the velocity of the roller coaster at <u>point C</u> is 19.91 m/s.

Read more here:

brainly.com/question/21288807?referrer=searchResults

I hope it helps you!

3 0

2 years ago

A uniform solid disk of mass 5.00 kg and diameter 47.0 cm starts from rest and rolls without slipping down a 40.0 ∘ incline that

saveliy_v [14]

Answer:

The speed it reaches the bottom is

$v=6.51m/s$

Explanation:

Given: $m=5.0kg$ , $r=47cm\frac{1m}{100cm}=0.47m$

Using the conservation of energy theorem

$U_i=K_E+K_{ER}$

$m*g*h=\frac{1}{2}*m*v^2+\frac{1}{2}*I*w^2$

$v=r*w$ , $I=\frac{1}{2}*m*r^2$

$m*g*h=\frac{1}{2}*m*(r*w)^2 +\frac{1}{2}*[\frac{1}{2} *m*r^2]*w^2$

$m*g*h=\frac{3}{4}*m*r^2*w^2$

$g*h=\frac{3}{4}*r^2*w^2$

Solve to w'

$w^2=\frac{4*g*h}{3*r^2}$

$h=x*sin(30)=6.5m*sin(30)=3.25m$

$w=\sqrt{\frac{4*9.8m/s^2*3.25m}{3*(0.235m)^2}}$

$w=27.74rad/s$

$v=27.74rad/s*0.235m=6.51m/s$

7 0

3 years ago

Consider f(x) = -4x2 + 24x + 3. Determine whether the function has a maximum or minimum value. Then find the

murzikaleks [220]

Answer:

The function has a maximum in $x=3$

The maximum is:

$f(3) = 39$

Explanation:

Find the first derivative of the function for the inflection point, then equal to zero and solve for x

$f(x)' = -4*2x + 24=0$

$-4*2x + 24=0$

$8x=24$

$x=3$

Now find the second derivative of the function and evaluate at x = 3.

If $f (3) ''< 0$ the function has a maximum

If $f (3) '' >0$ the function has a minimum

$f(x)''= 8$

Note that:

$f(3)''= -8$

the function has a maximum in $x=3$

The maximum is:

$f(3)=-4(3)^2+24(3) + 3\\\\f(3) = 39$

4 0

3 years ago