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3 years ago

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When charges qa, qb, and qc are placed respectively at the corners a, b, and c of a right triangle, the potential at the midpoin

t of the hypotenuse is 20 V. When the charge qa is removed, the potential at the midpoint becomes 15 V. When, instead, the charge qb is removed (qa and qc both in place), the potential at the midpoint becomes 12 V. What is the potential at the midpoint if only the charge qc is removed from the array of charges?

1 answer:

earnstyle [38]3 years ago

7 0

Answer:

8v

Explanation:

First we apply super position principle

Vt= v1 + v2+ v3

Remove qa

But vt= 20v

So V = v2+v3

V1= 20-15

= 5v

Remove qb

V= v1+v3

V=8v

So the potential when qa and qc are remove is the potential due to qb

Which is 8v

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May you help me answer this

Firdavs [7]

1) See three Kepler laws below

2a) Acceleration is $2.2 m/s^2$

2b) Tension in the string: 27.4 N

3a) Kinetic energy is the energy of motion, potential energy is the energy due to the position

3b) The kinetic energy of the object is 2.25 J

Explanation:

1)

There are three Kepler's law of planetary motion:

1st law: the planets orbit the sun in elliptical orbits, with the Sun located at one of the 2 focii
2nd law: a segment connecting the Sun with each planet sweeps out equal areas in equal time intervals. A direct consequence of this is that, when a planet is further from the sun, it travels slower, and when it is closer to the sun, it travels faster
3rd law: the square of the period of revolution of a planet around the sun is directly proportional to the cube of the semi-major axis of its orbit. Mathematically, $T^2 \propto r^3$ , where T is the period of revolution and r is the semi-major axis of the orbit

2a)

To solve the problem, we have to write the equation of motions for each block along the direction parallel to the incline.

For the block on the right, we have:

$M g sin \theta - T = Ma$ (1)

where

$Mg sin \theta$ is the component of the weight of the block parallel to the incline, with

M = 8.0 kg (mass of the block)

$g=9.8 m/s^2$ (acceleration of gravity)

$\theta=35^{\circ}$

T = tension in the string

a = acceleration of the block

For the block on the left, we have similarly

$T-mg sin \theta = ma$ (2)

where

m = 3.5 kg (mass of the block)

$\theta=35^{\circ}$

From (2) we get

$T=mg sin \theta + ma$

Substituting into (1),

$M g sin \theta - mg sin \theta - ma = Ma$

Solving for a,

$a=\frac{M-m}{M+m}g sin \theta=\frac{8.0-3.5}{8.0+3.5}(9.8)(sin 35^{\circ})=2.2 m/s^2$

2b)

The tension in the string can be calculated using the equation

$T=mg sin \theta + ma$

where

m = 3.5 kg (mass of lighter block)

$g=9.8 m/s^2$

$\theta=35^{\circ}$

$a=2.2 m/s^2$ (acceleration found in part 2)

Substituting,

$T=(3.5)(9.8)(sin 35^{\circ}) +(3.5)(2.2)=27.4 N$

3a)

The kinetic energy of an object is the energy due to its motion. It is calculated as

$K=\frac{1}{2}mv^2$

where

m is the mass of the object

v is its speed

The potential energy is the energy possessed by an object due to its position in a gravitational field. For an object near the Earth's surface, it is given by

$U=mgh$

where

m is the mass of the object

g is the strength of the gravitational field

h is the heigth of the object relative to the ground

3b)

The kinetic energy of an object is given by

$K=\frac{1}{2}mv^2$

where

m is the mass of the object

v is its speed

For the object in this problem,

m = 500 g = 0.5 kg

v = 3 m/s

Substituting, we find its kinetic energy:

$K=\frac{1}{2}(0.5)(3)^2=2.25 J$

Learn more about acceleration and forces:

brainly.com/question/11411375

brainly.com/question/1971321

brainly.com/question/2286502

brainly.com/question/2562700

And about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

7 0

4 years ago

What happens to the object if the line crosses the x-axis from the positive portion of a velocity versus time diagram?

stealth61 [152]

Answer:

A velocity time graph shows the change of velocity of an object with respect ot time. If the slope of the graph is increasing in the postive region, it means that the velocity is changing, if the slope is decreasing, it means the the velocity is decreasing, but the object is moving in the same direction (positve direction).

If this slope intersects the graph at x-axis, it means that the body has 0 velocity and has become still. After that, if the line enters in the negative region, it means that its velocity is started to increases again, but the body is movinging in the opposite direction (negative direction)

7 0

3 years ago

If a spacecraft in earth orbit is pushed by a thruster what will happen?

grin007 [14]

For the orbital speed of a spacecraft we know that

$v = \sqrt{\frac{GM}{r}}$

here

M = mass of the planet around which satellite is revolving

r = orbital radius

Now when thruster is used by the spacecraft its speed will change due to which orbital speed will change.

Since here while changing the speed mass of the planet will be same

we can say the speed of the spacecraft will changed by thruster due to which its orbital radius will change

so the correct answer must be

<em>b. the spacecraft will change motion and will maintain this new orbit until the thruster is fired again.</em>

8 0

3 years ago

How much force is needed to accelerate a 62 kg skier at 3 m/sec2?

Fittoniya [83]

Answer:

<h2>The answer is 186 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

<h3>force = mass × acceleration</h3>

From the question

mass = 62 kg

acceleration = 3 m/s²

We have the final answer as

force = 62 × 3

We have the final answer as

<h3>186 N</h3>

Hope this helps you

8 0

3 years ago

A 1000 kg car is rolling slowly across a level surface at 1m/s, heading toward a group of small innocent children.

Ne4ueva [31]

Answer:

The force is -1620.73 N.

Explanation:

Given that,

Mass of car = 1000 kg

Velocity = 1 m/s

Distance = 2 m

Angle = 30°

We need to calculate the force

Using formula of work done

$W_{net}=K_{f}-K_{i}$

$F\times d=K_{f}-K_{i}$

$Fd\cos\theta=-K_{i}$

$F=\dfrac{-K_{f}}{d\cos\theta}$

$F=-\dfrac{\dfrac{1}{2}mv^2}{d\cos\theta}$

Put the value into the formula

$F=-\dfrac{\dfrac{1}{2}\times1000\times(1)^2}{2\times\cos30}$

$F=-1620.73\ N$

Hence, The force is -1620.73 N.

7 0

4 years ago

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