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3 years ago

The ability to attract electrons is known as:

Physics

1 answer:

german3 years ago

8 0

The ability to attract electrons is called electronegativity.

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Does the tension of a string affect the speed of a wave

Nata [24]

Yes , increased tension suggests increased molecular attraction between the molecules of the ropes which affect the increase in the speed of wave.

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3 years ago

An astronaut is a short distance away from her space station without a tether rope. She has a large wrench. What should she do w

Hitman42 [59]

Answer: b. Throw it directly away from the space station.

Explanation:

According to <u>Newton's third law of motion</u>, <em>when two bodies interact between them, appear equal forces and opposite senses in each of them.</em>

To understand it better:

Each time a body or object exerts a force on a second body or object, it (the second body) will exert a force of equal magnitude but in the opposite direction on the first.

So, if the astronaut throws the wrench away from the space station (in the opposite direction of the space station), according to Newton's third law, she will be automatically moving towards the station and be safe.

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3 years ago

X-rays with frequency 3 ⨉ 10 18 Hz shine on a crystal, producing an interference pattern. If the first bright spot is observed a

FrozenT [24]

Answer:

$5.1645\times 10^{-11}\ m$

Explanation:

n = Order = 1

c = Speed of light = $3\times 10^8\ m/s$

f = Frequency = $3\times 10^{18}\ Hz$

$\theta$ = Angle = $75.5^{\circ}$

Lattice spacing is given by

$d=\dfrac{n\lambda}{2\sin\theta}\\\Rightarrow d=\dfrac{n\times c}{f2\sin\theta}\\\Rightarrow d=\dfrac{1\times 3\times 10^{8}}{3\times 10^{18}\times 2\times \sin75.5^{\circ}}\\\Rightarrow d=5.1645\times 10^{-11}\ m$

The lattice spacing of the crystal is $5.1645\times 10^{-11}\ m$

6 0

3 years ago

Which of earth sphere have the most mass

DochEvi [55]

The Geosphere which is 99.94% of earths mass

6 0

3 years ago

A solid conducting sphere of radius 2.00 cm has a charge of 6.88 μC. A conducting spherical shell of inner radius 4.00 cm and ou

zepelin [54]

Explanation:

Given that,

Radius R= 2.00

Charge = 6.88 μC

Inner radius = 4.00 cm

Outer radius = 5.00 cm

Charge = -2.96 μC

We need to calculate the electric field

Using formula of electric field

$E=\dfrac{kq}{r^2}$

(a). For, r = 1.00 cm

Here, r<R

So, E = 0

The electric field does not exist inside the sphere.

(b). For, r = 3.00 cm

Here, r >R

The electric field is

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times6.88\times10^{-6}}{(3.00\times10^{-2})^2}$

$E=6.88\times10^{7}\ N/C$

The electric field outside the solid conducting sphere and the direction is towards sphere.

(c). For, r = 4.50 cm

Here, r lies between R₁ and R₂.

So, E = 0

The electric field does not exist inside the conducting material

(d). For, r = 7.00 cm

The electric field is

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times(-2.96\times10^{-6})}{(7.00\times10^{-2})^2}$

$E=5.43\times10^{6}\ N/C$

The electric field outside the solid conducting sphere and direction is away of solid sphere.

Hence, This is the required solution.

6 0

3 years ago