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2 years ago

Explain briefly why metals are good conductors of electric current

2 answers:

7 0

The free electrons in metals can move through the metal, all while receiving and losing electrons, allowing metals to conduct electricity. Example: copper is a great conductor of electric current.

harina [27]2 years ago

7 0

Hello!

Just briefly, Metals are good conductors of electricity because of electrons moving freely in metals and metals like to give up electrons making them a great conductor of electricity.

Hope this helps! Thank you!!

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If two black holes collide and they send out the gravitational waves and ripples, then what happens when 3 black holes collide?

tester [92]

Answer:

I don’t know

Explanation:

6 0

2 years ago

Read 2 more answers

A purse at radius 2.00 m and a wallet at radius 3.00 m travel in uniform circular motion on the floor of a merry-go-round as the

djverab [1.8K]

Complete Question:

A purse at radius 2.00 m and a wallet at radius 3.00 m travel in uniform circular motion on the floor of a merry-go-round as the ride turns.

They are on the same radial line. At one instant, the acceleration of the purse is (2.00 m/s2 ) i + (4.00 m/s2 ) j .At that instant and in unit-vector notation, what is the acceleration of the wallet

Answer:

aw = 3 i + 6 j m/s2

Explanation:

Since both objects travel in uniform circular motion, the only acceleration that they suffer is the centripetal one, that keeps them rotating.
It can be showed that the centripetal acceleration is directly proportional to the square of the angular velocity, as follows:

$a_{c} = \omega^{2} * r (1)$

Since both objects are located on the same radial line, and they travel in uniform circular motion, by definition of angular velocity, both have the same angular velocity ω.

∴ ωp = ωw (2)

⇒ $a_{p} = \omega_{p} ^{2} * r_{p} (3)$

$a_{w} = \omega_{w}^{2} * r_{w} (4)$

Dividing (4) by (3), from (2), we have:

$\frac{a_{w} }{a_{p}} = \frac{r_{w} }{r_{p}}$

Solving for aw, we get:

$a_{w} = a_{p} *\frac{r_{w} }{r_{p} } = (2.0 i + 4.0 j) m/s2 * 1.5 = 3 i +6j m/s2$

7 0

3 years ago

A basketball has a coefficient of restitution of 0.821 in collisions with the wood floor of a basketball court. The ball is drop

Tanya [424]

Answer: The height of its fourth bounce = 0.43m

Explanation:

The coefficient of restitution denoted by (e), is the ratio that shows the final velocity to initial relative velocity between two objects after collision

IT is given by the formula in terms of height as

Coefficient of Restitution, e = √(2gh))/√(2gH) = √(h/H)

Where

Coefficient of Restitution, e= 0.821

H = 2.07 m

At fourth bounce , we have that

Coefficient of Restitution, e⁴ =√(h₄/H)

Putting the given values and solving , we have,

e⁴ =√(h₄/H)

= 0.821⁴ = √(h₄/2.07)

(0.821⁴ )² =h₄/2.07

0.2064 x 2.07 = 0.427 = 0.43

At fourth bounce, h₄ height = 0.43m

7 0

3 years ago

1) When making a digital animation of a person running on a sidewalk in a scene, which parameter would be an initial condition?

lbvjy [14]

Answer:

The position of the person's feet. a running shoe company studying how different surfaces affect the life of a shoe tread

Explanation:

100%

4 0

2 years ago

When 224-nm light falls on a metal, the current through a photoelectric circuit is brought to zero at a stopping voltage of 1.84

OLga [1]

Answer:

3.71 eV

Explanation:

λ = Wavelength of light = 224 nm = 224 x 10⁻⁹ m

c = speed of electromagnetic wave = 3 x 10⁸ m/s

V₀ = stopping potential = 1.84 volts

W₀ = Work function of the metal = ?

Using the equation

$\frac{hc}{\lambda } = eV_{o} + W_{o}$

$\frac{(6.63\times 10^{-34})(3\times 10^{8})}{224\times 10^{-9} } = (1.6\times 10^{-19})(1.84) + W_{o}$

$W_{o}$ = 5.94 x 10⁻¹⁹

$W_{o}$ = 3.71 eV

7 0

2 years ago