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3 years ago

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A total charge of 4.70 is distributed on two metal spheres. When the spheres are 10 cm apart, they each feel a repulsive force o

f 4.7 x 10^11 N. How much charge is on the sphere and which has the lower amount of charge?

1 answer:

Anna35 [415]3 years ago

3 0

Answer:0.114 C

Explanation:

Given

Total 4.7 C is distributed in two spheres

Let $q_1$ and $q_2$ be the charges such that

$q_1+q_2=4.7$

and Force between charge particles is given by

$F=\frac{kq_1q_2}{r^2}$

$4.7\times 10^11=\frac{9\times 10^9\times q_1\cdot q_2}{0.1^2}$

$q_1\cdot q_2=0.522$

put the value of $q_1$

$q_2\left ( 4.7-q_2\right )=0.522$

$q_2^2-4.7q_2+0.522=0$

$q_2=\frac{4.7\pm \sqrt{4.7^2-4\times 1\times 0.522}}{2}$

$q_2=0.114 C$

thus $q_1=4.586 C$

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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

3 years ago

The roller coaster is designed according to safety regulations that prohibit the speed of the car from exceeding 20 m/s. find th

hram777 [196]

maximum allowed value of the speed in roller coaster is given as

$v = 20 m/s$

now from kinematics we can say

$v^2 - v_i^2 = 2 a s$

here initial speed will be

$v_i = 0$

acceleration is due to gravity

$a = 9.8 m/s^2$

now we can use this to find the height

$20^2 - 0^2 = 2 * 9.8* h$

$400 = 19.6 *h$

$h = 20.4 m$

so maximum allowed height will be 20.4 m

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3 years ago

A light beam has speed c in vacuum and speed v in a certain plastic. The index of refraction n of this plastic is

Mnenie [13.5K]

Answer:

$n = \frac{c}{v}$

Explanation:

Refractive Index: It is a measure to find how fast the light travels through a medium. It is ration of the speed of light in vacuum to speed of light in the medium. Speed of light is not constant and varies depending on the density of the medium.

In vacuum the speed of light is 300000 km/s and is denoted by c. When the light beam enters any medium the speed will decrease. Here it is given that the speed in plastic is v. Thus the refractive index(n) is given as:

$n = \frac{c}{v}$

It is a dimensionless no.

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3 years ago

Who is responsible of refraction light 1/3 in eye

alexandr1967 [171]

The cornea is responsible of refraction light 1/3 in eye.

<h3>What is the function of the cornea?</h3>

In addition to the protective function, it plays a fundamental role in the formation of vision. Transparent, it works like a lens over the iris, focusing light from the pupil towards the retina.

Normally, the cornea and lens deflect (refract) incoming light rays, focusing them on the retina. The shape of the cornea is fixed, but the lens changes shape to focus on objects at different distances from the eye.

See more about cornea at brainly.com/question/2297282

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2 years ago

What is the term for hardening or curing the sealant or composite material by exposing the material to a curing light?

natka813 [3]

Answer:

Photopolymerization is the term for hardening or curing the sealant or composite material by exposing the material to a curing light

Explanation:

It the process through which a photopolymer reacts to the radiations in presence of the Ultra-violet light. Photopolymers refers to those materials and liquid or the plastic resins that are hardened in presence of light source such as the lasers or the lamps. When these materials are irradiated, they undergo chemical reactions and they are hardened and become solid, and hence the process called photopolymerization. Photopolymerization is majorly used in the coating and printing industries and in also in dentistry to fill the top of teeth and prevent the cavities

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3 years ago