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2 years ago

Which is described as the flow of charge particles?

Physics

1 answer:

Lady bird [3.3K]2 years ago

7 0

Electric current

For ex. The particles are in a strict

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A street lamp weighs 150N. It is supported by two wires that form an angle of 120° with each other. The tensions in each wire ar

____ [38]

Answer:

so you take 120÷2 wires

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2 years ago

Particle A and particle B are held together with a compressed spring between them. When they are released, the spring pushes the

leonid [27]

Answer:

$KE_A=33\ J$

$KE_B=99\ J$

Explanation:

Given:

Let mass of the particle B be, $m_B=m$

then the mass of particle A, $m_A=3m$

Energy stored in the compressed spring, $E=132\ J$

Now when the compression of the particles with the spring is released, the spring potential energy must get converted into the kinetic energy of the particles and their momentum must be conserved.

Kinetic energy:

$\frac{1}{2}m_A.v_A^2+\frac{1}{2}m_B.v_B^2=132$

$3m.v_A^2+m.v_B^2=264$ .............................(1)

Using the conservation of linear momentum:

$m_A.v_A+m_B.v_B=0$

$3m.v_A+m.v_B=0$ .............................(2)

Put the value of $v_A$ from eq. (2) into eq. (1)

$3m\times (\frac{-v_B}{3})^2+m.v_B^2=264$

$v_B^2=\frac{198}{m}$ ...........................(3)

Now the kinetic energy of particle B:

$KE_B=\frac{1}{2}\times m_B\times v_B^2$

$KE_B=\frac{1}{2}\times m\times \frac{198}{m}$

$KE_B=99\ J$

Put the value of $v_B^2$ form eq. (3) into eq. (1):

$v_A^2=\frac{22}{m}$

Now the kinetic energy of particle A:

$KE_A=\frac{1}{2}m_A.v_A^2$

$KE_B=\frac{1}{2}\times 3m\times \frac{22}{m}$

$KE_A=33\ J$

6 0

3 years ago

What is in the nucleus of an atom?

12345 [234]

The nucleus is the core of an atom which is made up of protons and neutrons. Every single type of atom has a dense center and the majority of the mass is located in here.

4 0

2 years ago

Read 2 more answers

Derive an expression for the gravitational potential energy of a system consisting of Earth and a brick of mass m placed at Eart

Arlecino [84]

Answer:

The gravitational potential energy of a system is -3/2 (GmE)(m)/RE

Explanation:

Given

mE = Mass of Earth

RE = Radius of Earth

G = Gravitational Constant

Let p = The mass density of the earth is

p = M/(4/3πRE³)

p = 3M/4πRE³

Taking for instance,a very thin spherical shell in the earth;

Let r = radius

dr = thickness

Its volume is given by;

dV = 4πr²dr

Since mass = density* volume;

It's mass would be

dm = p * 4πr²dr

The gravitational potential at the center due would equal;

dV = -Gdm/r

Substitute (p * 4πr²dr) for dm

dV = -G(p * 4πr²dr)/r

dV = -G(p * 4πrdr)

The gravitational potential at the center of the earth would equal;

V = ∫dV

V = ∫ -G(p * 4πrdr) {RE,0}

V = -4πGp∫rdr {RE,0}

V = -4πGp (r²/2) {RE,0}

V = -4πGp{RE²/2)

V = -4Gπ * 3M/4πRE³ * RE²/2

V = -3/2 GmE/RE

The gravitational potential energy of the system of the earth and the brick at the center equals

U = Vm

U = -3/2 GmE/RE * m

U = -3/2 (GmE)(m)/RE

5 0

2 years ago

Which mineral test was used if a geologist refers to a mineral as metallic?

natita [175]

They would use a magnet

4 0

3 years ago